Try this:
lapply(l, function(x)median(x[,sapply(x, is.numeric)]))
On Fri, Nov 7, 2008 at 8:42 AM, Ptit_Bleu <ptit_bleu@yahoo.fr> wrote:
>
> Hello,
>
> I have a list of data.frame
> rowsplit : List of 15
> $ (0,0.025] :'data.frame': 169 obs. of 7 variables:
> $ (0.025,0.05]:'data.frame': 174 obs. of 7 variables:
> $ (0.05,0.075]:'data.frame': 92 obs. of 7 variables:
> $ (0.075,0.1] :'data.frame': 76 obs. of 7 variables:
> $ (0.1,0.125] :'data.frame': 37 obs. of 7 variables:
> $ (0.125,0.15]:'data.frame': 32 obs. of 7 variables:
> $ (0.15,0.175]:'data.frame': 45 obs. of 7 variables:
> $ (0.175,0.2] :'data.frame': 56 obs. of 7 variables:
> $ (0.2,0.225] :'data.frame': 36 obs. of 7 variables:
> $ (0.225,0.25]:'data.frame': 47 obs. of 7 variables:
> $ (0.25,0.275]:'data.frame': 34 obs. of 7 variables:
> $ (0.275,0.3] :'data.frame': 43 obs. of 7 variables:
> $ (0.3,0.325] :'data.frame': 29 obs. of 7 variables:
> $ (0.325,0.35]:'data.frame': 29 obs. of 7 variables:
> $ (0.35,0.375]:'data.frame': 17 obs. of 7 variables:
>
> And I would like to get a data.frame gathering the median value of each
> variable for each intervall.
> as.data.frame(t(sapply(rowsplit, mean)) works well but I would prefer to
> use
> median but with sapply(rowsplit, median), I get the following message
> "Error
> in median.default(X[[1L]], ...) : need numeric data".
>
> Any idea ?
>
> Thanks in advance for your help,
> Have a nice week-end,
> Ptit Bleu.
>
> PS :I'm under XP and use R2.7.2.
> --
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>
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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