m = matrix(c(0,7,4,0,2,0,0,1,3,0,3,4),byrow = TRUE,ncol=3) colSum = apply( m, 2, sum )
#Need to deal with dividing by zero... m%*%diag(1/colSum) -----Original Message----- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Ted Harding Sent: Thursday, May 06, 2010 1:09 PM To: r-help@r-project.org Cc: n.via...@libero.it Subject: Re: [R] frequency On 06-May-10 17:06:26, n.via...@libero.it wrote: > > Dear list, > Im trying to do the following operation but im not able to do it > This is my table: > 1 2 3 > 1 0 7 4 > 2 0 2 0 > 3 0 1 3 > 4 0 3 4 > > what i would like to do is > > divide each row values with the corresponding column' sum,namely: > > 1 2 3 > 1 0 0.54 0.36 > 2 0 0.15 0 > 3 0 0.08 0.27 > 4 0 0.23 0.36 > > thanks for your attention There is a problem with the first columns, because it's sum is 0, so you would be doing "0/0", with result "NaN". So, for illustration (and to allow easy verification) I've replaced your first column with 1,2,3,4. Suppose your table is a matrix: M <- matrix(c(1,2,3,4,7,2,1,3,4,0,3,4),ncol=3) M # [,1] [,2] [,3] # [1,] 1 7 4 # [2,] 2 2 0 # [3,] 3 1 3 # [4,] 4 3 4 Note that the entries are given in the order of moving down successive columns. That is what a mtrix is: a vector of numbers, in that order, with in addition a "dim" attribute: dim(M) # [1] 4 3 Now you can do the clever bit. Transpose the matrix M: tM <- t(M) tM # [,1] [,2] [,3] [,4] # [1,] 1 2 3 4 # [2,] 7 2 1 3 # [3,] 4 0 3 4 This is a new matrix whose elements will again be read in the order of going down succesive columns, i.e. as c(1,7,4,2,2,0,3,1,3,4,3,3) Now compute the column sums of tM: Csums <- colSums(tM) Csums # [1] 10 13 11 Now, in the expression "tM/Csums", the successive elements of tM, in the above order, will be divided by the successive elements of Csums, with Csums being recycled until it is finished: tM/Csums # [,1] [,2] [,3] [,4] # [1,] 0.1000000 0.2000000 0.30000000 0.4000000 # [2,] 0.5384615 0.1538462 0.07692308 0.2307692 # [3,] 0.3636364 0.0000000 0.27272727 0.3636364 But this is the transpose of what you want, so transpose it back to t(tM/Rsums). So, just as a full check: M # [,1] [,2] [,3] # [1,] 1 7 4 # [2,] 2 2 0 # [3,] 3 1 3 # [4,] 4 3 4 Csums <- colSums(M) Csums # [1] 10 13 11 t(tM/Csums) # [,1] [,2] [,3] # [1,] 0.1 0.53846154 0.3636364 # [2,] 0.2 0.15384615 0.0000000 # [3,] 0.3 0.07692308 0.2727273 # [4,] 0.4 0.23076923 0.3636364 Explicitly: the division of t(M) = # [,1] [,2] [,3] [,4] # [1,] 1 2 3 4 # [2,] 7 2 1 3 # [3,] 4 0 3 4 by Csums=c(10,13,11) has been done as 1/10,7/13,4/11, 2/10,2/13,0/11, 3/10,1/13,3/13, 4/10,3/13,4/11 Hoping this helps, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk> Fax-to-email: +44 (0)870 094 0861 Date: 06-May-10 Time: 19:08:51 ------------------------------ XFMail ------------------------------ ______________________________________________ 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.
