cumulative sum by group and under some criteria
Hi,
If you don't want the m1=3, n1=3 combination in the final dataset:
library(plyr)
res2<- join(res1,d3,by=c("m1","n1"),type="inner")
tail(res2)
#??? m1 n1 x1 y1 m n x y cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H
#235? 3? 2? 2? 1 5 4 3 1?? 0.857375????? 0.512???? 0.9025?????? 0.64
#236? 3? 2? 2? 1 5 4 3 2?? 0.857375????? 0.512???? 0.9025?????? 0.64
#237? 3? 2? 2? 1 5 4 3 3?? 0.857375????? 0.512???? 0.9025?????? 0.64
#238? 3? 2? 2? 1 5 4 4 1?? 0.857375????? 0.512???? 0.9025?????? 0.64
#239? 3? 2? 2? 1 5 4 4 2?? 0.857375????? 0.512???? 0.9025?????? 0.64
#240? 3? 2? 2? 1 5 4 4 3?? 0.857375????? 0.512???? 0.9025?????? 0.64
A.K.
----- Original Message -----
From: Zjoanna <Zjoanna2013 at gmail.com>
To: r-help at r-project.org
Cc:
Sent: Monday, February 18, 2013 10:36 PM
Subject: Re: [R] cumulative sum by group and under some criteria
Thanks. I tried this code below, why does expanded dataset 'res1' has m1=3
and n1=3
, dataset 'd3' doesn't have m1=3, n1=3.
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")
d3
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
library(plyr)
res2<- join(res1,d3,by=c("m1","n1"),type="full")
res2
On Sun, Feb 17, 2013 at 11:10 PM, arun kirshna [via R] <
ml-node+s789695n4658895h56 at n4.nabble.com> wrote:
Hi,
"Yes, I wanted to expand directly from d. If there are other variables,
says "A", "B", "C" that I need to keep? in the final expanded data, how
to modify the code?"
d<-data.frame()
for (m1 in 2:3) {
? ? for (n1 in 2:2) {
? ? ? ? for (x1 in 0:(m1-1)) {
? ? ? ? ? ? for (y1 in 0:(n1-1)) {
d<-rbind(d,c(m1,n1,x1,y1))
}
}
}}
colnames(d)<-c("m1","n1","x1","y1")
d
res1<-do.call(rbind,lapply(unique(d$m1),function(m1)
do.call(rbind,lapply(unique(d$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)
expand.grid(m1,n1,x1,y1,m,n,x)) )))))))))))))
? names(res1)<- c("m1","n1","x1","y1","m","n","x")
set.seed(235)
? d$A<- sample(1:50,10,replace=TRUE)
? set.seed(23)
? d$B<- sample(1:50,10,replace=TRUE)
? d
? #? m1 n1 x1 y1? A? B
#1? 2? 2? 0? 0 50 29
#2? 2? 2? 0? 1 40 12
#3? 2? 2? 1? 0 31 17
#4? 2? 2? 1? 1? 7 36
#5? 3? 2? 0? 0 13 41
#6? 3? 2? 0? 1 27 22
#7? 3? 2? 1? 0 49 49
#8? 3? 2? 1? 1 47 49
#9? 3? 2? 2? 0 23 43
#10? 3? 2? 2? 1? 4 50
library(plyr)
res2<- join(res1,d,by=c("m1","n1","x1","y1"),type="full")
res2
? #? m1 n1 x1 y1 m n x? A? B
#1? 2? 2? 0? 0 4 4 0 50 29
#2? 2? 2? 0? 0 4 4 1 50 29
#3? 2? 2? 0? 0 4 4 2 50 29
#4? 2? 2? 0? 0 4 5 0 50 29
#5? 2? 2? 0? 0 4 5 1 50 29
#6? 2? 2? 0? 0 4 5 2 50 29
#7? 2? 2? 0? 0 5 4 0 50 29
#8? 2? 2? 0? 0 5 4 1 50 29
#9? 2? 2? 0? 0 5 4 2 50 29
#10? 2? 2? 0? 0 5 4 3 50 29
#11? 2? 2? 0? 1 4 4 0 40 12
#12? 2? 2? 0? 1 4 4 1 40 12
#13? 2? 2? 0? 1 4 4 2 40 12
#14? 2? 2? 0? 1 4 5 0 40 12
#15? 2? 2? 0? 1 4 5 1 40 12
#16? 2? 2? 0? 1 4 5 2 40 12
#17? 2? 2? 0? 1 5 4 0 40 12
#18? 2? 2? 0? 1 5 4 1 40 12
#19? 2? 2? 0? 1 5 4 2 40 12
#20? 2? 2? 0? 1 5 4 3 40 12
#21? 2? 2? 1? 0 4 4 1 31 17
#22? 2? 2? 1? 0 4 4 2 31 17
#23? 2? 2? 1? 0 4 4 3 31 17
#24? 2? 2? 1? 0 4 5 1 31 17
#25? 2? 2? 1? 0 4 5 2 31 17
#26? 2? 2? 1? 0 4 5 3 31 17
#27? 2? 2? 1? 0 5 4 1 31 17
#28? 2? 2? 1? 0 5 4 2 31 17
#29? 2? 2? 1? 0 5 4 3 31 17
#30? 2? 2? 1? 0 5 4 4 31 17
#31? 2? 2? 1? 1 4 4 1? 7 36
#32? 2? 2? 1? 1 4 4 2? 7 36
#33? 2? 2? 1? 1 4 4 3? 7 36
#34? 2? 2? 1? 1 4 5 1? 7 36
#35? 2? 2? 1? 1 4 5 2? 7 36
#36? 2? 2? 1? 1 4 5 3? 7 36
