cumulative sum by group and under some criteria
res<- do.call(rbind,lapply(result,function(x) x[which.max(x$N),])) res #??? m1 n1 x1 y1 m n x y? N #668? 2? 2? 2? 2 4 6 4 6 28 #144? 3? 3? 3? 3 5 5 5 5 32 A.K.
From: Joanna Zhang <zjoanna2013 at gmail.com>
To: arun <smartpink111 at yahoo.com>
Sent: Saturday, March 2, 2013 8:52 PM
Subject: Re: [R] cumulative sum by group and under some criteria
To: arun <smartpink111 at yahoo.com>
Sent: Saturday, March 2, 2013 8:52 PM
Subject: Re: [R] cumulative sum by group and under some criteria
this is an example of my data, the max value of N will not always be the last row of each data. It could be in any position in the data.
On Sat, Mar 2, 2013 at 7:39 PM, arun <smartpink111 at yahoo.com> wrote:
>
>
>Alight, but when you say the max value of N, it is the last row in the results[[3]] and results[[2]]
>tail(result[[2]],3)
>??? m1 n1 x1 y1 m n x y? N
>709? 2? 2? 2? 2 6 4 6 2 26
>710? 2? 2? 2? 2 6 4 6 3 27
>711? 2? 2? 2? 2 6 4 6 4 28
>> tail(result[[3]],3)
>??? m1 n1 x1 y1 m n x y? N
>142? 3? 3? 3? 3 5 5 5 3 30
>143? 3? 3? 3? 3 5 5 5 4 31
>144? 3? 3? 3? 3 5 5 5 5 32
>
>
>So, are you saying that you want only row #711 and #144? as the final result.
>
>
>A.K.
>________________________________
>From: Joanna Zhang <zjoanna2013 at gmail.com>
>To: arun <smartpink111 at yahoo.com>
>Sent: Saturday, March 2, 2013 7:37 PM
>
>Subject: Re: [R] cumulative sum by group and under some criteria
>
>
>Change maxN to 10, you will get result[2] and [3]
>
>
>
>On Sat, Mar 2, 2013 at 6:35 PM, arun <smartpink111 at yahoo.com> wrote:
>
>
>>
>>AFter running this code, I get result[[1]], [[3]],[[4]], [[5]] as NULL.? The only list element is [[2]].? what do you mean by putting all data result[] in one dataset?
>>
>>
>>
>>
>>
>>________________________________
>>From: Joanna Zhang <zjoanna2013 at gmail.com>
>>To: arun <smartpink111 at yahoo.com>
>>Sent: Saturday, March 2, 2013 7:27 PM
>>
>>Subject: Re: [R] cumulative sum by group and under some criteria
>>
>>
>>Hi, is there a way to put all data "result[] " ?in one data set and extract the row with ?the max value of N? I am almost there. Really appreciate your help. I have to finish it this weekend.
>>
>>
>>
>>maxN<-9
>>c11<-0.4
>>c12<-0.4
>>c1<-0.5
>>c2<-0.5
>>
>>p0L<-0.05
>>p0H<-0.05
>>p1L<-0.25
>>p1H<-0.25
>>
>>result <- vector("list",5)
>>
>>for (a in 2: (maxN-6)) {
>>d <- data.frame ()
>>for ( m1 in a:a) {
>>? ? ? ?for (n1 in 2: (maxN-m1-4)){
>>? ? ? ? ? for (x1 in 0: m1) {
>>? ? ? ? ? ? ? ?for (y1 in 0: n1) {
>>? ? ? ? ? ? ? ? ? ? ? ?#p11<- (x1/m1)
>>? ? ? ? ? ? ? ? ? ? ? ?#p12<- (y1/n1) ? ? ? ? ? ? ? ? ? ? ? ? ? ??
>>
>>? ? ? ? ? ? ? ? ? ? ? ?term1_p0 = dbinom(x1,m1, p0L, log=FALSE)* dbinom(y1,n1,p0H, log=FALSE)
>>? ? ? ? ? ? ? ? ? ? ? ?term1_p1 = dbinom(x1,m1, p1L, log=FALSE)* dbinom(y1,n1,p1H, log=FALSE) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??
>>
>>? ? ? ? ? ? ? ? ? ? ? ?d<-rbind(d, c(m1,n1,x1,y1,term1_p0,term1_p1))
>>}}
>>}}
>>colnames(d)<-c("m1","n1","x1","y1","term1_p0","term1_p1")?
>>tail(d)
>>
>>########## add Qm Qn ##################################
>>set.seed(8)
>>d1<-do.call(rbind,lapply(seq_len(nrow(d)),function(i){
>>Pm<- rbeta(1000,0.2+d[i,"x1"],0.8+d[i,"m1"]-d[i,"x1"]);
>>Pn<- rbeta(1000,0.2+d[i,"y1"],0.8+d[i,"n1"]-d[i,"y1"]);
>>Fm<- ecdf(Pm);
>>Fn<- ecdf(Pn);
>>
>>Fmm<- Fm(p1L);
>>Fnn<- Fn(p1H);
>>
>>R<- (Fmm+Fnn)/2;?
>>Fmm_f<- max(R, Fmm);
>>Fnn_f<- min(R, Fnn);
>>Qm<- 1-Fmm_f;
>>Qn<- 1-Fnn_f;
>>data.frame(Qm,Qn)}))
>>d2<-cbind(d,d1)
>>head(d2)
>>
>>
>>library(zoo)
>>lst1<- split(d2,list(d$m1,d$n1))?
>>d2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){?
>>x[,9:14]<-NA;?
>>x[,9:10][x$Qm<=c11,]<-cumsum(x[,5:6][x$Qm<=c11,]);?
>>x[,11:12][x$Qn<=c12,]<-cumsum(x[,5:6][x$Qn<=c12,]);
>>x[,13:14]<-cumsum(x[,5:6]);?
>>colnames(x)[9:14]<- c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H","sumTerm1_p0","sumTerm1_p1");?
>>x1<-na.locf(x);?
>>x1[,9:14][is.na(x1[,9:14])]<-0;?
>>x1}
>>))?
>>row.names(d2)<-1:nrow(d2)?
>>tail(d2)
>>
>>res1<-aggregate(.~m1+n1,data=d2[,c(1:2,9:12)],max)
>>head(res1)
>>
>>d3<-res1[res1[,4]<=0.60 & res1[,6]<0.40,]?
>>tail(d3)
>>
>>library(plyr)?
>>d4<- join(d3,d,by=c("m1","n1"),type="inner")
>>head(d4)?
>>
>>if ( nrow(d4)==0) break
>>
>>res2<-do.call(rbind,lapply(unique(d4$m1),function(m1)?
>>do.call(rbind,lapply(unique(d4$n1),function(n1)
>>do.call(rbind,lapply(unique(d4$x1),function(x1)
>>do.call(rbind,lapply(unique(d4$y1),function(y1)
>>
>>#do.call(rbind,lapply(0:m1,function(x1)?
>>#do.call(rbind,lapply(0:n1,function(y1)?
>>do.call(rbind,lapply((m1+2):(maxN-2-n1),function(m)?
>>do.call(rbind,lapply((n1+2):(maxN-m),function(n)?
>>do.call(rbind,lapply(x1:(x1+m-m1), function(x)?
>>do.call(rbind,lapply(y1:(y1+n-n1), function(y)
>>expand.grid(m1,n1,x1,y1,m,n,x,y)) )))))))))))))))
>>
>>
>>names(res2)<- c("m1","n1","x1","y1","m","n","x","y")
>>attr(res2,"out.attrs")<-NULL?
>>tail(res2)
>>res2$N<-res2$m1+res2$n1+res2$x1+res2$y1+res2$m+res2$n+res2$x+res2$y
>>result[[a]]<-res2
>>
>>}
>>
>>result
>>
>>
>>
>>
>>On Sat, Mar 2, 2013 at 4:56 PM, Joanna Zhang <zjoanna2013 at gmail.com> wrote:
>>
>>I just used if (nrow(d4)==0) break to solve the problem.
>>>
>>>
>>>
>>>On Sat, Mar 2, 2013 at 4:00 PM, arun <smartpink111 at yahoo.com> wrote:
>>>
>>>
>>>>
>>>>f1<- function(dat){
>>>>stopifnot(nrow(dat)!=0)
>>>>do.call(rbind,lapply(unique(dat$m1),function(m1)
>>>>?do.call(rbind,lapply(unique(dat$n1),function(n1)
>>>>?do.call(rbind,lapply(unique(dat$x1),function(x1)
>>>>?do.call(rbind,lapply(unique(dat$y1),function(y1)
>>>>
>>>>?
>>>>?#do.call(rbind,lapply(0:m1,function(x1)
>>>>?#do.call(rbind,lapply(0:n1,function(y1)
>>>>?do.call(rbind,lapply((m1+2):(maxN-2-n1),function(m)
>>>>?do.call(rbind,lapply((n1+2):(maxN-m),function(n)
>>>>?do.call(rbind,lapply(x1:(x1+m-m1), function(x)
>>>>?do.call(rbind,lapply(y1:(y1+n-n1), function(y)
>>>>?expand.grid(m1,n1,x1,y1,m,n,x,y)) )))))))))))))))
>>>>}
>>>>
>>>>f1(d4)
>>>>#Error: nrow(dat) != 0 is not TRUE
>>>>?head(f1(d3),2)
>>>>#? Var1 Var2 Var3 Var4 Var5 Var6 Var7 Var8
>>>>#1??? 3??? 2??? 0??? 0??? 5??? 4??? 0??? 0
>>>>#2??? 3??? 2??? 0??? 0??? 5??? 4??? 0??? 1
>>>>
>>>>A.K.
>>>>
>>>>
>>>>
>>>>________________________________
>>>>From: Joanna Zhang <zjoanna2013 at gmail.com>
>>>>To: arun <smartpink111 at yahoo.com>
>>>>Sent: Saturday, March 2, 2013 4:47 PM
>>>>
>>>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>
>>>>
>>>>got it. is there a way to exit the loop if there is no row (empty data) in d3new?
>>>>
>>>>
>>>>
>>>>On Sat, Mar 2, 2013 at 3:38 PM, arun <smartpink111 at yahoo.com> wrote:
>>>>
>>>>HI,
>>>>>d
>>>>>
>>>>>#?? m1 n1 x1 y1???? term1_p0???? term1_p1
>>>>>#1?? 3? 2? 0? 0 0.7737809375 0.2373046875
>>>>>#2?? 3? 2? 0? 1 0.0814506250 0.1582031250
>>>>>#3?? 3? 2? 0? 2 0.0021434375 0.0263671875
>>>>>#4?? 3? 2? 1? 0 0.1221759375 0.2373046875
>>>>>#5?? 3? 2? 1? 1 0.0128606250 0.1582031250
>>>>>#6?? 3? 2? 1? 2 0.0003384375 0.0263671875
>>>>>#7?? 3? 2? 2? 0 0.0064303125 0.0791015625
>>>>>#8?? 3? 2? 2? 1 0.0006768750 0.0527343750
>>>>>#9?? 3? 2? 2? 2 0.0000178125 0.0087890625
>>>>>#10? 3? 2? 3? 0 0.0001128125 0.0087890625
>>>>>#11? 3? 2? 3? 1 0.0000118750 0.0058593750
>>>>>#12? 3? 2? 3? 2 0.0000003125 0.0009765625
>>>>>
>>>>>d2
>>>>>#?? m1 n1 x1 y1???? term1_p0???? term1_p1???? Qm???? Qn
>>>>>#1?? 3? 2? 0? 0 0.7737809375 0.2373046875 0.0500 0.0810
>>>>>#2?? 3? 2? 0? 1 0.0814506250 0.1582031250 0.0560 0.6690
>>>>>#3?? 3? 2? 0? 2 0.0021434375 0.0263671875 0.0410 0.9680
>>>>>#4?? 3? 2? 1? 0 0.1221759375 0.2373046875 0.3150 0.3150
>>>>>#5?? 3? 2? 1? 1 0.0128606250 0.1582031250 0.5200 0.6580
>>>>>#6?? 3? 2? 1? 2 0.0003384375 0.0263671875 0.5360 0.9640
>>>>>#7?? 3? 2? 2? 0 0.0064303125 0.0791015625 0.4945 0.4945
>>>>>#8?? 3? 2? 2? 1 0.0006768750 0.0527343750 0.7815 0.7815
>>>>>#9?? 3? 2? 2? 2 0.0000178125 0.0087890625 0.9030 0.9650
>>>>>#10? 3? 2? 3? 0 0.0001128125 0.0087890625 0.5350 0.5350
>>>>>#11? 3? 2? 3? 1 0.0000118750 0.0058593750 0.8195 0.8195
>>>>>#12? 3? 2? 3? 2 0.0000003125 0.0009765625 0.9835 0.9835
>>>>>
>>>>>?lst1<- split(d2,list(d2$m1,d2$n1))
>>>>>d3<- do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
>>>>>
>>>>>x[,9:14]<-NA;
>>>>>x[,9:10][x$Qm<=c11,]<-cumsum(x[,5:6][x$Qm<=c11,]);
>>>>>x[,11:12][x$Qn<=c12,]<-cumsum(x[,5:6][x$Qn<=c12,]);
>>>>>x[,13:14]<-cumsum(x[,5:6]);
>>>>>colnames(x)[9:14]<- c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H","sumTerm1_p0","sumTerm1_p1");
>>>>>x1<-na.locf(x);
>>>>>x1[,9:14][is.na(x1[,9:14])]<-0;
>>>>>x1}
>>>>>))
>>>>>?row.names(d3)<-1:nrow(d3)
>>>>>
>>>>>res1<-aggregate(.~m1+n1,data=d3[,c(1:2,9:12)],max)
>>>>>res1
>>>>># m1 n1 cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H
>>>>>#1? 3? 2? 0.9795509? 0.6591797? 0.8959569? 0.4746094
>>>>>?d3New<- res1[res1[,4]<=0.60 & res1[,6]<0.40,]
>>>>>d3New
>>>>>#[1] m1???????? n1???????? cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H? # zero 0 rows
>>>>>#<0 rows> (or 0-length row.names)
>>>>>?library(plyr)
>>>>>?d4<-join(d3New,d,by=c("m1","n1"),type="inner")
>>>>>?d4
>>>>># [1] m1???????? n1???????? cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H #zero 0 rows
>>>>># [7] x1???????? y1???????? term1_p0?? term1_p1?
