Hi:

Here are a few one-liners. Calling your data frame dd,

aggregate(cbind(height, weight, age) ~ sample, data = dd, FUN = mean)
  sample   height    weight      age
1      A 12.20000 0.5033333 6.000000
2      B 12.75000 0.7150000 4.500000
3      C 11.35250 0.5125000 3.750000
4      D 14.99333 0.6733333 5.333333

With package doBy:

library(doBy)
summaryBy(height + weight + age ~ sample, data = dd, FUN = mean)
  sample height.mean weight.mean age.mean
1      A    12.20000   0.5033333 6.000000
2      B    12.75000   0.7150000 4.500000
3      C    11.35250   0.5125000 3.750000
4      D    14.99333   0.6733333 5.333333

With package plyr:

library(plyr)
ddply(dd, .(sample), colwise(mean, .(height, weight, age)))
  sample   height    weight      age
1      A 12.20000 0.5033333 6.000000
2      B 12.75000 0.7150000 4.500000
3      C 11.35250 0.5125000 3.750000
4      D 14.99333 0.6733333 5.333333

Dennis

On Fri, Mar 11, 2011 at 1:32 AM, Aline Santos <aline...@gmail.com> wrote:

> Hello R-helpers:
>
> I have data like this:
>
> sample    replicate    height    weight    age
> A    1.00    12.0    0.64    6.00
> A    2.00    12.2    0.38    6.00
> A    3.00    12.4    0.49    6.00
> B    1.00    12.7    0.65    4.00
> B    2.00    12.8    0.78    5.00
> C    1.00    11.9    0.45    6.00
> C    2.00    11.84    0.44    2.00
> C    3.00    11.43    0.32    3.00
> C    4.00    10.24    0.84    4.00
> D    1.00    14.2    0.54    2.00
> D    2.00    15.67    0.67    7.00
> D    3.00    15.11    0.81    7.00
>
> Now, how can I calculate the mean for each condition (heigth, weigth, age)
> in each sample, considering the samples have different number of
> replicates?
>
>
> The final matrix should look like:
>
> sample    height    weight    age
> A    12.20    0.50    6.00
> B     12.75      0.72      4.50
> C     11.35      0.51      3.75
> D     14.99      0.67      5.33
>
> This is a simplified version of my dataset, which consist of 100 samples
> (unequally distributed in 530 replicates) for 600 different conditions.
>
> I appreciate all the help.
>
> A.S.
>
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>
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>

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