Hi, You can try this: n<- 3 j3<-n*seq_len(nrow(A)/n) vec1<- rep("j3",n) eval(parse(text=paste0("A","[",paste0(vec1,"-",seq(n)-1),",]",collapse="*"))) # [,1] [,2] [,3] #[1,] 28 80 162 #[2,] 162 80 28
Just saw Bert's new solution: n<-3 j <- seq_len(nrow(A))%%n b <- A[j==0,] for(i in seq_len(n-1))b <- b*A[j==i,] b # [,1] [,2] [,3] #[1,] 28 80 162 #[2,] 162 80 28 #For the bigger dataset with large "n, these methods may not work: set.seed(28) mat1<- matrix(sample(1:20,1e5*3,replace=TRUE),ncol=3) n<- 40 nrow(mat1)%%40 #[1] 0 j <- seq_len(nrow(mat1))%%n b <- mat1[j==0,] for(i in seq_len(n-1))b <- b*mat1[j==i,] Warning messages: 1: In b * mat1[j == i, ] : NAs produced by integer overflow 2: In b * mat1[j == i, ] : NAs produced by integer overflow 3: In b * mat1[j == i, ] : NAs produced by integer overflow 4: In b * mat1[j == i, ] : NAs produced by integer overflow 5: In b * mat1[j == i, ] : NAs produced by integer overflow 6: In b * mat1[j == i, ] : NAs produced by integer overflow 7: In b * mat1[j == i, ] : NAs produced by integer overflow 8: In b * mat1[j == i, ] : NAs produced by integer overflow 9: In b * mat1[j == i, ] : NAs produced by integer overflow 10: In b * mat1[j == i, ] : NAs produced by integer overflow head(b,3) # [,1] [,2] [,3] #[1,] NA NA NA #[2,] NA NA NA #[3,] NA NA NA j40<- n*seq_len(nrow(mat1)/n) vec1<- rep("j40",n) vec1<- rep("j22",n) res<- eval(parse(text= paste(paste0("mat1","[",paste0(vec1,"-",seq(n)-1),",]"),collapse="*") )) Warning messages: 1: In mat1[j22 - 0, ] * mat1[j22 - 1, ] * mat1[j22 - 2, ] * mat1[j22 - : NAs produced by integer overflow 2: In mat1[j22 - 0, ] * mat1[j22 - 1, ] * mat1[j22 - 2, ] * mat1[j22 - : NAs produced by integer overflow 3: In mat1[j22 - 0, ] * mat1[j22 - 1, ] * mat1[j22 - 2, ] * mat1[j22 - : NAs produced by integer overflow 4: In mat1[j22 - 0, ] * mat1[j22 - 1, ] * mat1[j22 - 2, ] * mat1[j22 - : head(res,3) # [,1] [,2] [,3] #[1,] NA NA NA #[2,] NA NA NA #[3,] NA NA NA A.K. ________________________________ From: Edouard Hardy <hardy.edou...@gmail.com> To: arun <smartpink...@yahoo.com> Sent: Monday, September 2, 2013 2:32 PM Subject: Re: [R] Product of certain rows in a matrix Yes, n is 250 or more... Edouard Hardy On Mon, Sep 2, 2013 at 8:31 PM, arun <smartpink...@yahoo.com> wrote: Also, BTW, are you looking for n>100? > > > > > > On Mon, Sep 2, 2013 at 8:27 PM, arun <smartpink...@yahoo.com> wrote: > >Hi, > >Not sure I understand your question. If you don't know "n", then how are you >applying other solutions also.. >A.K. > > > >Again, thank you for your help. > >I understand Bert's solution but this is possible only if I know n. > >How can I do A[j3,]*A[j3-1,]*A[j3-2,] (n=3) for n terms ? > > > >----- Original Message ----- >From: arun <smartpink...