>>>>>#<0 rows> (or 0-length row.names)
>>>>>
>>>>>do.call(rbind,lapply(unique(d4$m1),function(m1)
>>>>>
>>>>>?do.call(rbind,lapply(unique(d4$n1),function(n1)
>>>>>?do.call(rbind,lapply(unique(d4$x1),function(x1)
>>>>>?do.call(rbind,lapply(unique(d4$y1),function(y1)
>>>>>?
>>>>>?#do.call(rbind,lapply(0:m1,function(x1)
>>>>>?#do.call(rbind,lapply(0:n1,function(y1)
>>>>>?do.call(rbind,lapply((m1+2):(maxN-2-n1),function(m)
>>>>>?do.call(rbind,lapply((n1+2):(maxN-m),function(n)
>>>>>?do.call(rbind,lapply(x1:(x1+m-m1), function(x)
>>>>>?do.call(rbind,lapply(y1:(y1+n-n1), function(y)
>>>>>?expand.grid(m1,n1,x1,y1,m,n,x,y)) )))))))))))))))
>>>>>#NULL
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>A.K.
>>>>>________________________________
>>>>>From: Joanna Zhang <zjoanna2013 at gmail.com>
>>>>>To: arun <smartpink111 at yahoo.com>
>>>>>Sent: Saturday, March 2, 2013 3:59 PM
>>>>>
>>>>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>
>>>>>
>>>>>I can't figure out why there is an error for names (): Error in names(res2) <- c("m1", "n1", "x1", "y1", "m", "n", "x", "y") :?
>>>>>? attempt to set an attribute on NULL
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>maxN<-9
>>>>>c11<-0.4
>>>>>c12<-0.4
>>>>>c1<-0.5
>>>>>c2<-0.5
>>>>>
>>>>>p0L<-0.05
>>>>>p0H<-0.05
>>>>>p1L<-0.25
>>>>>p1H<-0.25
>>>>>
>>>>>alpha<-0.20
>>>>>beta<-0.80
>>>>>
>>>>>result <- vector("list",5)
>>>>>result
>>>>>
>>>>>for (a in 2: (maxN-6)) {
>>>>>d <- data.frame ()
>>>>>for ( m1 in a:a) {
>>>>>? ? ? ?for (n1 in 2: (maxN-m1-4)){
>>>>>? ? ? ? ? for (x1 in 0: m1) {
>>>>>? ? ? ? ? ? ? ?for (y1 in 0: n1) {
>>>>>
>>>>>
>>>>>? ? ? ? ? ? ? ? ? ? ? ?term1_p0 = dbinom(x1,m1, p0L, log=FALSE)* dbinom(y1,n1,p0H, log=FALSE)
>>>>>? ? ? ? ? ? ? ? ? ? ? ?term1_p1 = dbinom(x1,m1, p1L, log=FALSE)* dbinom(y1,n1,p1H, log=FALSE) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??
>>>>>
>>>>>? ? ? ? ? ? ? ? ? ? ? ?d<-rbind(d, c(m1,n1,x1,y1,term1_p0,term1_p1))
>>>>>}}
>>>>>}}
>>>>>colnames(d)<-c("m1","n1","x1","y1","term1_p0","term1_p1")?
>>>>>tail(d)
>>>>>
>>>>>set.seed(8)
>>>>>d1<-do.call(rbind,lapply(seq_len(nrow(d)),function(i){
>>>>>Pm<- rbeta(1000,0.2+d[i,"x1"],0.8+d[i,"m1"]-d[i,"x1"]);
>>>>>Pn<- rbeta(1000,0.2+d[i,"y1"],0.8+d[i,"n1"]-d[i,"y1"]);
>>>>>Fm<- ecdf(Pm);
>>>>>Fn<- ecdf(Pn);
>>>>>#Fmm<- Fm(d[i,"p11"]);
>>>>>#Fnn<- Fn(d[i,"p12"]);
>>>>>
>>>>>Fmm<- Fm(p1L);
>>>>>Fnn<- Fn(p1H);
>>>>>
>>>>>R<- (Fmm+Fnn)/2;?
>>>>>Fmm_f<- max(R, Fmm);
>>>>>Fnn_f<- min(R, Fnn);
>>>>>Qm<- 1-Fmm_f;
>>>>>Qn<- 1-Fnn_f;
>>>>>data.frame(Qm,Qn)}))
>>>>>d2<-cbind(d,d1)
>>>>>head(d2)
>>>>>
>>>>>
>>>>>library(zoo)
>>>>>lst1<- split(d2,list(d$m1,d$n1))?
>>>>>d2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){?
>>>>>x[,9:14]<-NA;?
>>>>>x[,9:10][x$Qm<=c11,]<-cumsum(x[,5:6][x$Qm<=c11,]);?
>>>>>x[,11:12][x$Qn<=c12,]<-cumsum(x[,5:6][x$Qn<=c12,]);
>>>>>x[,13:14]<-cumsum(x[,5:6]);?
>>>>>colnames(x)[9:14]<- c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H","sumTerm1_p0","sumTerm1_p1");?
>>>>>x1<-na.locf(x);?
>>>>>x1[,9:14][is.na(x1[,9:14])]<-0;?
>>>>>x1}
>>>>>))?
>>>>>row.names(d2)<-1:nrow(d2)?
>>>>>tail(d2)
>>>>>
>>>>>res1<-aggregate(.~m1+n1,data=d2[,c(1:2,9:12)],max)
>>>>>head(res1)
>>>>>
>>>>>d3<-res1[res1[,4]<=0.60 & res1[,6]<0.40,]?
>>>>>tail(d3)
>>>>>
>>>>>library(plyr)?
>>>>>d4<- join(d3,d,by=c("m1","n1"),type="inner")
>>>>>head(d4)?
>>>>>tail(d4)
>>>>>
>>>>>res2<-do.call(rbind,lapply(unique(d4$m1),function(m1)?
>>>>>do.call(rbind,lapply(unique(d4$n1),function(n1)
>>>>>do.call(rbind,lapply(unique(d4$x1),function(x1)
>>>>>do.call(rbind,lapply(unique(d4$y1),function(y1)
>>>>>
>>>>>#do.call(rbind,lapply(0:m1,function(x1)?
>>>>>#do.call(rbind,lapply(0:n1,function(y1)?
>>>>>do.call(rbind,lapply((m1+2):(maxN-2-n1),function(m)?
>>>>>do.call(rbind,lapply((n1+2):(maxN-m),function(n)?
>>>>>do.call(rbind,lapply(x1:(x1+m-m1), function(x)?
>>>>>do.call(rbind,lapply(y1:(y1+n-n1), function(y)
>>>>>expand.grid(m1,n1,x1,y1,m,n,x,y)) )))))))))))))))
>>>>>
>>>>>names(res2)<- c("m1","n1","x1","y1","m","n","x","y")
>>>>>attr(res2,"out.attrs")<-NULL?
>>>>>tail(res2)
>>>>>result[[a]]<-res2
>>>>>}
>>>>>result
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>On Sat, Mar 2, 2013 at 1:05 PM, arun <smartpink111 at yahoo.com> wrote:
>>>>>
>>>>>Alright, then go ahead and use a loop.
>>>>>>
>>>>>>A.K.
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>________________________________
>>>>>>From: Joanna Zhang <zjoanna2013 at gmail.com>
>>>>>>To: arun <smartpink111 at yahoo.com>
>>>>>>Sent: Saturday, March 2, 2013 1:53 PM
>>>>>>
>>>>>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>
>>>>>>
>>>>>>Thank you! When I run my real data, there was a warning massage 'lack of memory'. I am thinking that I can use a loop to run the same code for each part of the data. But I need to keep each output data "d2" and combine them.?
>>>>>>
>>>>>>for example:
>>>>>>
>>>>>>maxN<-9
>>>>>>c11<-0.4
>>>>>>c12<-0.4
>>>>>>
>>>>>>p0L<-0.05
>>>>>>p0H<-0.05
>>>>>>p1L<-0.25
>>>>>>p1H<-0.25
>>>>>>
>>>>>>
>>>>>>for (a in 2: (maxN-6)) {
>>>>>>d <- data.frame ()
>>>>>>for ( m1 in a:a) {
>>>>>>? ? ?for (n1 in 2: (maxN-m1-4)){
>>>>>>? ? ? ? ? for (x1 in 0: m1) {
>>>>>>? ? ? ? ? ? ? ?for (y1 in 0: n1) {
>>>>>>? ? ? ? ? ? ? ? ? ? ? ?p11<- (x1/m1)
>>>>>>? ? ? ? ? ? ? ? ? ? ? ?p12<- (y1/n1) ? ? ? ? ? ? ? ? ? ? ? ? ? ??
>>>>>>
>>>>>>? ? ? ? ? ? ? ? ? ? ? ?term1_p0 = dbinom(x1,m1, p0L, log=FALSE)* dbinom(y1,n1,p0H, log=FALSE)
>>>>>>? ? ? ? ? ? ? ? ? ? ? ?term1_p1 = dbinom(x1,m1, p1L, log=FALSE)* dbinom(y1,n1,p1H, log=FALSE) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??
>>>>>>
>>>>>>? ? ? ? ? ? ? ? ? ? ? ?d<-rbind(d, c(m1,n1,x1,y1,p11,p12,term1_p0,term1_p1))
>>>>>>}}
>>>>>>}}
>>>>>>colnames(d)<-c("m1","n1","x1","y1","p11","p12","term1_p0","term1_p1")?
>>>>>>tail(d)
>>>>>>
>>>>>>set.seed(8)
>>>>>>d1<-do.call(rbind,lapply(seq_len(nrow(d)),function(i){
>>>>>>Pm<- rbeta(1000,0.2+d[i,"x1"],0.8+d[i,"m1"]-d[i,"x1"]);
>>>>>>Pn<- rbeta(1000,0.2+d[i,"y1"],0.8+d[i,"n1"]-d[i,"y1"]);
>>>>>>Fm<- ecdf(Pm);
>>>>>>Fn<- ecdf(Pn);
>>>>>>
>>>>>>Fmm<- Fm(p1L);
>>>>>>Fnn<- Fn(p1H);
>>>>>>
>>>>>>R<- (Fmm+Fnn)/2;?
