[R] sapply and median, possible or not ?

2008-11-07 Thread Ptit_Bleu

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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Re: [R] sapply and median, possible or not ?

2008-11-07 Thread Ptit_Bleu

Unfortunately, I have the same error message.
lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
with median.
Strange, isn't it?

Any other idea?

Thanks in advance,
Ptit Bleu.


Henrique Dallazuanna wrote:
> 
> Try this:
> 
> lapply(l, function(x)median(x[,sapply(x, is.numeric)]))
> 
> 
> -- 
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
> 
>   [[alternative HTML version deleted]]
> 
> 
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
> 
> 

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Re: [R] sapply and median, possible or not ?

2008-11-07 Thread Henrique Dallazuanna
You can provide a example of your data?

On Fri, Nov 7, 2008 at 9:14 AM, Ptit_Bleu <[EMAIL PROTECTED]> wrote:

>
> Unfortunately, I have the same error message.
> lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
> with median.
> Strange, isn't it?
>
> Any other idea?
>
> Thanks in advance,
> Ptit Bleu.
>
>
> Henrique Dallazuanna wrote:
> >
> > Try this:
> >
> > lapply(l, function(x)median(x[,sapply(x, is.numeric)]))
> >
> >
> > --
> > Henrique Dallazuanna
> > Curitiba-Paraná-Brasil
> > 25° 25' 40" S 49° 16' 22" O
> >
> >   [[alternative HTML version deleted]]
> >
> >
> > __
> > 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.
> >
> >
>
> --
> View this message in context:
> http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378663.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

[[alternative HTML version deleted]]

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Re: [R] sapply and median, possible or not ?

2008-11-07 Thread Keith Jewell
I haven't looked at the detail, but I guess the answer is that mean works on 
a data frame while median doesn't.

?mean

For a data frame, a named vector with the appropriate method being applied 
column by column.
-

I guess to use median you'll need nested '[l/s]apply's, the outer working 
through the list of dataframes and the inner working through the columns of 
each dataframe.

Or perhaps, by analogy with mean.data.frame you could just define
-
median.data.frame <- function(x, ...)
sapply(x, median, ...)


I haven't tried it, but it might work

hth

Keith J
-
"Ptit_Bleu" <[EMAIL PROTECTED]> wrote in message 
news:[EMAIL PROTECTED]
>
> Unfortunately, I have the same error message.
> lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
> with median.
> Strange, isn't it?
>
> Any other idea?
>
> Thanks in advance,
> Ptit Bleu.
>
>
> Henrique Dallazuanna wrote:
>>
>> Try this:
>>
>> lapply(l, function(x)median(x[,sapply(x, is.numeric)]))
>>
>>
>> -- 
>> Henrique Dallazuanna
>> Curitiba-Paraná-Brasil
>> 25° 25' 40" S 49° 16' 22" O
>>
>> [[alternative HTML version deleted]]
>>
>>
>> __
>> 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.
>>
>>
>
> -- 
> View this message in context: 
> http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378663.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>

__
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Re: [R] sapply and median, possible or not ?

2008-11-07 Thread Henrique Dallazuanna
Try this:

lapply(l, function(x)median(x[,sapply(x, is.numeric)]))

On Fri, Nov 7, 2008 at 8:42 AM, Ptit_Bleu <[EMAIL PROTECTED]> 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.
> --
> View this message in context:
> http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378222.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

[[alternative HTML version deleted]]

__
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.