@yahoo.com> >To: Edouard Hardy <hardy.edou...@gmail.com> >Cc: R help <r-help@r-project.org>; Bert Gunter <gunter.ber...@gene.com> > >Sent: Monday, September 2, 2013 1:57 PM >Subject: Re: [R] Product of certain rows in a matrix > >HI, >You could modify Bert's solution: >n<-3 > >j3<-n*seq_len(nrow(A)/n) >A[j3,]*A[j3-1,]*A[j3-2,] ##assuming that nrow(dataset)%%n==0 ># [,1] [,2] [,3] >#[1,] 28 80 162 >#[2,] 162 80 28 > > >#Speed comparison > > >set.seed(28) >mat1<- matrix(sample(1:20,1e5*3,replace=TRUE),ncol=3) > >n<-4 >system.time({res1<- >t(sapply(split(as.data.frame(mat1),as.numeric(gl(nrow(mat1),n,nrow(mat1)))),function(x) > apply(x,2,prod))) }) ># user system elapsed ># 8.508 0.620 9.146 >system.time({res2<- >t(sapply(split(as.data.frame(mat1),as.numeric(gl(nrow(mat1),n,nrow(mat1)))),function(x) > Reduce("*",as.data.frame(t(x))))) }) ># user system elapsed ># 8.556 0.000 8.566 > >A1<- data.frame(mat1,ID=as.numeric(gl(nrow(mat1),n,nrow(mat1)))) > system.time({res3<- aggregate(A1[,-4],list(A1[,4]),colProds)[,-1]}) ># user system elapsed ># 11.536 0.000 11.553 > > >nrow(mat1)%%n >#[1] 0 >system.time({j4<- n*seq_len(nrow(mat1)/n) > res5<- mat1[j4,]*mat1[j4-1,]*mat1[j4-2,]*mat1[j4-3,] > }) > ># user system elapsed ># 0.004 0.000 0.004 > > dimnames(res2)<- dimnames(res5) >identical(res2,res5) >#[1] TRUE > > >#if >n<-6 > nrow(mat1)%%6 >#[1] 4 > > >system.time({ > mat2<-mat1[seq(nrow(mat1)-4),] >j6<- n*seq_len(nrow(mat2)/n) > res6<- mat2[j6,]*mat2[j6-1,]*mat2[j6-2,]*mat2[j6-3,]*mat2[j6-4,]*mat2[j6-5,] >res6New<-rbind(res6,apply(tail(mat1,4),2,prod) >)}) > ># user system elapsed > # 0.004 0.000 0.006 > > > >system.time({res6Alt<- >t(sapply(split(as.data.frame(mat1),as.numeric(gl(nrow(mat1),n,nrow(mat1)))),function(x) > Reduce("*",as.data.frame(t(x))))) }) >#user system elapsed > # 5.576 0.000 5.583 >dimnames(res6Alt)<- dimnames(res6New) > > >all.equal(res6New,res6Alt) >#[1] TRUE > > >A.K. > > > >As you said, this is very loooong. >Do you have a better solution on big data ? > > > >----- Original Message ----- >From: arun <smartpink...@yahoo.com> >To: Edouard Hardy <hardy.edou...@gmail.com> >Cc: R help <r-help@r-project.org>; Bert Gunter <gunter.ber...@gene.com> >Sent: Monday, September 2, 2013 12:07 PM >Subject: Re: [R] Product of certain rows in a matrix > > > >Hi, >No problem. >n<- 4 > >t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew)))),function(x) > apply(x,2,prod))) > ># V1 V2 V3 >#1 252 640 1134 >#2 18 30 20 > > >This could be a bit slow if you have big dataset. > > >A.K. > > > >________________________________ >From: Edouard Hardy <hardy.edou...@gmail.com> >To: arun <smartpink...