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 <zjoanna2013 at gmail.com>
To: arun <smartpink111 at yahoo.com>
Sent: Saturday, February 16, 2013 8:46 PM
Subject: Re: [R] cumulative sum by group and under some criteria
To: arun <smartpink111 at yahoo.com>
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 <smartpink111 at yahoo.com> 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 <Zjoanna2013 at gmail.com>
>To: r-help at r-project.org
>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] <
>ml-node+s789695n4657773h74 at n4.nabble.com> 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
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>> >>>
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