#37? 2? 2? 1? 1 5 4 1? 7 36
#38? 2? 2? 1? 1 5 4 2? 7 36
#39? 2? 2? 1? 1 5 4 3? 7 36
#40? 2? 2? 1? 1 5 4 4? 7 36
#41? 3? 2? 0? 0 5 4 0 13 41
#42? 3? 2? 0? 0 5 4 1 13 41
#43? 3? 2? 0? 0 5 4 2 13 41
#44? 3? 2? 0? 1 5 4 0 27 22
#45? 3? 2? 0? 1 5 4 1 27 22
#46? 3? 2? 0? 1 5 4 2 27 22
#47? 3? 2? 1? 0 5 4 1 49 49
#48? 3? 2? 1? 0 5 4 2 49 49
#49? 3? 2? 1? 0 5 4 3 49 49
#50? 3? 2? 1? 1 5 4 1 47 49
#51? 3? 2? 1? 1 5 4 2 47 49
#52? 3? 2? 1? 1 5 4 3 47 49
#53? 3? 2? 2? 0 5 4 2 23 43
#54? 3? 2? 2? 0 5 4 3 23 43
#55? 3? 2? 2? 0 5 4 4 23 43
#56? 3? 2? 2? 1 5 4 2? 4 50
#57? 3? 2? 2? 1 5 4 3? 4 50
#58? 3? 2? 2? 1 5 4 4? 4 50
A.K.
----- Original Message -----
From: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4658895&i=0>>
To: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4658895&i=1>>
Cc: R help <[hidden email]<http://user/SendEmail.jtp?type=node&node=4658895&i=2>>
Sent: Saturday, February 16, 2013 11:49 PM
Subject: Re: [R] cumulative sum by group and under some criteria
d2<- data.frame()
for (m1 in 2:3) {
? ? for (n1 in 2:2) {
? ? ? ? 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)){
? d2<- rbind(d2,c(m1,n1,x1,y1,m,n,x))
? }}}}}}}
colnames(d2)<-c("m1","n1","x1","y1","m","n","x")
res<-do.call(rbind,lapply(2:3,function(m1)
do.call(rbind,lapply(2:2,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)
expand.grid(m1,n1,x1,y1,m,n,x)) )))))))))))))
names(res)<- c("m1","n1","x1","y1","m","n","x")
? attr(res,"out.attrs")<-NULL
? identical(d2,res)
#[1] TRUE
A.K.
________________________________ From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4658895&i=3>> To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4658895&i=4>> ? Sent: Saturday, February 16, 2013 8:46 PM Subject: Re: [R] cumulative sum by group and under some criteria Hi, What I need is to expand each row by adding several columns and . Let me restate the question. I have a dataset d, I want to expand it to d2 showed below. d<-data.frame() for (m1 in 2:3) { ? ? for (n1 in 2:2) { ? ? ? ? for (x1 in 0:(m1-1)) { ? ? ? ? ? ? for (y1 in 0:(n1-1)) { d<-rbind(d,c(m1,n1,x1,y1)) } } }} colnames(d)<-c("m1","n1","x1","y1") d ? ? m1 n1 x1 y1 1? 2? 2? 0? 0 2? 2? 2? 0? 1 3? 2? 2? 1? 0 4? 2? 2? 1? 1 5? 3? 2? 0? 0 6? 3? 2? 0? 1 7? 3? 2? 1? 0 8? 3? 2? 1? 1 9? 3? 2? 2? 0 10? 3? 2? 2? 1 I want to expand it as follows: for (m in (m1+2): (7-n1){ ? ? for (n in (n1+2):(9-m){ ? ? ? ? for (x in x1:(x1+m-m1){ }}} so for the first row, ? ? m1 n1 x1 y1 1? 2? 2? 0? 0 it should be expanded as ? ? m1 n1 x1 y1? m? n? x ? ? 2? 2? 0? 0? 4? 4? 0 ? ? 2? 2? 0? 0? 4? 4? 1 ? ? 2? 2? 0? 0? 4? 4? 2 ? ? 2? 2? 0? 0? 4? 5? 0 ? ? 2? 2? 0? 0? 4? 5? 1 ? ? 2? 2? 0? 0? 4? 5? 2 ? On Tue, Feb 12, 2013 at 8:19 PM, arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4658895&i=5>> wrote: Hi, Saw your reply again in Nabble.? I thought I sent you the solution previously. res3new<-? aggregate(.~m1+n1,data=res2[,c(1:2,12:13)],max) d2<-res3new[res3new[,3]<0.01 & res3new[,4]<0.01,] m1<- 3 #from d2 maxN<- 9 n1<- 2 #from d2 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=4658895&i=6>> To: [hidden email] <http://user/SendEmail.jtp?type=node&node=4658895&i=7> 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=4658895&i=8>> 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. ------------------------------ ? 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