HI,
If you want to try other ways:
fun1 <- function(mat, rowN) {
dm <- dim(mat)[1]
rowN1 <- rowN - 1
indx <- rep(1:rowN, dm - rowN1) + rep(seq(0, dm - rowN), each = rowN)
indx1 <- (seq_along(indx)-1)%/%rowN+1
as.vector(tapply(indx, list(indx1), FUN = function(i) sum(mat[i, ])))
}
#or
fun2 <- function(mat, rowN) {
dm <- dim(mat)[1]
rowN1 <- rowN - 1
indx <- rep(1:rowN, dm - rowN1) + rep(seq(0, dm - rowN), each = rowN)
mat1 <- mat[indx, ]
indx1 <- rep((seq_along(indx) - 1)%/%rowN + 1, ncol(mat))
colSums(matrix(mat1[sort.int(indx1, method = "quick", index.return =
TRUE)$ix],
ncol = dm - rowN1))
}
##But the above methods are slower in large datasets.
##Rui's function
funOld <- function(mat, rowN) {
dm <- dim(mat)[1]
rowN1 <- rowN - 1
sapply(1:(dm - rowN1), function(i) sum(mat[i:(i + rowN1), ]))
}
###Modified Rui's function using ?for() loop instead of ?sapply()
funNew <- function(mat, rowN) {
dm <- dim(mat)[1]
rowN1 <- rowN - 1
vec <- vector(mode = "numeric", length = dm - rowN1)
for (i in 1:(dm - rowN1)) {
vec[i] <- sum(mat[i:(i + rowN1), ])
}
vec
}
##Speed comparison
set.seed(348)
el1 <- matrix(sample(1:300, 1e6*50,replace=TRUE),ncol=50)
system.time(r1 <- fun1(el1,3))
# user system elapsed
# 15.888 0.120 16.042
system.time(r2 <- funOld(el1,3))
# user system elapsed
# 5.061 0.004 5.076
system.time(r3 <- fun2(el1,3))
# user system elapsed
#12.371 1.329 13.735
system.time(r4 <- funNew(el1,3))
# user system elapsed
# 4.716 0.000 4.727
library(compiler)
funOldc <- cmpfun(funOld)
funNewc <- cmpfun(funNew)
fun2c <- cmpfun(fun2)
system.time(r5 <- funOldc(el1,3))
# user system elapsed
# 7.458 0.000 7.476
system.time(r6 <- funNewc(el1,3))
# user system elapsed
# 3.529 0.000 3.536
sapply(paste0("r",2:6),function(x) all.equal(r1,get(x) ))
# r2 r3 r4 r5 r6
#TRUE TRUE TRUE TRUE TRUE
A.K.
On Friday, May 9, 2014 5:56 PM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
Hello,
Try the following.
sapply(1:(30 - 2), function(i) sum(el[i:(i+2), ]))
but with number of rows instead of 30.
Hope this helps,
Rui Barradas
Em 09-05-2014 22:35, eliza botto escreveu:
> Dear useRs,
> I have a matrix, say "el" of 30 rows and 10 columns, as
> el<-matrix(sample(1:300),ncol=10)
> I want to sum up various sets of three rows of each column in the following
> manner
> sum(el[c(1,2,3),]) ##adding row number 1, 2 and 3 of each column
> sum(el[c(2,3,4),])##adding row number 2, 3 and 4 of each column
> sum(el[c(3,4,5),])##adding row number 3, 4 and 5 of each column
> sum(el[c(4,5,6),])
> sum(el[c(5,6,7),])
> sum(el[c(6,7,8),])
> sum(el[c(7,8,9),])
> sum(el[c(8,9,10),])
> sum(el[c(9,10,11),])
> ................
> ......
> so on
> ..............
> I know how to do it manually, but since my original matrix has 2000 rows, I
> therefore want to figure out a more conveinient way.
> Thankyou so very much in advance,
>
> Eliza
> [[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
>
______________________________________________
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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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
