He everybody, I want to add 1 to some elements of a vector: x is a vector u is a vector of idices, that is, integers assumed to be within the range 1..length(x) and I want to add 1 to the elements of x each time their index appears in u x[u]<-x[u]+1 works only when there are no duplicated values in u I found this solution: tu <- table(u) indices <- as.numeric(names(tu)) x[indices] <- x[indices]+tu but it looks ugly to me and I would prefer to avoid calling the function 'table' since this is to be done millions of times as part of a simulation program. Eric Elguero G?n?tique & Adaptation des Plasmodium IRD Montpellier - FRance
should be able to: u <- unique(u) x[u] <- x[u] + 1 On Thu, Feb 10, 2011 at 6:50 AM, Eric Elguero <Eric.Elguero at ird.fr> wrote:> He everybody, > > I want to add 1 to some elements of ?a vector: > > x is a vector > u is a vector of idices, that is, integers > assumed to be within the range 1..length(x) > and I want to add 1 to the elements of x > each time their index appears in u > > x[u]<-x[u]+1 works only when there are no > duplicated values in u > > I found this solution: > > tu <- table(u) > indices <- as.numeric(names(tu)) > x[indices] <- x[indices]+tu > > but it looks ugly to me and I would > prefer to avoid calling the function 'table' > since this is to be done millions of times > as part of a simulation program. > > Eric Elguero > G?n?tique & Adaptation des Plasmodium > IRD > Montpellier - FRance > > ______________________________________________ > R-help at 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. >-- Jim Holtman Data Munger Guru What is the problem that you are trying to solve?
On Thu, Feb 10, 2011 at 12:50:03PM +0100, Eric Elguero wrote:> He everybody, > > I want to add 1 to some elements of a vector: > > x is a vector > u is a vector of idices, that is, integers > assumed to be within the range 1..length(x) > and I want to add 1 to the elements of x > each time their index appears in u > > x[u]<-x[u]+1 works only when there are no > duplicated values in u > > I found this solution: > > tu <- table(u) > indices <- as.numeric(names(tu)) > x[indices] <- x[indices]+tu > > but it looks ugly to me and I would > prefer to avoid calling the function 'table' > since this is to be done millions of times > as part of a simulation program.If we start with a zero vector x, then the operation, which you ask for, is equivalent to computing the frequency table of u. So, i think, using function table() may be adequate. However, let me suggest a slightly different code n <- 8 x <- rep(0, times=8) u <- c(1, 2, 5, 2, 5, 1, 2, 8, 2, 1) x <- x + unclass(table(factor(u, levels=1:n))) x 1 2 3 4 5 6 7 8 3 4 0 0 2 0 0 1 Another solution may be used, if the vectors u all have the same length (not only the same range of values). Then it is possible to compute first the frequency table for each component of u separately in different rows of a matrix x m <- 10 x <- matrix(0, nrow=m, ncol=n) x[cbind(1:m, u)] <- x[cbind(1:m, u)] + 1 This may be repeated for different vectors u and when all repetitions are finished, then the final result may be obtained as colSums(x) [1] 3 4 0 0 2 0 0 1 Hope this helps. Petr Savicky.
Try using tabulate() instead of table(). E.g.,
compare your original
f0 <- function (x, u) {
tu <- table(u)
indices <- as.numeric(names(tu))
x[indices] <- x[indices] + tu
x
}
to
f1 <- function (x, u) {
x + tabulate(u, nbins = length(x))
}
I tried it for a 20-long x and 30-long u. f0 and f1
gave identical results. 10^5 iterations of f0 took
62 seconds, f1 1.72 seconds.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of Eric Elguero
> Sent: Thursday, February 10, 2011 3:50 AM
> To: r-help at r-project.org
> Subject: [R] modifynig some elements of a vector
>
> He everybody,
>
> I want to add 1 to some elements of a vector:
>
> x is a vector
> u is a vector of idices, that is, integers
> assumed to be within the range 1..length(x)
> and I want to add 1 to the elements of x
> each time their index appears in u
>
> x[u]<-x[u]+1 works only when there are no
> duplicated values in u
>
> I found this solution:
>
> tu <- table(u)
> indices <- as.numeric(names(tu))
> x[indices] <- x[indices]+tu
>
> but it looks ugly to me and I would
> prefer to avoid calling the function 'table'
> since this is to be done millions of times
> as part of a simulation program.
>
> Eric Elguero
> G?n?tique & Adaptation des Plasmodium
> IRD
> Montpellier - FRance
>
> ______________________________________________
> R-help at 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.
>