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 <zjoanna2...@gmail.com> To: arun <smartpink...@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 <smartpink...@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 <zjoanna2...@gmail.com> >To: r-help@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+s789695n4657773...@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 >> 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< >> 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the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ R-help@r-project.org mailing list 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.