@yahoo.com> >Cc: R help <r-help@r-project.org> >Sent: Monday, September 2, 2013 11:58 AM >Subject: Re: [R] Product of certain rows in a matrix > > > >Thank you A.K. >And do you have a solution without installing any package ? >Thank you in advance. >E.H. > > > >Edouard Hardy > > > >On Mon, Sep 2, 2013 at 5:56 PM, arun <smartpink...@yahoo.com> wrote: > > >> >>HI, >>In my first solutions: >> n<-3 >> >>t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew)))),colProds)) >># [,1] [,2] [,3] >>#1 28 80 162 >>#2 162 80 28 >>#3 1 3 5 >> n<-4 >> >>t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew)))),colProds)) >># [,1] [,2] [,3] >>#1 252 640 1134 >>#2 18 30 20 >> >>A.K. >> >>________________________________ >>From: Edouard Hardy <hardy.edou...@gmail.com> >>To: arun <smartpink...@yahoo.com> >>Cc: Bert Gunter <gunter.ber...@gene.com>; R help <r-help@r-project.org> >>Sent: Monday, September 2, 2013 11:46 AM >> >>Subject: Re: [R] Product of certain rows in a matrix >> >> >> >>Thank you all for your responses. >>The real problem is that all your answer work for products 2 by 2. >>I now have to do the product n by n row. >>Do you have a solution ? >>Thank you in advance, >>E.H. >> >> >> >>Edouard Hardy >> >> >> >>On Mon, Sep 2, 2013 at 5:43 PM, arun <smartpink...@yahoo.com> wrote: >> >>I guess in such situations, >>> >>> >>>fun1<- function(mat){ >>> if(nrow(mat)%%2==0){ >>> j<- 2*seq_len(nrow(mat)/2) >>> b<- mat[j,]* mat[j-1,] >>> } >>> else {mat1<- mat[-nrow(mat),] >>> j<- 2*seq_len(nrow(mat1)/2) >>> b<- rbind(mat1[j,]*mat1[j-1,],mat[nrow(mat),]) >>> } >>>b >>>} >>>fun1(A) >>># [,1] [,2] [,3] >>> >>>#[1,] 4 10 18 >>>#[2,] 63 64 63 >>>#[3,] 18 10 4 >>> fun1(Anew) >>># [,1] [,2] [,3] >>> >>>#[1,] 4 10 18 >>>#[2,] 63 64 63 >>>#[3,] 18 10 4 >>>#[4,] 1 3 5 >>> >>> >>>A.K. >>> >>> >>> >>> >>>----- Original Message ----- >>>From: arun <smartpink...@yahoo.com> >>>To: Bert Gunter <gunter.ber...@gene.com> >>>Cc: R help <r-help@r-project.org> >>> >>>Sent: Monday, September 2, 2013 11:26 AM >>>Subject: Re: [R] Product of certain rows in a matrix >>> >>>Hi Bert, >>>Thanks. It is a better solution. >>> >>>If nrow() is not even. >>> >>>Anew<- rbind(A,c(1,3,5)) >>>j<-seq_len(nrow(Anew)/2)### >>> Anew[j,]*Anew[j-1,] >>>#Error in Anew[j, ] * Anew[j - 1, ] : non-conformable arrays >>> >>>t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),2,7))),colProds)) >>> [,1] [,2] [,3] >>>1 4 10 18 >>>2 63 64 63 >>>3 18 10 4 >>>4 1 3 5 >>> >>>A.K. >>> >>> >>> >>> >>> >>> >>>________________________________ >>>From: Bert Gunter <gunter.ber...@gene.com> >>>To: arun <smartpink...