>>>>>>Fmm_f<- max(R, Fmm);
>>>>>>Fnn_f<- min(R, Fnn);
>>>>>>Qm<- 1-Fmm_f;
>>>>>>Qn<- 1-Fnn_f;
>>>>>>data.frame(Qm,Qn)}))
>>>>>>d2<-cbind(d,d1)
>>>>>>head(d2) ? ? ? ? ? ? ? ? ? ? ? ?# need to name "d2" using the value of a (or another way) to?distinguish from each other??
>>>>>>}
>>>>>>
>>>>>># combine all "d2" here
>>>>>>
>>>>>>
>>>>>>
>>>>>>On Fri, Mar 1, 2013 at 12:51 PM, arun <smartpink111 at yahoo.com> wrote:
>>>>>>
>>>>>>
>>>>>>>
>>>>>>>Hi,
>>>>>>>
>>>>>>>It seems like you haven't even looked at the output of d2, (first d2)
>>>>>>>d2<- cbind(d,d1)
>>>>>>>head(d2,3)
>>>>>>>#? m1 n1 x1 y1 p11 p12?? term1_p0?? term1_p1??? Qm??? Qn
>>>>>>>#1? 2? 2? 0? 0?? 0 0.0 0.81450625 0.31640625 1.000 1.000
>>>>>>>#2? 2? 2? 0? 1?? 0 0.5 0.08573750 0.21093750 0.666 0.666
>>>>>>>#3? 2? 2? 0? 2?? 0 1.0 0.00225625 0.03515625 0.500 0.500
>>>>>>>?ncol(d2)? # you have only 10 columns in d2
>>>>>>>#[1] 10
>>>>>>>
>>>>>>>
>>>>>>>#2nd problem:
>>>>>>>You are splitting using d
>>>>>>>
>>>>>>>###############Your code
>>>>>>>
>>>>>>>library(zoo)
>>>>>>>lst1<- split(d,list(d$m1,d$n1)) # should split byd2, because `d` doesn't have Qm or Qn columns?
>>>>>>>d2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
>>>>>>>x[,13:18]<-NA; #### this code was created for another dataset which obviously had 12 columns
>>>>>>>x[,13:14][x$Qm<=c11,]<-cumsum(x[,11:12][x$Qm<=c11,]); #### here your term1_p0 and term1_p1 are columns 7 and 8.
>>>>>>>
>>>>>>>x[,15:16][x$Qn<=c12,]<-cumsum(x[,11:12][x$Qn<=c12,]);
>>>>>>>x[,17:18]<-cumsum(x[,11:12]);
>>>>>>>colnames(x)[13:18]<- c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H","sumTerm1_p0","sumTerm1_p1");
>>>>>>>x1<-na.locf(x);
>>>>>>>x1[,13:18][is.na(x1[,13:18])]<-0;
>>>>>>>x1}
>>>>>>>))
>>>>>>>##########################################
>>>>>>>
>>>>>>>
>>>>>>>#corrected codes:
>>>>>>>
>>>>>>>lst1<- split(d2,list(d2$m1,d2$n1))
>>>>>>>
>>>>>>>dNew<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
>>>>>>>x[,11:16]<-NA;
>>>>>>>x[,11:12][x$Qm<=c11,]<-cumsum(x[,7:8][x$Qm<=c11,]);
>>>>>>>x[,13:14][x$Qn<=c12,]<-cumsum(x[,7:8][x$Qn<=c12,]);
>>>>>>>x[,15:16]<-cumsum(x[,7:8]);
>>>>>>>colnames(x)[11:16]<- c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H","sumTerm1_p0","sumTerm1_p1");
>>>>>>>x1<-na.locf(x);
>>>>>>>x1[,11:16][is.na(x1[,11:16])]<-0;
>>>>>>>x1}
>>>>>>>))
>>>>>>>?row.names(dNew)<- 1:nrow(dNew)
>>>>>>>?head(dNew,3)
>>>>>>>#? m1 n1 x1 y1 p11 p12?? term1_p0?? term1_p1??? Qm??? Qn cterm1_P0L cterm1_P1L
>>>>>>>#1? 2? 2? 0? 0?? 0 0.0 0.81450625 0.31640625 1.000 1.000????????? 0????????? 0
>>>>>>>#2? 2? 2? 0? 1?? 0 0.5 0.08573750 0.21093750 0.666 0.666????????? 0????????? 0
>>>>>>>#3? 2? 2? 0? 2?? 0 1.0 0.00225625 0.03515625 0.500 0.500????????? 0????????? 0
>>>>>>>#? cterm1_P0H cterm1_P1H sumTerm1_p0 sumTerm1_p1
>>>>>>>#1????????? 0????????? 0?? 0.8145062?? 0.3164062
>>>>>>>#2????????? 0????????? 0?? 0.9002438?? 0.5273438
>>>>>>>#3????????? 0????????? 0?? 0.9025000?? 0.5625000
>>>>>>>
>>>>>>>
>>>>>>>A.K.
>>>>>>>
>>>>>>>
>>>>>>>________________________________
>>>>>>>
>>>>>>>From: Joanna Zhang <zjoanna2013 at gmail.com>
>>>>>>>To: arun <smartpink111 at yahoo.com>
>>>>>>>Sent: Friday, March 1, 2013 11:40 AM
>>>>>>>
>>>>>>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>
>>>>>>>
>>>>>>>Hi, why there is an error when I run the cumulative sum code below?
>>>>>>>
>>>>>>>Error in `[<-.data.frame`(`*tmp*`, , 16:21, value = NA) :
>>>>>>>? new columns would leave holes after existing columns
>>>>>>>
>>>>>>>
>>>>>>>maxN<-9
>>>>>>>c11<-0.4
>>>>>>>c12<-0.4
>>>>>>>c1<-0.5
>>>>>>>c2<-0.5
>>>>>>>p0L<-0.05
>>>>>>>p0H<-0.05
>>>>>>>p1L<-0.25
>>>>>>>p1H<-0.25
>>>>>>>
>>>>>>>d <- data.frame ()
>>>>>>>for ( m1 in 2: (maxN-6)) {
>>>>>>>???? for (n1 in 2: (maxN-m1-4)){
>>>>>>>????????? for (x1 in 0: m1) {
>>>>>>>?????????????? for (y1 in 0: n1) {
>>>>>>>?????????????????????? p11<- (x1/m1)
>>>>>>>?????????????????????? p12<- (y1/n1)????????????????????????????
>>>>>>>
>>>>>>>?????????????????????? term1_p0 = dbinom(x1,m1, p0L, log=FALSE)* dbinom(y1,n1,p0H, log=FALSE)
>>>>>>>?????????????????????? term1_p1 = dbinom(x1,m1, p1L, log=FALSE)* dbinom(y1,n1,p1H, log=FALSE)????????????????????????????????????????????????????????
>>>>>>>
>>>>>>>?????????????????????? d<-rbind(d, c(m1,n1,x1,y1,p11,p12,term1_p0,term1_p1))
>>>>>>>}}
>>>>>>>}}
>>>>>>>colnames(d)<-c("m1","n1","x1","y1","p11","p12","term1_p0","term1_p1")
>>>>>>>d
>>>>>>>tail(d)
>>>>>>>set.seed(8)
>>>>>>>d1<-do.call(rbind,lapply(seq_len(nrow(d)),function(i){
>>>>>>>Pm<- rbeta(1000,0.2+d[i,"x1"],0.8+d[i,"m1"]-d[i,"x1"]);
>>>>>>>Pn<- rbeta(1000,0.2+d[i,"y1"],0.8+d[i,"n1"]-d[i,"y1"]);
>>>>>>>Fm<- ecdf(Pm);
>>>>>>>Fn<- ecdf(Pn);
>>>>>>>Fmm<- Fm(d[i,"p11"]);
>>>>>>>Fnn<- Fn(d[i,"p12"]);
>>>>>>>R<- (Fmm+Fnn)/2;
>>>>>>>Fmm_f<- max(R, Fmm);
>>>>>>>Fnn_f<- min(R, Fnn);
>>>>>>>Qm<- 1-Fmm_f;
>>>>>>>Qn<- 1-Fnn_f;
>>>>>>>data.frame(Qm,Qn)}))
>>>>>>>d2<-cbind(d,d1)
>>>>>>>d2
>>>>>>>
>>>>>>>library(zoo)
>>>>>>>lst1<- split(d,list(d$m1,d$n1))
>>>>>>>d2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
>>>>>>>x[,13:18]<-NA;
>>>>>>>x[,13:14][x$Qm<=c11,]<-cumsum(x[,11:12][x$Qm<=c11,]);
>>>>>>>x[,15:16][x$Qn<=c12,]<-cumsum(x[,11:12][x$Qn<=c12,]);
>>>>>>>x[,17:18]<-cumsum(x[,11:12]);
>>>>>>>colnames(x)[13:18]<- c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H","sumTerm1_p0","sumTerm1_p1");
>>>>>>>x1<-na.locf(x);
>>>>>>>x1[,13:18][is.na(x1[,13:18])]<-0;
>>>>>>>x1}
>>>>>>>))
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>On Tue, Feb 26, 2013 at 8:56 PM, arun <smartpink111 at yahoo.com> wrote:
>>>>>>>
>>>>>>>??
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>________________________________
>>>>>>>> From: Joanna Zhang <zjoanna2013 at gmail.com>
>>>>>>>>To: arun <smartpink111 at yahoo.com>
>>>>>>>>Sent: Tuesday, February 26, 2013 9:51 PM
>>>>>>>>
>>>>>>>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>Hi,
>>>>>>>>
>>>>>>>>#
>>>>>>>>Pm2<-rbeta(1000, 0.2+1, 0.8+3) #obs4?
>>>>>>>>this is for x=1, m=2
>>>>>>>>
>>>>>>>>?length(Pm2)
>>>>>>>>>#[1] 1000
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>Pn2<-rbeta(1000, 0.2, 0.8+4)
>>>>>>>>>?length(Pn2)
>>>>>>>>>#[1] 1000
>>>>>>>>>Here, you are creating Pm2 or Pn2 from a single observation.
>>>>>>>>>
>>>>>>>>>In the code, it is creating 1000 values in total from the combination of values from x, m,
>>>>>>>>>?Pm2<-rbeta(1000, 0.2+res2$x, 0.8+res2$m-res2$x)
>>>>>>>>>?length(Pm2)
>>>>>>>>>#[1] 1000
>>>>>>>>>
>>>>>>>>>I don't get it here. What values of x and m are used here? I thought it should create 1000 observations for each combination of x,m in the data and this is what I want.
>>>>>>>>>
>>>>>>>>?
>>>>>>>>A.K.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>----- Original Message -----
>>>>>>>>>
>>>>>>>>>From: Zjoanna <Zjoanna2013 at gmail.com>
>>>>>>>>>To: r-help at r-project.org
>>>>>>>>>Cc:
>>>>>>>>>
>>>>>>>>>Sent: Tuesday, February 26, 2013 3:13 PM
>>>>>>>>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>Hi Arun
>>>>>>>>>
>>>>>>>>>I noticed that the values of Fmm, Fnn, and other corresponding variables
>>>>>>>>>are not correct, for example,? for the 4th obs after you run this code, the
>>>>>>>>>Fmm is 0.40,? but if you use the x, m, y, n in the 4th row to calculate
>>>>>>>>>them, the results are not consistent, same for the 5th obs.
