Instead of colSums(t(aMatrix)), why not the more
direct rowSums(aMatrix)?
If time is an issue (which it won't be unless
the number of columns of M is big), compare:
> M <- matrix(2e5:1, nrow=2)
> v <- 1:ncol(M)
> system.time(z1 <- sapply(seq_along(v), function(i) sum(M[,i] < v[i])))
user system elapsed
0.532 0.000 0.532
> system.time(z2 <- colSums(t(apply(M, 1, "<", v))))
user system elapsed
0.004 0.000 0.006
> system.time(z3 <- rowSums(apply(M, 1, "<", v)))
user system elapsed
0.008 0.000 0.005
> system.time(z4 <- colSums(M < matrix(v, nrow=nrow(M), ncol=ncol(M),
byrow=TRUE)))
user system elapsed
0.000 0.000 0.002
> isTRUE(all.equal(z1, z2)) && isTRUE(all.equal(z1,z3)) &&
isTRUE(all.equal(z1,z4))
[1] TRUE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Milan Bouchet-
> Valat
> Sent: Friday, February 03, 2012 10:17 AM
> To: Ernest Adrogué
> Cc: r-help@r-project.org
> Subject: Re: [R] sapply help
>
> Le vendredi 03 février 2012 à 18:27 +0100, Ernest Adrogué a écrit :
> > 3-02-2012, 08:37 (-0800); Filoche escriu:
> > > Hi every one.
> > >
> > > I'm learning how to use sapply (and other function of this family).
> > >
> > > Here's what I'm trying to do.
> > >
> > > I have a vector of lets say 5 elements. I also have a matrix of nX5. I
> > > would
> > > like to know how many element by column are inferior to each element of my
> > > vector.
> > >
> > > On this example:
> > > v = c(1:5)
> > > M = matrix(3,2,5)
> > >
> > > I would like to have a vector at the end which give me
> > >
> > > 0 0 0 2 2
> > >
> >
> > This does that:
> >
> > > sapply(1:5, function(i) sum(M[,i] < v[i]))
> > [1] 0 0 0 2 2
> >
> > Basically, it's like a loop where at each iteration the function is
> > called with one element of the vector 1:5 as argument, so what this
> > really does is
> >
> > sum(M[,1] < v[1]))
> > sum(M[,2] < v[2]))
> > ...
> >
> > and then the results are put all together in a vector.
> Though in your case, I think there are shorter solutions. For example:
> > colSums(t(apply(M, 1, "<", v)))
> [1] 0 0 0 2 2
>
> apply() is more suited to matrices. Here, it takes each row separately,
> and compares it with v. Then, you can just sum the result to count the
> number of cases that fulfill the condition.
>
>
> Cheers
>
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