sweep(x, 2, m, FUN='/')
On Sun, Dec 27, 2009 at 2:26 PM, Muhammad Rahiz <
muhammad.ra...@ouce.ox.ac.uk> wrote:
> Hi useRs,
>
> I ran into an inconsistent output problem again. Here is the simplify
> illustration
>
> I've got a matrix as follows
>
> > x
> V1 V2 V3
> [1,] 1 2 3
> [2,] 4 5 6
> [3,] 7 8 9
>
> Associated with the matrix is a scaling factor, sca, derived from, say the
> mean of the respective columns, V1-V3;
>
> > sca
> V1 V2 V3
> 2.5 1.7 3.6
>
> The idea is that the scaling factor gets applied to each element in the
> column matrix. So
>
> > out <- x / sca
>
> would give me
>
> V1 V2 V3
> 1 *2.5 2 *1.7 3 *3.6
> 4 *2.5 5 *1.7 6 *3.6
> 7 *2.5 8 *1.7 9 *3.6
>
> But what actually happen is this,
>
> V1 V2 V3
> 1 *2.5 2 *2.5 3 *2.5
> 4 *1.7 5 *1.7 6 *1.7
> 7 *3.6 8 *3.6 9 *3.6
>
> I can do the following;
>
> < x[,1] / sca[1]
> < x[,2] / sca[2]
>
> which is OK for a set of test data but not for my actual dataset.
>
> At the moment, I'm thinking of something in the lines of a for loop
> function i.e.
>
> for (i in ...){
> statement...
> }
>
> Is there a syntax in the for loop that allows me to select the column/row
> in the file?
>
> Thanks.
>
> --
> Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
>
> Climate Research Laboratory, School of Geography & the Environment
> Oxford University Centre for the Environment
> South Parks Road, Oxford, OX1 3QY, United Kingdom Tel: +44 (0)1865-285194
> Mobile: +44 (0)7854-625974
> Email: muhammad.ra...@ouce.ox.ac.uk
>
>
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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