>>>>>>>>>
>>>>>>>>>#check
>>>>>>>>>#
>>>>>>>>>Pm2<-rbeta(1000, 0.2+1, 0.8+3) #obs4
>>>>>>>>>Pn2<-rbeta(1000, 0.2, 0.8+4)
>>>>>>>>>Fm2<- ecdf(Pm2)
>>>>>>>>>Fn2<- ecdf(Pn2)
>>>>>>>>>Fmm2<-Fm2(1/4)
>>>>>>>>>Fnn2<-Fn2(0)
>>>>>>>>>Fmm2? #0.582
>>>>>>>>>Fnn2? ?#0
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>Pm2<-rbeta(1000, 0.2+1, 0.8+3) #obs5
>>>>>>>>>Pn2<-rbeta(1000, 0.2+1, 0.8+3)
>>>>>>>>>Fm2<- ecdf(Pm2)
>>>>>>>>>Fn2<- ecdf(Pn2)
>>>>>>>>>Fmm2<-Fm2(1/4)
>>>>>>>>>Fnn2<-Fn2(1/4)
>>>>>>>>>Fmm2 #0.404
>>>>>>>>>Fnn2? #0.416
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>On Sat, Feb 23, 2013 at 10:53 PM, arun kirshna [via R] <
>>>>>>>>>ml-node+s789695n4659514h45 at n4.nabble.com> wrote:
>>>>>>>>>
>>>>>>>>>> Hi,
>>>>>>>>>> d3<-structure(list(m1 = c(2, 3, 2), n1 = c(2, 2, 3), cterm1_P0L =
>>>>>>>>>> c(0.9025,
>>>>>>>>>> 0.857375, 0.9025), cterm1_P1L = c(0.64, 0.512, 0.64), cterm1_P0H =
>>>>>>>>>> c(0.9025,
>>>>>>>>>> 0.9025, 0.857375), cterm1_P1H = c(0.64, 0.64, 0.512)), .Names = c("m1",
>>>>>>>>>> "n1", "cterm1_P0L", "cterm1_P1L", "cterm1_P0H", "cterm1_P1H"), row.names =
>>>>>>>>>> c(NA,
>>>>>>>>>> 3L), class = "data.frame")
>>>>>>>>>> d2<- data.frame()
>>>>>>>>>> for (m1 in 2:3) {
>>>>>>>>>>? ? ?for (n1 in 2:3) {
>>>>>>>>>>? ? ? ? ?for (x1 in 0:(m1-1)) {
>>>>>>>>>>? ? ? ? ? ? ?for (y1 in 0:(n1-1)) {
>>>>>>>>>>? ? ? ? ?for (m in (m1+2): (7-n1)){
>>>>>>>>>>? ? ? ? ? ? ? ? for (n in (n1+2):(9-m)){
>>>>>>>>>>? ? ? ? ? ? ? ? for (x in x1:(x1+m-m1)){
>>>>>>>>>>? ? ? ? ? ? ? for(y in y1:(y1+n-n1)){
>>>>>>>>>>? d2<- rbind(d2,c(m1,n1,x1,y1,m,n,x,y))
>>>>>>>>>>? }}}}}}}}
>>>>>>>>>> colnames(d2)<-c("m1","n1","x1","y1","m","n","x","y")
>>>>>>>>>>? #or
>>>>>>>>>>
>>>>>>>>>> res1<-do.call(rbind,lapply(unique(d3$m1),function(m1)
>>>>>>>>>>? do.call(rbind,lapply(unique(d3$n1),function(n1)
>>>>>>>>>>? do.call(rbind,lapply(0:(m1-1),function(x1)
>>>>>>>>>>? do.call(rbind,lapply(0:(n1-1),function(y1)
>>>>>>>>>>? do.call(rbind,lapply((m1+2):(7-n1),function(m)
>>>>>>>>>>? do.call(rbind,lapply((n1+2):(9-m),function(n)
>>>>>>>>>>? do.call(rbind,lapply(x1:(x1+m-m1), function(x)
>>>>>>>>>>? do.call(rbind,lapply(y1:(y1+n-n1), function(y)
>>>>>>>>>>? expand.grid(m1,n1,x1,y1,m,n,x,y)) )))))))))))))))
>>>>>>>>>>? names(res1)<- c("m1","n1","x1","y1","m","n","x","y")
>>>>>>>>>>? attr(res1,"out.attrs")<-NULL
>>>>>>>>>> res1[]<- sapply(res1,as.integer)
>>>>>>>>>>
>>>>>>>>>> library(plyr)
>>>>>>>>>> res2<- join(res1,d3,by=c("m1","n1"),type="inner")
>>>>>>>>>>
>>>>>>>>>> #Assuming that these are the values you used:
>>>>>>>>>>
>>>>>>>>>> p0L<-0.05
>>>>>>>>>> p0H<-0.05
>>>>>>>>>> p1L<-0.20
>>>>>>>>>> p1H<-0.20
>>>>>>>>>> res2<- within(res2,{p1<- x/m; p2<- y/n;term2_p0<-dbinom(x1,m1, p0L,
>>>>>>>>>> log=FALSE)* dbinom(y1,n1,p0H, log=FALSE)*dbinom(x-x1,m-m1, p0L, log=FALSE)*
>>>>>>>>>> dbinom(y-y1,n-n1,p0H, log=FALSE);term2_p1<- dbinom(x1,m1, p1L, log=FALSE)*
>>>>>>>>>> dbinom(y1,n1,p1H, log=FALSE)*dbinom(x-x1,m-m1, p1L, log=FALSE)*
>>>>>>>>>> dbinom(y-y1,n-n1,p1H, log=FALSE);Pm2<-rbeta(240, 0.2+x,
>>>>>>>>>> 0.8+m-x);Pn2<-rbeta(240, 0.2+y, 0.8+n-y)})
>>>>>>>>>> Fm2<- ecdf(res2$Pm2)
>>>>>>>>>> Fn2<- ecdf(res2$Pn2)
>>>>>>>>>>
>>>>>>>>>> res3<- within(res2,{Fmm2<-Fm2(p1);Fnn2<- Fn2(p2);R2<- (Fmm2+Fnn2)/2}) #not
>>>>>>>>>> sure about this step especially the Fm2() or Fn2()
>>>>>>>>>> res3$Fmm_f2<-apply(res3[,c("R2","Fmm2")],1,min)
>>>>>>>>>>? res3$Fnn_f2<-apply(res3[,c("R2","Fnn2")],1,max)
>>>>>>>>>> res3<- within(res3,{Qm2<- 1-Fmm_f2;Qn2<- 1-Fnn_f2})
>>>>>>>>>> head(res3)
>>>>>>>>>> #? m1 n1 x1 y1 m n x y cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H
>>>>>>>>>> Pn2
>>>>>>>>>> #1? 2? 2? 0? 0 4 4 0 0? ? ?0.9025? ? ? ?0.64? ? ?0.9025? ? ? ?0.64
>>>>>>>>>> 0.001084648
>>>>>>>>>> #2? 2? 2? 0? 0 4 4 0 1? ? ?0.9025? ? ? ?0.64? ? ?0.9025? ? ? ?0.64
>>>>>>>>>> 0.504593909
>>>>>>>>>> #3? 2? 2? 0? 0 4 4 0 2? ? ?0.9025? ? ? ?0.64? ? ?0.9025? ? ? ?0.64
>>>>>>>>>> 0.541379357
>>>>>>>>>> #4? 2? 2? 0? 0 4 4 1 0? ? ?0.9025? ? ? ?0.64? ? ?0.9025? ? ? ?0.64
>>>>>>>>>> 0.138947785
>>>>>>>>>> #5? 2? 2? 0? 0 4 4 1 1? ? ?0.9025? ? ? ?0.64? ? ?0.9025? ? ? ?0.64
>>>>>>>>>> 0.272364957
>>>>>>>>>> #6? 2? 2? 0? 0 4 4 1 2? ? ?0.9025? ? ? ?0.64? ? ?0.9025? ? ? ?0.64
>>>>>>>>>> 0.761635059
>>>>>>>>>> #? ? ? ? ? ?Pm2? ?term2_p1? ? ?term2_p0? ?p2? ?p1? ? ? ? R2? ? ? Fnn2 Fmm2
>>>>>>>>>> #1 1.212348e-05 0.16777216 0.6634204313 0.00 0.00 0.0000000 0.0000000? 0.0
>>>>>>>>>> #2 1.007697e-03 0.08388608 0.0698337296 0.25 0.00 0.1791667 0.3583333? 0.0
>>>>>>>>>> #3 1.106946e-05 0.01048576 0.0018377297 0.50 0.00 0.3479167 0.6958333? 0.0
>>>>>>>>>> # 2.086758e-01 0.08388608 0.0698337296 0.00 0.25 0.2000000 0.0000000? 0.4
>>>>>>>>>> #5 2.382179e-01 0.04194304 0.0073509189 0.25 0.25 0.3791667 0.3583333? 0.4
>>>>>>>>>> #6 4.494673e-01 0.00524288 0.0001934452 0.50 0.25 0.5479167 0.6958333? 0.4
>>>>>>>>>> #? ? ?Fmm_f2? ? Fnn_f2? ? ? ?Qn2? ? ? ?Qm2
>>>>>>>>>> #1 0.0000000 0.0000000 1.0000000 1.0000000
>>>>>>>>>> #2 0.0000000 0.3583333 0.6416667 1.0000000
>>>>>>>>>> #3 0.0000000 0.6958333 0.3041667 1.0000000
>>>>>>>>>> #4 0.2000000 0.2000000 0.8000000 0.8000000
>>>>>>>>>> #5 0.3791667 0.3791667 0.6208333 0.6208333
>>>>>>>>>> #6 0.4000000 0.6958333 0.3041667 0.6000000
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> A.K.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> ________________________________
>>>>>>>>>> From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=0>>
>>>>>>>>>>
>>>>>>>>>> To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=1>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Sent: Friday, February 22, 2013 11:02 AM
>>>>>>>>>> Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Thanks!? Then I need to create new variables based on the res2.? I can't
>>>>>>>>>> find Fmm_f1, Fnn_f2, R2, Qm2, Qn2 until? running the code several times and
>>>>>>>>>> the values of Fnn_f2, Fmm_f2 are correct.
