Thanks to both of you! I learned something new from both of your posts :-)
And you were both right, my example numbers were wrong! I accidentally
computed them based on 0.25 * mean instead of 0.2 * mean.
Thanks again!
Simon
On Wed, Dec 5, 2012 at 4:07 AM, arun <smartpink...@yahoo.com> wrote:
> HI,
>
> I am not sure the output you wanted is correct:
>
> "
> sample1 sample2 sample3
> 1 1.0 0 0.5
> "
>
> because
> 0.2*colMeans(x[,-4])
> sample1 sample2 sample3
> # 28.40 24.08 21.36
>
>
> This might help you:
> apply(x[-4],2,function(y) length(y[y <0.2*mean(y) &
> x$class=="a"])/length(x[x$class=="a"]))
> #sample1 sample2 sample3
> # 0.0 0.0 0.5
> A.K.
>
>
>
> ----- Original Message -----
> From: Simon <simonzm...@gmail.com>
> To: r-help@r-project.org
> Cc:
> Sent: Tuesday, December 4, 2012 4:49 AM
> Subject: [R] computing marginal values based on multiple columns?
>
> Hello all,
>
> I have what feels like a simple problem, but I can't find an simple
> answer. Consider this data frame:
>
> > x <- data.frame(sample1=c(35,176,182,193,124),
> sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
> class=c('a','a','c','b','c'))
>
> > x
> sample1 sample2 sample3 class
> 1 35 198 12 a
> 2 176 176 154 a
> 3 182 190 21 c
> 4 193 23 191 b
> 5 124 15 156 c
>
> Now I wish to know: for each sample, for values < 20% of the sample mean,
> what percentage of those are class a?
>
> I want to end up with a table like:
>
> sample1 sample2 sample3
> 1 1.0 0 0.5
>
> I can calculate this for an individual sample using this rather clumsy
> expression:
>
> length(which(x$sample1 < mean(x$sample1) & x$class=='a')) /
> length(which(x$sample1 < mean(x$sample1)))
>
> I'd normally propagate it across the data frame using apply, but I
> can't because it depends on more than one column.
>
> Any help much appreciated!
>
> Cheers,
>
> Simon
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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>
>
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