@yahoo.com> >>>Cc: R help <r-help@r-project.org> >>>Sent: Monday, September 2, 2013 10:55 AM >>>Subject: Re: [R] Product of certain rows in a matrix >>> >>> >>> >>>These elaborate manipulations are unnecessary and inefficient. Use indexing >>>instead: >>> >>>j <- 2*seq_len(nrow(A)/2) >>>b <- A[j,]*A[j-1,] >>>b >>>[,1] [,2] [,3] >>>[1,] 4 10 18 >>>[2,] 63 64 63 >>>[3,] 18 10 4 >>> >>>[,1] [,2] [,3] >>>[1,] 4 10 18 >>>[2,] 63 64 63 >>>[3,] 18 10 4 >>>[,1] [,2] [,3] >>>[1,] 4 10 18 >>>[2,] 63 64 63 >>>[3,] 18 10 4[,1] [,2] [,3] >>>[1,] 4 10 18 >>>[2,] 63 64 63 >>>[3,] 18 10 4 >>>[,1] [,2] [,3] >>>[1,] 4 10 18 >>>[2,] 63 64 63 >>>[3,] 18 10 4 >>> >>> >>> >>> >>> >>>On Mon, Sep 2, 2013 at 7:25 AM, arun <smartpink...@yahoo.com> wrote: >>> >>>Hi, >>>>You could try: >>>> >>>>A<- matrix(unlist(read.table(text=" >>>>1 2 3 >>>>4 5 6 >>>>7 8 9 >>>>9 8 7 >>>>6 5 4 >>>>3 2 1 >>>>",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) >>>> >>>>library(matrixStats) >>>> >>>>res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) >>>> res1 >>>># [,1] [,2] [,3] >>>>#1 4 10 18 >>>>#2 63 64 63 >>>>#3 18 10 4 >>>> >>>> >>>> >>>>res2<-t(sapply(split(as.data.frame(A),((seq_len(nrow(A))-1)%/%2)+1),colProds)) >>>> identical(res1,res2) >>>>#[1] TRUE >>>> >>>>#or >>>> t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),function(x) >>>>apply(x,2,prod))) >>>> >>>>#or >>>>library(plyr) >>>> >>>>as.matrix(ddply(as.data.frame(A),.(as.numeric(gl(nrow(A),2,6))),colProds)[,-1]) >>>># V1 V2 V3 >>>>#[1,] 4 10 18 >>>>#[2,] 63 64 63 >>>>#[3,] 18 10 4 >>>> >>>>#or >>>>do.call(rbind,tapply(seq_len(nrow(A)),list(as.numeric(gl(nrow(A),2,6))),FUN=function(x) >>>> colProds(A[x,]))) >>>>#or >>>>A1<- data.frame(A,ID=as.numeric(gl(nrow(At),2,6))) >>>> aggregate(A1[,-4],list(A1[,4]),colProds)[,-1] >>>># X1 X2 X3 >>>>#1 4 10 18 >>>>#2 63 64 63 >>>>#3 18 10 4 >>>> >>>>#or >>>>library(data.table) >>>>At<- data.table(A1,key='ID') >>>>subset(At[,lapply(.SD,colProds),by=ID],select=-1) >>>># X1 X2 X3 >>>>#1: 4 10 18 >>>>#2: 63 64 63 >>>>#3: 18 10 4 >>>> >>>>A.K. >>>> >>>> >>>> >>>> >>>>Hello, >>>> >>>>I have this matrix : >>>>A = >>>>1 2 3 >>>>4 5 6 >>>>7 8 9 >>>>9 8 7 >>>>6 5 4 >>>>3 2 1 >>>> >>>>I would like to have this matrix (product of rows 2 by 2) : >>>>A = >>>>4 10 18 >>>>63 64 63 >>>>18 10 4 >>>> >>>>Is it possible to do that without a loop ? >>>> >>>>Thank you in advance ! >>>> >>>>______________________________________________ >>>>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. >>>> >>> >>> >>>-- >>> >>> >>>Bert Gunter >>>Genentech Nonclinical Biostatistics >>> >>>Internal Contact Info: >>>Phone: 467-7374 >>>Website: >>> >>>http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm >>> >>>______________________________________________ >>>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. >>> >> > ______________________________________________ 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.