>>>>>>>>>>
>>>>>>>>>> attach(res2)
>>>>>>>>>> res2$p1<-x/m
>>>>>>>>>> res2$p2<-y/n
>>>>>>>>>> res2$term2_p0 <- dbinom(x1,m1, p0L, log=FALSE)* dbinom(y1,n1,p0H,
>>>>>>>>>> log=FALSE)*dbinom(x-x1,m-m1, p0L, log=FALSE)* dbinom(y-y1,n-n1,p0H,
>>>>>>>>>> log=FALSE)
>>>>>>>>>> res2$term2_p1 <- dbinom(x1,m1, p1L, log=FALSE)* dbinom(y1,n1,p1H,
>>>>>>>>>> log=FALSE)*dbinom(x-x1,m-m1, p1L, log=FALSE)* dbinom(y-y1,n-n1,p1H,
>>>>>>>>>> log=FALSE)
>>>>>>>>>> Pm2<-rbeta(1000, 0.2+x, 0.8+m-x)
>>>>>>>>>> Fm2<-ecdf(Pm2)
>>>>>>>>>> res2$Fmm2<-Fm2(x/m)? #not correct, it comes out after running code two
>>>>>>>>>> times
>>>>>>>>>> Pn2<-rbeta(1000, 0.2+y, 0.8+n-y)
>>>>>>>>>> Fn2<-ecdf(Pn2)
>>>>>>>>>> res2$Fnn2<-Fn2(y/n)
>>>>>>>>>> res2$R2<-(Fmm2+Fnn2)/2
>>>>>>>>>> res2$Fmm_f2<-min(R2,Fmm2)? # not correct
>>>>>>>>>> res2$Fnn_f2<-max(R2,Fnn2)
>>>>>>>>>> res2$Qm2<-(1-Fmm_f2)
>>>>>>>>>> res2$Qn2<-(1-Fnn_f2)
>>>>>>>>>> detach(res2)
>>>>>>>>>> res2
>>>>>>>>>> head(res2)
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Tue, Feb 19, 2013 at 4:09 PM, arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=2>>
>>>>>>>>>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>> Hi,
>>>>>>>>>>
>>>>>>>>>> >
>>>>>>>>>> >""suppose that I have a dataset 'd'
>>>>>>>>>> >? ?m1? n1? ? A? ? ? ? ? ? ?B? ? ?C? ? ? ? ?D
>>>>>>>>>> >1? 2? 2? ?0.902500? ? 0.640? ?0.9025? ? 0.64
>>>>>>>>>> >2? 3? 2? ?0.857375? ? 0.512? ?0.9025? ? 0.64
>>>>>>>>>> >I want to add? x1 (from 0 to m1), y1(from 0 to n1), m (range from
>>>>>>>>>> >m1+2 to 7-n1), n(from n1+2 to 9-m), x (x1 to x1+m-m1), y(y1 to y1+n-n1),
>>>>>>>>>> expanding to another dataset 'd2' based on each row (combination of m1
>>>>>>>>>> >and n1)""
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >Try:
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> > d<-read.table(text="
>>>>>>>>>> >
>>>>>>>>>> >m1? n1? ? A? ? ? ? ? ? ?B? ? ?C? ? ? ? ?D
>>>>>>>>>> >1? 2? 2? ?0.902500? ? 0.640? ?0.9025? ? 0.64
>>>>>>>>>> >2? 3? 2? ?0.857375? ? 0.512? ?0.9025? ? 0.64
>>>>>>>>>> >",sep="",header=TRUE)
>>>>>>>>>> >
>>>>>>>>>> >vec1<- paste(d[,1],d[,2],d[,3],d[,4],d[,5],d[,6])
>>>>>>>>>> >res1<- do.call(rbind,lapply(vec1,function(m1)
>>>>>>>>>> do.call(rbind,lapply(0:(as.numeric(substr(m1,1,1))),function(x1)
>>>>>>>>>> do.call(rbind,lapply(0:(as.numeric(substr(m1,3,3))),function(y1)
>>>>>>>>>> do.call(rbind,lapply((as.numeric(substr(m1,1,1))+2):(7-as.numeric(substr(m1,3,3))),function(m)
>>>>>>>>>> do.call(rbind,lapply((as.numeric(substr(m1,3,3))+2):(9-m),function(n)
>>>>>>>>>> >
>>>>>>>>>> > do.call(rbind,lapply(x1:(x1+m-as.numeric(substr(m1,1,1))), function(x)
>>>>>>>>>> > do.call(rbind,lapply(y1:(y1+n-as.numeric(substr(m1,3,3))), function(y)
>>>>>>>>>> > expand.grid(m1,x1,y1,m,n,x,y)) )))))))))))))
>>>>>>>>>> >
>>>>>>>>>> names(res1)<- c("group","x1","y1","m","n","x","y")
>>>>>>>>>>
>>>>>>>>>> > res1$m1<- NA; res1$n1<- NA; res1$A<- NA; res1$B<- NA; res1$C<- NA;res1$D
>>>>>>>>>> <- NA
>>>>>>>>>> >res1[,8:13]<-do.call(rbind,lapply(strsplit(as.character(res1$group),"
>>>>>>>>>> "),as.numeric))
>>>>>>>>>> >res2<- res1[,c(8:9,2:7,10:13)]
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> > head(res2)
>>>>>>>>>> >#? m1 n1 x1 y1 m n x y? ? ? A? ? B? ? ? C? ? D
>>>>>>>>>> >#1? 2? 2? 0? 0 4 4 0 0 0.9025 0.64 0.9025 0.64
>>>>>>>>>> >#2? 2? 2? 0? 0 4 4 0 1 0.9025 0.64 0.9025 0.64
>>>>>>>>>> >#3? 2? 2? 0? 0 4 4 0 2 0.9025 0.64 0.9025 0.64
>>>>>>>>>> >#4? 2? 2? 0? 0 4 4 1 0 0.9025 0.64 0.9025 0.64
>>>>>>>>>> >#5? 2? 2? 0? 0 4 4 1 1 0.9025 0.64 0.9025 0.64
>>>>>>>>>> >#6? 2? 2? 0? 0 4 4 1 2 0.9025 0.64 0.9025 0.64
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >________________________________
>>>>>>>>>> >From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=3>>
>>>>>>>>>>
>>>>>>>>>> >To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=4>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> >Sent: Tuesday, February 19, 2013 11:43 AM
>>>>>>>>>> >
>>>>>>>>>> >Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >Thanks. I can get the data I expected (get rid of the m1=3, n1=3) using
>>>>>>>>>> the join and 'inner' code, but just curious about the way to expand the
>>>>>>>>>> data. There should be a way to expand the data based on each row
>>>>>>>>>> (combination of the variables), unique(d3$m1 & d3$n1) ?.
>>>>>>>>>> >
>>>>>>>>>> >or is there a way to use 'data.frame' and 'for' loop to expand directly
>>>>>>>>>> from the data? like res1<-data.frame (d3) for () {....
>>>>>>>>>> >
>>>>>>>>>> >
>>>>>>>>>> >On Tue, Feb 19, 2013 at 9:55 AM, arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=5>>
>>>>>>>>>
>>>>>>>>>> wrote:
>>>>>>>>>> >
>>>>>>>>>> >If you can provide me the output that you expect with all the rows of the
>>>>>>>>>> combination in the res2, I can take a look.
>>>>>>>>>> >>
>>>>>>>>>> >>
>>>>>>>>>> >>
>>>>>>>>>> >>
>>>>>>>>>> >>
>>>>>>>>>> >>
>>>>>>>>>> >>________________________________
>>>>>>>>>> >>
>>>>>>>>>> >>From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=6>>
>>>>>>>>>>
>>>>>>>>>> >>To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=7>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> >>
>>>>>>>>>>? >>Sent: Tuesday, February 19, 2013 10:42 AM
>>>>>>>>>> >>
>>>>>>>>>> >>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>>> >>
>>>>>>>>>> >>
>>>>>>>>>> >>Thanks. But I thougth the expanded dataset 'res1' should not have
>>>>>>>>>> combination of m1=3 and n1=3 because it is based on dataset 'd3' which
>>>>>>>>>> doesn't have m1=3 and n1=3, right?>
>>>>>>>>>> >>>In the example that you provided:
>>>>>>>>>> >>> (m1+2):(maxN-(n1+2))
>>>>>>>>>> >>>#[1] 5
>>>>>>>>>> >>> (n1+2):(maxN-5)
>>>>>>>>>> >>>#[1] 4
>>>>>>>>>> >>>#Suppose
>>>>>>>>>> >>> x1<- 4
>>>>>>>>>> >>> y1<- 2
>>>>>>>>>> >>> x1:(x1+5-m1)
>>>>>>>>>> >>>#[1] 4 5 6
>>>>>>>>>> >>> y1:(y1+4-n1)
>>>>>>>>>> >>>#[1] 2 3 4
>>>>>>>>>> >>>
>>>>>>>>>> >>> datnew<-expand.grid(5,4,4:6,2:4)
>>>>>>>>>> >>> colnames(datnew)<- c("m","n","x","y")
>>>>>>>>>> >>>datnew<-within(datnew,{p1<- x/m;p2<-y/n})
>>>>>>>>>> >>>res<-cbind(datnew,d2[rep(1:nrow(d2),nrow(datnew)),])
>>>>>>>>>> >>> row.names(res)<- 1:nrow(res)
>>>>>>>>>> >>> res
>>>>>>>>>> >>>#? m n x y? ?p2? p1 m1 n1 cterm1_P1L cterm1_P0H
>>>>>>>>>> >>>#1 5 4 4 2 0.50 0.8? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#2 5 4 5 2 0.50 1.0? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#3 5 4 6 2 0.50 1.2? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#4 5 4 4 3 0.75 0.8? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#5 5 4 5 3 0.75 1.0? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#6 5 4 6 3 0.75 1.2? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#7 5 4 4 4 1.00 0.8? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#8 5 4 5 4 1.00 1.0? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>#9 5 4 6 4 1.00 1.2? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>
>>>>>>>>>> >>>A.K.
>>>>>>>>>> >>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>----- Original Message -----
>>>>>>>>>> >>>From: Zjoanna <[hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=8>>
>>>>>>>>>>
>>>>>>>>>> >>>To: [hidden email]<http://user/SendEmail.jtp?type=node&node=4659514&i=9>
>>>>>>>>>
>>>>>>>>>> >>>Cc:
>>>>>>>>>> >>>
>>>>>>>>>> >>>Sent: Sunday, February 10, 2013 6:04 PM
>>>>>>>>>> >>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>>> >>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>Hi,
>>>>>>>>>> >>>How to expand or loop for one variable n based on another variable? for
>>>>>>>>>> >>>example, I want to add m (from m1 to maxN- n1-2) and for each m, I want
>>>>>>>>>> to
>>>>>>>>>> >>>add n (n1+2 to maxN-m), and similarly add x and y, then I need to do
>>>>>>>>>> some
>>>>>>>>>> >>>calculations.
>>>>>>>>>> >>>
>>>>>>>>>> >>>d3<-data.frame(d2)
>>>>>>>>>> >>>? ? for (m in (m1+2):(maxN-(n1+2)){
>>>>>>>>>> >>>? ? ? ?for (n in (n1+2):(maxN-m)){
>>>>>>>>>> >>>? ? ? ? ? ? ?for (x in x1:(x1+m-m1)){
>>>>>>>>>> >>>? ? ? ? ? ? ? ? ? for (y in y1:(y1+n-n1)){
>>>>>>>>>> >>>? ? ? ? ? ? ? ? ? ? ? ?p1<- x/m
>>>>>>>>>> >>>? ? ? ? ? ? ? ? ? ? ? ?p2<- y/n
>>>>>>>>>> >>>}}}}
>>>>>>>>>> >>>
>>>>>>>>>> >>>On Thu, Feb 7, 2013 at 12:16 AM, arun kirshna [via R] <
>>>>>>>>>>? >>>[hidden email] <http://user/SendEmail.jtp?type=node&node=4659514&i=10>>
>>>>>>>>>
>>>>>>>>>> wrote:
>>>>>>>>>> >>>
>>>>>>>>>> >>>> Hi,
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> Anyway, just using some random combinations:
>>>>>>>>>> >>>>? dnew<- expand.grid(4:10,5:10,6:10,3:7,4:5,6:8)
>>>>>>>>>> >>>> names(dnew)<-c("m","n","x1","y1","x","y")
>>>>>>>>>> >>>> resF<- cbind(dnew,d2[rep(1:nrow(d2),nrow(dnew)),])
>>>>>>>>>> >>>>
>>>>>>>>>> >>>>? row.names(resF)<- 1:nrow(resF)
>>>>>>>>>> >>>>? head(resF)
>>>>>>>>>> >>>> #? m n x1 y1 x y m1 n1 cterm1_P1L cterm1_P0H
>>>>>>>>>> >>>> #1 4 5? 6? 3 4 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> #2 5 5? 6? 3 4 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> #3 6 5? 6? 3 4 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> #4 7 5? 6? 3 4 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> #5 8 5? 6? 3 4 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> #6 9 5? 6? 3 4 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>>
>>>>>>>>>> >>>>? nrow(resF)
>>>>>>>>>> >>>> #[1] 6300
>>>>>>>>>> >>>> I am not sure what you want to do with this.
>>>>>>>>>> >>>> A.K.
>>>>>>>>>> >>>> ________________________________
>>>>>>>>>> >>>> From: Joanna Zhang <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=0>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> To: arun <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=1>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> Sent: Wednesday, February 6, 2013 10:29 AM
>>>>>>>>>> >>>> Subject: Re: cumulative sum by group and under some criteria
>>>>>>>>>> >>>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> Hi,
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> Thanks! I need to do some calculations in the expended data, the
>>>>>>>>>> expended
>>>>>>>>>> >>>> data would be very large, what is an efficient way, doing
>>>>>>>>>> calculations
>>>>>>>>>> >>>> while expending the data, something similiar with the following, or
>>>>>>>>>> >>>> expending data using the code in your message and then add
>>>>>>>>>> calculations in
>>>>>>>>>> >>>> the expended data?
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> d3<-data.frame(d2)
>>>>>>>>>> >>>>? ? for .......{
>>>>>>>>>> >>>>? ? ? ? ? for {
>>>>>>>>>> >>>>? ? ? ? ? ? ? ?for .... {
>>>>>>>>>> >>>>? ? ? ? ? ? ? ? ? ?for .....{
>>>>>>>>>> >>>>? ? ? ? ? ? ? ? ? ? ? ? p1<- x/m
>>>>>>>>>> >>>>? ? ? ? ? ? ? ? ? ? ? ? p2<- y/n
>>>>>>>>>> >>>>? ? ? ? ? ? ? ? ? ? ? ?..........
>>>>>>>>>> >>>> }}
>>>>>>>>>> >>>> }}
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> I also modified your code for expending data:
>>>>>>>>>> >>>> dnew<-expand.grid((m1+2):(maxN-(n1+2)),(n1+2):(maxN-m),0:m1,0:n1,
>>>>>>>>>> >>>> x1:(x1+m-m1),y1:(y1+n-n1))
>>>>>>>>>> >>>> names(dnew)<-c("m","n","x1","y1","x","y")
>>>>>>>>>> >>>> dnew
>>>>>>>>>> >>>> resF<-cbind(dnew[,c(2,1)],d2[rep(1:nrow(d2),nrow(dnew)),])? ? # this
>>>>>>>>>> is
>>>>>>>>>> >>>> not correct, how to modify it.
>>>>>>>>>> >>>> resF
>>>>>>>>>> >>>> row.names(resF)<-1:nrow(resF)
>>>>>>>>>> >>>> resF
>>>>>>>>>> >>>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> On Tue, Feb 5, 2013 at 2:46 PM, arun <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=2>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> wrote:
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> Hi,
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >You can reduce the steps to reach d2:
>>>>>>>>>> >>>> >res3<-
>>>>>>>>>> >>>> with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >#Change it to:
>>>>>>>>>> >>>> >res3new<-? aggregate(.~m1+n1,data=res2[,c(1:2,12:13)],max)
>>>>>>>>>> >>>> >res3new
>>>>>>>>>> >>>> > m1 n1 cterm1_P1L cterm1_P0H
>>>>>>>>>> >>>> >1? 2? 2? ? 0.01440 0.00273750
>>>>>>>>>> >>>> >2? 3? 2? ? 0.00032 0.00250000
>>>>>>>>>> >>>> >3? 2? 3? ? 0.01952 0.00048125
>>>>>>>>>> >>>> >d2<-res3new[res3new[,3]<0.01 & res3new[,4]<0.01,]
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> > dnew<-expand.grid(4:10,5:10)
>>>>>>>>>> >>>> > names(dnew)<-c("n","m")
>>>>>>>>>> >>>> >resF<-cbind(dnew[,c(2,1)],d2[rep(1:nrow(d2),nrow(dnew)),])
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >row.names(resF)<-1:nrow(resF)
>>>>>>>>>> >>>> > head(resF)
>>>>>>>>>> >>>> >#? m n m1 n1 cterm1_P1L cterm1_P0H
>>>>>>>>>> >>>> >#1 5 4? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> >#2 5 5? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> >#3 5 6? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> >#4 5 7? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> >#5 5 8? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> >#6 5 9? 3? 2? ? 0.00032? ? ?0.0025
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >A.K.
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >________________________________
>>>>>>>>>> >>>> >From: Joanna Zhang <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=3>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >To: arun <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=4>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >Sent: Tuesday, February 5, 2013 2:48 PM
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >Subject: Re: cumulative sum by group and under some criteria
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >? Hi ,
>>>>>>>>>> >>>> >what I want is :
>>>>>>>>>> >>>> >m? ?n? ? m1? ? n1 cterm1_P1L? ?cterm1_P0H
>>>>>>>>>> >>>> > 5? ?4? ? 3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> > 5? ?5? ? 3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> > 5? ?6? ? 3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> > 5? ?7? ? 3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> > 5? ?8? ?3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> > 5? ?9? ?3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >5? ?10? 3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >6? ? 4? ?3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >6? ? 5? ?3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >6? ? 6? ?3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >6? ? 7? ?3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >.....
>>>>>>>>>> >>>> >6? ? 10? 3? ? ? ?2? ? 0.00032? ? ? ? ?0.00250000
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >On Tue, Feb 5, 2013 at 1:12 PM, arun <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=5>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> wrote:
>>>>>>>>>> >>>> >
>>>>>>>>>> >>>> >Hi,
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>Saw your message on Nabble.
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>"I want to add some more columns based on the results. Is the
>>>>>>>>>> following
>>>>>>>>>> >>>> code good way to create such a data frame and How to see the column m
>>>>>>>>>> and n
>>>>>>>>>> >>>> in the updated data?
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>d2<- reres3[res3[,3]<0.01 & res3[,4]<0.01,]
>>>>>>>>>> >>>> >># should be a typo
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>colnames(d2)[1:2]<- c("m1","n1");
>>>>>>>>>> >>>> >>d2 #already a data.frame
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>d3<-data.frame(d2)
>>>>>>>>>> >>>> >>? ?for (m in (m1+2):10){
>>>>>>>>>> >>>> >>? ? ? ? for (n in (n1+2):10){
>>>>>>>>>> >>>> >> d3<-rbind(d3, c(d2))}}" #this is not making much sense to me.
>>>>>>>>>> >>>>? Especially, you mentioned you wanted add more columns.
>>>>>>>>>> >>>> >>#Running this step gave error
>>>>>>>>>> >>>> >>#Error: object 'm1' not found
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>Not sure what you want as output.
>>>>>>>>>> >>>> >>Could you show the ouput that is expected:
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>A.K.
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>________________________________
>>>>>>>>>> >>>> >>From: Joanna Zhang <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=6>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >>To: arun <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=7>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >>Sent: Tuesday, February 5, 2013 10:23 AM
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>Subject: Re: cumulative sum by group and under some criteria
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>Hi,
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>Yes, I changed code. You answered the questions. But how can I put
>>>>>>>>>> two
>>>>>>>>>> >>>> criteria in the code, if both the maximum value of cterm1_p1L <= 0.01
>>>>>>>>>> and
>>>>>>>>>> >>>> cterm1_p1H <=0.01, the output the m1,n1.
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>>? >>On Tue, Feb 5, 2013 at 8:47 AM, arun <[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=8>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> wrote:
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>> HI,
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>I am not getting the same results as yours:? You must have changed
>>>>>>>>>> the
>>>>>>>>>> >>>> dataset.
>>>>>>>>>> >>>> >>> res2[,1:2][res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95,]
>>>>>>>>>> >>>> >>>? ?m1 n1
>>>>>>>>>> >>>> >>>1? ?2? 2
>>>>>>>>>> >>>> >>>2? ?2? 2
>>>>>>>>>> >>>> >>>3? ?2? 2
>>>>>>>>>> >>>> >>>4? ?2? 2
>>>>>>>>>> >>>> >>>5? ?2? 2
>>>>>>>>>> >>>> >>>6? ?2? 2
>>>>>>>>>> >>>> >>>7? ?2? 2
>>>>>>>>>> >>>> >>>8? ?2? 2
>>>>>>>>>> >>>> >>>9? ?2? 2
>>>>>>>>>> >>>> >>>10? 3? 2
>>>>>>>>>> >>>> >>>11? 3? 2
>>>>>>>>>> >>>> >>>12? 3? 2
>>>>>>>>>> >>>> >>>13? 3? 2
>>>>>>>>>> >>>> >>>14? 3? 2
>>>>>>>>>> >>>> >>>15? 3? 2
>>>>>>>>>> >>>> >>>16? 3? 2
>>>>>>>>>> >>>> >>>17? 3? 2
>>>>>>>>>> >>>> >>>18? 3? 2
>>>>>>>>>> >>>> >>>19? 3? 2
>>>>>>>>>> >>>> >>>20? 3? 2
>>>>>>>>>> >>>> >>>21? 3? 2
>>>>>>>>>> >>>> >>>22? 2? 3
>>>>>>>>>> >>>> >>>23? 2? 3
>>>>>>>>>> >>>> >>>24? 2? 3
>>>>>>>>>> >>>> >>>25? 2? 3
>>>>>>>>>> >>>> >>>26? 2? 3
>>>>>>>>>> >>>> >>>27? 2? 3
>>>>>>>>>> >>>> >>>28? 2? 3
>>>>>>>>>> >>>> >>>29? 2? 3
>>>>>>>>>> >>>> >>>30? 2? 3
>>>>>>>>>> >>>> >>>31? 2? 3
>>>>>>>>>> >>>> >>>32? 2? 3
>>>>>>>>>> >>>> >>>33? 2? 3
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Regarding the maximum value within each block, haven't I answered
>>>>>>>>>> in
>>>>>>>>>> >>>> the earlier post.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>aggregate(cterm1_P1L~m1+n1,data=res2,max)
>>>>>>>>>> >>>> >>>#? m1 n1 cterm1_P1L
>>>>>>>>>> >>>> >>>#1? 2? 2? ? 0.01440
>>>>>>>>>> >>>> >>>#2? 3? 2? ? 0.00032
>>>>>>>>>> >>>> >>>#3? 2? 3? ? 0.01952
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
>>>>>>>>>> >>>> >>>#? Group.1 Group.2 cterm1_P1L cterm1_P0H
>>>>>>>>>> >>>> >>>#1? ? ? ?2? ? ? ?2? ? 0.01440 0.00273750
>>>>>>>>>> >>>> >>>#2? ? ? ?3? ? ? ?2? ? 0.00032 0.00250000
>>>>>>>>>> >>>> >>>#3? ? ? ?2? ? ? ?3? ? 0.01952 0.00048125
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>A.K.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>----- Original Message -----
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>From: "[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=9>";;;;;;;;;;
>>>>>>>>>> >>>> <[hidden email] <
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=10>>
>>>>>>>>>> >>>> >>>To: [hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=11>
>>>>>>>>>> >>>>? >>>Cc:
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Sent: Tuesday, February 5, 2013 9:33 AM
>>>>>>>>>> >>>> >>>Subject: Re: cumulative sum by group and under some criteria
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Hi,
>>>>>>>>>> >>>> >>>If use this
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>res2[,1:2][res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95,]
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>the results are the following, but actually only m1=3, n1=2
>>>>>>>>>> sastify the
>>>>>>>>>> >>>> criteria, as I need to look at the row with maximum value within each
>>>>>>>>>> >>>> block,not every row.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>? ?m1 n1
>>>>>>>>>> >>>> >>>1? ?2? 2
>>>>>>>>>> >>>> >>>10? 3? 2
>>>>>>>>>> >>>> >>>11? 3? 2
>>>>>>>>>> >>>> >>>12? 3? 2
>>>>>>>>>> >>>> >>>13? 3? 2
>>>>>>>>>> >>>> >>>14? 3? 2
>>>>>>>>>> >>>> >>>15? 3? 2
>>>>>>>>>> >>>> >>>16? 3? 2
>>>>>>>>>> >>>> >>>17? 3? 2
>>>>>>>>>> >>>> >>>18? 3? 2
>>>>>>>>>> >>>> >>>19? 3? 2
>>>>>>>>>> >>>> >>>20? 3? 2
>>>>>>>>>> >>>> >>>21? 3? 2
>>>>>>>>>> >>>> >>>22? 2? 3
>>>>>>>>>> >>>> >>>23? 2? 3
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>><quote author='arun kirshna'>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Hi,
>>>>>>>>>> >>>> >>>Thanks. This extract every row that satisfy the condition, but I
>>>>>>>>>> need
>>>>>>>>>> >>>> look
>>>>>>>>>> >>>> >>>at the last row (the maximum of cumulative sum) for each block
>>>>>>>>>> (m1,n1).
>>>>>>>>>> >>>> for
>>>>>>>>>> >>>> >>>example, if I set the criteria
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95, this should extract
>>>>>>>>>> m1= 3,
>>>>>>>>>> >>>> n1 =
>>>>>>>>>> >>>> >>>2.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Hi,
>>>>>>>>>> >>>> >>>I am not sure I understand your question.
>>>>>>>>>> >>>> >>>res2$cterm1_P1L<0.6 & res2$cterm1_P0H<0.95
>>>>>>>>>> >>>> >>> #[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>>>>>>>>>> TRUE
>>>>>>>>>> >>>> TRUE
>>>>>>>>>> >>>> >>>TRUE
>>>>>>>>>> >>>> >>>#[16] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>>>>>>>>>> TRUE
>>>>>>>>>> >>>> TRUE
>>>>>>>>>> >>>> >>>TRUE
>>>>>>>>>> >>>> >>>#[31] TRUE TRUE TRUE
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>This will extract all the rows.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>res2[,1:2][res2$cterm1_P1L<0.01 & res2$cterm1_P1L!=0,]
>>>>>>>>>> >>>> >>>#? ?m1 n1
>>>>>>>>>> >>>> >>>#21? 3? 2
>>>>>>>>>> >>>> >>>This extract only the row you wanted.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>For the different groups:
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>aggregate(cterm1_P1L~m1+n1,data=res2,max)
>>>>>>>>>> >>>> >>>#? m1 n1 cterm1_P1L
>>>>>>>>>> >>>> >>>#1? 2? 2? ? 0.01440
>>>>>>>>>> >>>> >>>#2? 3? 2? ? 0.00032
>>>>>>>>>> >>>> >>>#3? 2? 3? ? 0.01952
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>> aggregate(cterm1_P1L~m1+n1,data=res2,function(x) max(x)<0.01)
>>>>>>>>>> >>>> >>> # m1 n1 cterm1_P1L
>>>>>>>>>> >>>> >>>#1? 2? 2? ? ? FALSE
>>>>>>>>>> >>>> >>>#2? 3? 2? ? ? ?TRUE
>>>>>>>>>> >>>> >>>#3? 2? 3? ? ? FALSE
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>res4<-aggregate(cterm1_P1L~m1+n1,data=res2,function(x)
>>>>>>>>>> max(x)<0.01)
>>>>>>>>>> >>>> >>>res4[,1:2][res4[,3],]
>>>>>>>>>> >>>> >>>#? m1 n1
>>>>>>>>>> >>>> >>>#2? 3? 2
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>A.K.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>----- Original Message -----
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>From: "[hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=12>";;;;;;;;;;
>>>>>>>>>> >>>> <[hidden email] <
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=13>>
>>>>>>>>>> >>>> >>>To: [hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=14>
>>>>>>>>>> >>>>? >>>Cc:
>>>>>>>>>> >>>> >>>Sent: Sunday, February 3, 2013 3:58 PM
>>>>>>>>>> >>>> >>>Subject: Re: cumulative sum by group and under some criteria
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Hi,
>>>>>>>>>> >>>> >>>Let me restate my questions. I need to get the m1 and n1 that
>>>>>>>>>> satisfy
>>>>>>>>>> >>>> some
>>>>>>>>>> >>>> >>>criteria, for example in this case, within each group, the maximum
>>>>>>>>>> >>>> >>>cterm1_p1L ( the last row in this group) <0.01. I need to extract
>>>>>>>>>> m1=3,
>>>>>>>>>> >>>> >>>n1=2, I only need m1, n1 in the row.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Also, how to create the structure from the data.frame, I am new to
>>>>>>>>>> R, I
>>>>>>>>>> >>>> need
>>>>>>>>>> >>>> >>>to change the maxN and run the loop to different data.
>>>>>>>>>> >>>> >>>Thanks very much for your help!
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>><quote author='arun kirshna'>
>>>>>>>>>> >>>> >>>HI,
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>I think this should be more correct:
>>>>>>>>>> >>>> >>>maxN<-9
>>>>>>>>>> >>>> >>>c11<-0.2
>>>>>>>>>> >>>> >>>c12<-0.2
>>>>>>>>>> >>>> >>>p0L<-0.05
>>>>>>>>>> >>>> >>>p0H<-0.05
>>>>>>>>>> >>>> >>>p1L<-0.20
>>>>>>>>>> >>>> >>>p1H<-0.20
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>d <- structure(list(m1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
>>>>>>>>>> >>>> >>>2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
>>>>>>>>>> >>>> >>>? ? n1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
>>>>>>>>>> >>>> >>>? ? 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), x1 = c(0,
>>>>>>>>>> >>>> >>>? ? 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2,
>>>>>>>>>> >>>> >>>? ? 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3), y1 = c(0, 1, 2, 0,
>>>>>>>>>> >>>> >>>? ? 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1,
>>>>>>>>>> >>>> >>>? ? 2, 0, 1, 2, 0, 1, 2, 0, 1, 2), Fmm = c(0, 0, 0, 0.7, 0.59,
>>>>>>>>>> >>>> >>>? ? 0.64, 1, 1, 1, 0, 0, 0, 0, 0.63, 0.7, 0.74, 0.68, 1, 1, 1,
>>>>>>>>>> >>>> >>>? ? 1, 0, 0, 0, 0.62, 0.63, 0.6, 0.63, 0.6, 0.68, 1, 1, 1), Fnn =
>>>>>>>>>> c(0,
>>>>>>>>>> >>>> >>>? ? 0.64, 1, 0, 0.51, 1, 0, 0.67, 1, 0, 0.62, 0.69, 1, 0, 0.54,
>>>>>>>>>> >>>> >>>? ? 0.62, 1, 0, 0.63, 0.73, 1, 0, 0.63, 1, 0, 0.7, 1, 0, 0.7,
>>>>>>>>>> >>>> >>>? ? 1, 0, 0.58, 1), Qm = c(1, 1, 1, 0.65, 0.45, 0.36, 0.5, 0.165,
>>>>>>>>>> >>>> >>>? ? 0, 1, 1, 1, 1, 0.685, 0.38, 0.32, 0.32, 0.5, 0.185, 0.135,
>>>>>>>>>> >>>> >>>? ? 0, 1, 1, 1, 0.69, 0.37, 0.4, 0.685, 0.4, 0.32, 0.5, 0.21,
>>>>>>>>>> >>>> >>>? ? 0), Qn = c(1, 0.36, 0, 0.65, 0.45, 0, 0.5, 0.165, 0, 1, 0.38,
>>>>>>>>>> >>>> >>>? ? 0.31, 0, 0.685, 0.38, 0.32, 0, 0.5, 0.185, 0.135, 0, 1, 0.37,
>>>>>>>>>> >>>> >>>? ? 0, 0.69, 0.3, 0, 0.685, 0.3, 0, 0.5, 0.21, 0), term1_p0 =
>>>>>>>>>> >>>> c(0.81450625,
>>>>>>>>>> >>>> >>>? ? 0.0857375, 0.00225625, 0.0857375, 0.009025, 0.0002375,
>>>>>>>>>> 0.00225625,
>>>>>>>>>> >>>> >>>? ? 0.0002375, 6.25e-06, 0.7737809375, 0.1221759375,
>>>>>>>>>> >>>> 0.00643031249999999,
>>>>>>>>>> >>>> >>>? ? 0.0001128125, 0.081450625, 0.012860625, 0.000676875,
>>>>>>>>>> 1.1875e-05,
>>>>>>>>>> >>>> >>>? ? 0.0021434375, 0.0003384375, 1.78125e-05, 3.125e-07,
>>>>>>>>>> 0.7737809375,
>>>>>>>>>> >>>> >>>? ? 0.081450625, 0.0021434375, 0.1221759375, 0.012860625,
>>>>>>>>>> >>>> 0.0003384375,
>>>>>>>>>> >>>> >>>? ? 0.00643031249999999, 0.000676875, 1.78125e-05, 0.0001128125,
>>>>>>>>>> >>>> >>>? ? 1.1875e-05, 3.125e-07), term1_p1 = c(0.4096, 0.2048, 0.0256,
>>>>>>>>>> >>>> >>>? ? 0.2048, 0.1024, 0.0128, 0.0256, 0.0128, 0.0016, 0.32768,
>>>>>>>>>> >>>> >>>? ? 0.24576, 0.06144, 0.00512, 0.16384, 0.12288, 0.03072, 0.00256,
>>>>>>>>>> >>>> >>>? ? 0.02048, 0.01536, 0.00384, 0.00032, 0.32768, 0.16384, 0.02048,
>>>>>>>>>> >>>> >>>? ? 0.24576, 0.12288, 0.01536, 0.06144, 0.03072, 0.00384, 0.00512,
>>>>>>>>>> >>>> >>>? ? 0.00256, 0.00032)), .Names = c("m1", "n1", "x1", "y1", "Fmm",
>>>>>>>>>> >>>> >>>"Fnn", "Qm", "Qn", "term1_p0", "term1_p1"), row.names = c(NA,
>>>>>>>>>> >>>> >>>33L), class = "data.frame")
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>library(zoo)
>>>>>>>>>> >>>> >>>lst1<- split(d,list(d$m1,d$n1))
>>>>>>>>>> >>>> >>>res2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
>>>>>>>>>> >>>> >>>x[,11:14]<-NA;
>>>>>>>>>> >>>> >>>x[,11:12][x$Qm<=c11,]<-cumsum(x[,9:10][x$Qm<=c11,]);
>>>>>>>>>> >>>> >>>x[,13:14][x$Qn<=c12,]<-cumsum(x[,9:10][x$Qn<=c12,]);
>>>>>>>>>> >>>> >>>colnames(x)[11:14]<-
>>>>>>>>>> >>>> c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H");
>>>>>>>>>> >>>> >>>x1<-na.locf(x);
>>>>>>>>>> >>>> >>>x1[,11:14][is.na(x1[,11:14])]<-0;
>
>>>>>>>>>> >>>> >>>x1}))
>>>>>>>>>> >>>> >>>row.names(res2)<- 1:nrow(res2)
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>> res2
>>>>>>>>>> >>>> >>> #? m1 n1 x1 y1? Fmm? Fnn? ? Qm? ? Qn? ? ?term1_p0 term1_p1
>>>>>>>>>> >>>> cterm1_P0L
>>>>>>>>>> >>>> >>>cterm1_P1L? ?cterm1_P0H cterm1_P1H
>>>>>>>>>> >>>> >>>
>
>>>>>>>>>> >>>> >>>#1? ?2? 2? 0? 0 0.00 0.00 1.000 1.000 0.8145062500? 0.40960
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#2? ?2? 2? 0? 1 0.00 0.64 1.000 0.360 0.0857375000? 0.20480
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#3? ?2? 2? 0? 2 0.00 1.00 1.000 0.000 0.0022562500? 0.02560
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0022562500? ? 0.02560
>>>>>>>>>> >>>> >>>#4? ?2? 2? 1? 0 0.70 0.00 0.650 0.650 0.0857375000? 0.20480
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0022562500? ? 0.02560
>>>>>>>>>> >>>> >>>#5? ?2? 2? 1? 1 0.59 0.51 0.450 0.450 0.0090250000? 0.10240
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0022562500? ? 0.02560
>>>>>>>>>> >>>> >>>#6? ?2? 2? 1? 2 0.64 1.00 0.360 0.000 0.0002375000? 0.01280
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024937500? ? 0.03840
>>>>>>>>>> >>>> >>>#7? ?2? 2? 2? 0 1.00 0.00 0.500 0.500 0.0022562500? 0.02560
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024937500? ? 0.03840
>>>>>>>>>> >>>> >>>#8? ?2? 2? 2? 1 1.00 0.67 0.165 0.165 0.0002375000? 0.01280
>>>>>>>>>> >>>> 0.0002375000
>>>>>>>>>> >>>> >>> 0.01280 0.0027312500? ? 0.05120
>>>>>>>>>> >>>> >>>#9? ?2? 2? 2? 2 1.00 1.00 0.000 0.000 0.0000062500? 0.00160
>>>>>>>>>> >>>> 0.0002437500
>>>>>>>>>> >>>> >>> 0.01440 0.0027375000? ? 0.05280
>>>>>>>>>> >>>> >>>#10? 3? 2? 0? 0 0.00 0.00 1.000 1.000 0.7737809375? 0.32768
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#11? 3? 2? 0? 1 0.00 0.63 1.000 0.370 0.0814506250? 0.16384
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#12? 3? 2? 0? 2 0.00 1.00 1.000 0.000 0.0021434375? 0.02048
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0021434375? ? 0.02048
>>>>>>>>>> >>>> >>>#13? 3? 2? 1? 0 0.62 0.00 0.690 0.690 0.1221759375? 0.24576
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0021434375? ? 0.02048
>>>>>>>>>> >>>> >>>#14? 3? 2? 1? 1 0.63 0.70 0.370 0.300 0.0128606250? 0.12288
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0021434375? ? 0.02048
>>>>>>>>>> >>>> >>>#15? 3? 2? 1? 2 0.60 1.00 0.400 0.000 0.0003384375? 0.01536
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024818750? ? 0.03584
>>>>>>>>>> >>>> >>>#16? 3? 2? 2? 0 0.63 0.00 0.685 0.685 0.0064303125? 0.06144
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024818750? ? 0.03584
>>>>>>>>>> >>>> >>>#17? 3? 2? 2? 1 0.60 0.70 0.400 0.300 0.0006768750? 0.03072
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024818750? ? 0.03584
>>>>>>>>>> >>>> >>>#18? 3? 2? 2? 2 0.68 1.00 0.320 0.000 0.0000178125? 0.00384
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024996875? ? 0.03968
>>>>>>>>>> >>>> >>>#19? 3? 2? 3? 0 1.00 0.00 0.500 0.500 0.0001128125? 0.00512
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024996875? ? 0.03968
>>>>>>>>>> >>>> >>>#20? 3? 2? 3? 1 1.00 0.58 0.210 0.210 0.0000118750? 0.00256
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0024996875? ? 0.03968
>>>>>>>>>> >>>> >>>#21? 3? 2? 3? 2 1.00 1.00 0.000 0.000 0.0000003125? 0.00032
>>>>>>>>>> >>>> 0.0000003125
>>>>>>>>>> >>>> >>> 0.00032 0.0025000000? ? 0.04000
>>>>>>>>>> >>>> >>>#22? 2? 3? 0? 0 0.00 0.00 1.000 1.000 0.7737809375? 0.32768
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#23? 2? 3? 0? 1 0.00 0.62 1.000 0.380 0.1221759375? 0.24576
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#24? 2? 3? 0? 2 0.00 0.69 1.000 0.310 0.0064303125? 0.06144
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0000000000? ? 0.00000
>>>>>>>>>> >>>> >>>#25? 2? 3? 0? 3 0.00 1.00 1.000 0.000 0.0001128125? 0.00512
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0001128125? ? 0.00512
>>>>>>>>>> >>>> >>>#26? 2? 3? 1? 0 0.63 0.00 0.685 0.685 0.0814506250? 0.16384
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0001128125? ? 0.00512
>>>>>>>>>> >>>> >>>#27? 2? 3? 1? 1 0.70 0.54 0.380 0.380 0.0128606250? 0.12288
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0001128125? ? 0.00512
>>>>>>>>>> >>>> >>>#28? 2? 3? 1? 2 0.74 0.62 0.320 0.320 0.0006768750? 0.03072
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0001128125? ? 0.00512
>>>>>>>>>> >>>> >>>#29? 2? 3? 1? 3 0.68 1.00 0.320 0.000 0.0000118750? 0.00256
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0001246875? ? 0.00768
>>>>>>>>>> >>>> >>>#30? 2? 3? 2? 0 1.00 0.00 0.500 0.500 0.0021434375? 0.02048
>>>>>>>>>> >>>> 0.0000000000
>>>>>>>>>> >>>> >>> 0.00000 0.0001246875? ? 0.00768
>>>>>>>>>> >>>> >>>#31? 2? 3? 2? 1 1.00 0.63 0.185 0.185 0.0003384375? 0.01536
>>>>>>>>>> >>>> 0.0003384375
>>>>>>>>>> >>>> >>> 0.01536 0.0004631250? ? 0.02304
>>>>>>>>>> >>>> >>>#32? 2? 3? 2? 2 1.00 0.73 0.135 0.135 0.0000178125? 0.00384
>>>>>>>>>> >>>> 0.0003562500
>>>>>>>>>> >>>> >>> 0.01920 0.0004809375? ? 0.02688
>>>>>>>>>> >>>> >>>#33? 2? 3? 2? 3 1.00 1.00 0.000 0.000 0.0000003125? 0.00032
>>>>>>>>>> >>>> 0.0003565625
>>>>>>>>>> >>>> >>> 0.01952 0.0004812500? ? 0.02720
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>#Sorry, some values in my previous solution didn't look right. I
>>>>>>>>>> >>>> didn't
>
>>>>>>>>>> >>>> >>>A.K.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>----- Original Message -----
>>>>>>>>>> >>>> >>>From: Zjoanna <[hidden email]<
>>
>
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=15>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >>>To: [hidden email]<
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=16>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>Cc:
>>>>>>>>>> >>>> >>>Sent: Friday, February 1, 2013 12:19 PM
>
>>
>>>>>>>>>> >>>> >>>Subject: Re: [R] cumulative sum by group and under some criteria
>>>>>>>>>> >>>> >>>
>>
>
>>>>>>>>>> >>>> >>>Thank you very much for your reply. Your code work well with this
>>>>>>>>>> >>>> example.
>>>>>>>>>> >>>> >>>I modified a little to fit my real data, I got an error massage.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop,
>>>>>>>>>> ...) :
>>>>>>>>>> >>>> >>>? Group length is 0 but data length > 0
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] <
>>>>>>>>>> >>>>? >>>[hidden email] <
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=17>>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> wrote:
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>> Hi,
>>>>>>>>>> >>>> >>>> Try this:
>>>>>>>>>> >>>> >>>> colnames(d)<-c("m1","n1","x1","y1","p11","p12")
>>>>>>>>>> >>>> >>>> library(zoo)
>>>>>>>>>> >>>> >>>> res1<-
>>>>>>>>>> >>>> do.call(rbind,lapply(lapply(split(d,list(d$m1,d$n1)),function(x)
>>>>>>>>>> >>>> >>>> {x$cp11[x$x1>1]<- cumsum(x$p11[x$x1>1]);x$cp12[x$y1>1]<-
>>>>>>>>>> >>>> >>>> cumsum(x$p12[x$y1>1]);x}),function(x)
>>>>>>>>>> >>>> >>>> {x$cp11<-na.locf(x$cp11,na.rm=F);x$cp12<-
>>>>>>>>>> >>>> na.locf(x$cp12,na.rm=F);x}))
>>>>>>>>>> >>>> >>>> #there would be a warning here as one of the list element is
>>>>>>>>>> NULL.
>>>>>>>>>> >>>> The,
>>>>>>>>>> >>>> >>>> warning is okay
>>>>>>>>>> >>>> >>>> row.names(res1)<- 1:nrow(res1)
>>>>>>>>>> >>>> >>>> res1[,7:8][is.na(res1[,7:8])]<- 0
>>>>>>>>>> >>>> >>>> res1
>>>>>>>>>> >>>> >>>>? #? m1 n1 x1 y1? p11? p12 cp11 cp12
>>>>>>>>>> >>>> >>>> #1? ?2? 2? 0? 0 0.00 0.00 0.00 0.00
>>>>>>>>>> >>>> >>>> #2? ?2? 2? 0? 1 0.00 0.50 0.00 0.00
>>>>>>>>>> >>>> >>>> #3? ?2? 2? 0? 2 0.00 1.00 0.00 1.00
>>>>>>>>>> >>>> >>>> #4? ?2? 2? 1? 0 0.50 0.00 0.00 1.00
>>>>>>>>>> >>>> >>>> #5? ?2? 2? 1? 1 0.50 0.50 0.00 1.00
>>>>>>>>>> >>>> >>>> #6? ?2? 2? 1? 2 0.50 1.00 0.00 2.00
>>>>>>>>>> >>>> >>>> #7? ?2? 2? 2? 0 1.00 0.00 1.00 2.00
>>>>>>>>>> >>>> >>>> #8? ?2? 2? 2? 1 1.00 0.50 2.00 2.00
>>>>>>>>>> >>>> >>>> #9? ?2? 2? 2? 2 1.00 1.00 3.00 3.00
>>>>>>>>>> >>>> >>>> #10? 3? 2? 0? 0 0.00 0.00 0.00 0.00
>>>>>>>>>> >>>> >>>> #11? 3? 2? 0? 1 0.00 0.50 0.00 0.00
>>>>>>>>>> >>>> >>>> #12? 3? 2? 0? 2 0.00 1.00 0.00 1.00
>>>>>>>>>> >>>> >>>> #13? 3? 2? 1? 0 0.33 0.00 0.00 1.00
>>>>>>>>>> >>>> >>>> #14? 3? 2? 1? 1 0.33 0.50 0.00 1.00
>>>>>>>>>> >>>> >>>> #15? 3? 2? 1? 2 0.33 1.00 0.00 2.00
>>>>>>>>>> >>>> >>>> #16? 3? 2? 2? 0 0.67 0.00 0.67 2.00
>>>>>>>>>> >>>> >>>> #17? 3? 2? 2? 1 0.67 0.50 1.34 2.00
>>>>>>>>>> >>>> >>>> #18? 3? 2? 2? 2 0.67 1.00 2.01 3.00
>>>>>>>>>> >>>> >>>> #19? 3? 2? 3? 0 1.00 0.00 3.01 3.00
>>>>>>>>>> >>>> >>>> #20? 3? 2? 3? 1 1.00 0.50 4.01 3.00
>>>>>>>>>> >>>> >>>> #21? 3? 2? 3? 2 1.00 1.00 5.01 4.00
>>>>>>>>>> >>>> >>>> #22? 2? 3? 0? 0 0.00 0.00 0.00 0.00
>>>>>>>>>> >>>> >>>> #23? 2? 3? 0? 1 0.00 0.33 0.00 0.00
>>>>>>>>>> >>>> >>>> #24? 2? 3? 0? 2 0.00 0.67 0.00 0.67
>>>>>>>>>> >>>> >>>> #25? 2? 3? 0? 3 0.00 1.00 0.00 1.67
>>>>>>>>>> >>>> >>>> #26? 2? 3? 1? 0 0.50 0.00 0.00 1.67
>>>>>>>>>> >>>> >>>> #27? 2? 3? 1? 1 0.50 0.33 0.00 1.67
>>>>>>>>>> >>>> >>>> #28? 2? 3? 1? 2 0.50 0.67 0.00 2.34
>>>>>>>>>> >>>> >>>> #29? 2? 3? 1? 3 0.50 1.00 0.00 3.34
>>>>>>>>>> >>>> >>>> #30? 2? 3? 2? 0 1.00 0.00 1.00 3.34
>>>>>>>>>> >>>> >>>> #31? 2? 3? 2? 1 1.00 0.33 2.00 3.34
>>>>>>>>>> >>>> >>>> #32? 2? 3? 2? 2 1.00 0.67 3.00 4.01
>>
>>>>>>>>>> >>>> >>>> #33? 2? 3? 2? 3 1.00 1.00 4.00 5.01
>>>>>>>>>> >>>> >>>> A.K.
>>>>>>>>>> >>>> >>>>
>>>>>>>>>> >>>> >>>> ------------------------------
>>>>>>>>>> >>>> >>>>? If you reply to this email, your message will be added to the
>>>>>>>>>> >>>> discussion
>>>>>>>>>> >>>> >>>> below:
>>>>>>>>>> >>>> >>>>
>>>>>>>>>> >>>> >>>>
>>>>>>>>>> >>>>
>>>>>>>>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657196.html
>>>>>>>>>> >>>> >>>> To unsubscribe from cumulative sum by group and under some
>>
>>>>>>>>>> criteria,
>>>>>>>>>> >>>> click
>>>>>>>>>> >>>> >>>> here<
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >>>> .
>>>>>>>>>> >>>> >>>> NAML<
>>>>>>>>>> >>>>
>>>>>>>>>> http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewer&id=instant_html%21nabble%3Aemail.naml&base=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespace&breadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml>
>>>>>>>>>>
>>>>>>>>>> >>>>
>>>>>>>>>> >>>> >>>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>--
>>>>>>>>>> >>>> >>>View this message in context:
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>>
>>>>>>>>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657315.html
>>>>>>>>>> >>>> >>>Sent from the R help mailing list archive at Nabble.com.
>>>>>>>>>> >>>> >>>? ? [[alternative HTML version deleted]]
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>______________________________________________
>>>>>>>>>> >>>> >>>[hidden email] <
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=18>mailing list
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>https://stat.ethz.ch/mailman/listinfo/r-help
>>>>>>>>>> >>>> >>>PLEASE do read the posting guide
>>>>>>>>>> >>>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>>>>>>>>>> <http://www.r-project.org/posting-guide.html>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>and provide commented, minimal, self-contained, reproducible code.
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>
>>>>>>>>>> >>>> >>>______________________________________________
>>>>>>>>>> >>>> >>>[hidden email] <
>>>>>>>>>> http://user/SendEmail.jtp?type=node&node=4657773&i=19>mailing list
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>https://stat.ethz.ch/mailman/listinfo/r-help
>>>>>>>>>> >>>> >>>PLEASE do read the posting guide
>>>>>>>>>> >>>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>>>>>>>>>> <http://www.r-project.org/posting-guide.html>
>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>and provide commented, minimal, self-contained, reproducible code.
>>>>>>>>>> >>>> >>>
>>>>
>>>>>>>>>> >>>> >>></quote>
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>>>>>>>>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657360.html
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>>>>>>>>>> >>>
>>>>>>>>>> >>>> >>>and provide commented, minimal, self-contained, reproducible code.
>>>>>>>>>> >>>> >>>
>>>>
>>>>>>>>>> >>>> >>></quote>
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>>>>>>>>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657582.html
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>>>>>>>>>> >>>
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>>
>>>>
>>>>>
>>>>>>>>>
>>>>>>>>>https://stat.ethz.ch/mailman/listinfo/r-help
>>>>>>>>>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>>>>
>>>>
>>>>>>>>>and provide commented, minimal, self-contained, reproducible code.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>??????????????????????????????????????????? ?????????????????????? ?
>>>>>>>? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ?
>>>>>>?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ???
>>>>>
>>>>
>>>? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ???
>>
>