R help - May 2011 - Different behavior of median and mean function - Why?

Hi together,

below is a small example which produces outcome I do not understand,
namely that the median function works fine on a data.frame without
negative numbers, but doesn't work on a data.frame with one negative
number. I'm sure there is a reasonable explanation for that or better,
that I'm doing something wrong and someone could guide me how to solve
it. I tried googling it, but couldn't find a solution:
> #Set up data frame
> df <- data.frame(V1=c(1,2,3,4),V2=c(2,3,4,5))
> #Both work fine
> mean(df)
 V1  V2
2.5 3.5
> median(df)
[1] 2.5 3.5
>
> #Now, I just make one number negative in the data.frame
> df <- data.frame(V1=c(1,2,3,-4),V2=c(2,3,4,5))
> mean(df)#Works fine
 V1  V2
0.5 3.5
> median(df)#Why do I  get that error?
[1]  NA 0.5
Warnmeldung:
In mean.default(X[[1L]], ...) :
  argument is not numeric or logical: returning NA
> #It works fine on both columns seperately
> median(df$V1)
[1] 1.5
> median(df$V2)
[1] 3.5
>
> sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: x86_64-pc-mingw32/x64 (64-bit)

locale:
[1] C

attached base packages:
[1] graphics  grDevices datasets  stats     utils     methods   base

other attached packages:
[1] urca_1.2-5      zoo_1.6-5       svSocket_0.9-51

loaded via a namespace (and not attached):
[1] grid_2.13.0     lattice_0.19-23 svMisc_0.9-61   tcltk_2.13.0
tools_2.13.0


Thanks Christoph
--
--------------------------------------------------------------------------------------------------------------------------------------------------------------------

Christoph J?ckel (Dipl.-Kfm.)

--------------------------------------------------------------------------------------------------------------------------------------------------------------------

Research Assistant

Chair for Financial Management and Capital Markets | Lehrstuhl f?r
Finanzmanagement und Kapitalm?rkte

TUM School of Management | Technische Universit?t M?nchen

Arcisstr. 21 | D-80333 M?nchen | Germany

Summary (df) will also work.

On 5/26/11, Christoph J?ckel <christoph.jaeckel at wi.tum.de>
wrote:
> Hi together,
>
> below is a small example which produces outcome I do not understand,
> namely that the median function works fine on a data.frame without
> negative numbers, but doesn't work on a data.frame with one negative
> number. I'm sure there is a reasonable explanation for that or better,
> that I'm doing something wrong and someone could guide me how to solve
> it. I tried googling it, but couldn't find a solution:
>
>> #Set up data frame
>> df <- data.frame(V1=c(1,2,3,4),V2=c(2,3,4,5))
>> #Both work fine
>> mean(df)
>  V1  V2
> 2.5 3.5
>> median(df)
> [1] 2.5 3.5
>>
>> #Now, I just make one number negative in the data.frame
>> df <- data.frame(V1=c(1,2,3,-4),V2=c(2,3,4,5))
>> mean(df)#Works fine
>  V1  V2
> 0.5 3.5
>> median(df)#Why do I  get that error?
> [1]  NA 0.5
> Warnmeldung:
> In mean.default(X[[1L]], ...) :
>   argument is not numeric or logical: returning NA
>> #It works fine on both columns seperately
>> median(df$V1)
> [1] 1.5
>> median(df$V2)
> [1] 3.5
>>
>> sessionInfo()
> R version 2.13.0 (2011-04-13)
> Platform: x86_64-pc-mingw32/x64 (64-bit)
>
> locale:
> [1] C
>
> attached base packages:
> [1] graphics  grDevices datasets  stats     utils     methods   base
>
> other attached packages:
> [1] urca_1.2-5      zoo_1.6-5       svSocket_0.9-51
>
> loaded via a namespace (and not attached):
> [1] grid_2.13.0     lattice_0.19-23 svMisc_0.9-61   tcltk_2.13.0
> tools_2.13.0
>
>
> Thanks Christoph
> --
>
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
>
> Christoph J?ckel (Dipl.-Kfm.)
>
>
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
>
> Research Assistant
>
> Chair for Financial Management and Capital Markets | Lehrstuhl f?r
> Finanzmanagement und Kapitalm?rkte
>
> TUM School of Management | Technische Universit?t M?nchen
>
> Arcisstr. 21 | D-80333 M?nchen | Germany
>
> ______________________________________________
> R-help at 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.
>
-- 
Sent from my mobile device

Due to the recession, requests for instant gratification will be
deferred until arrears in scheduled gratification have been satisfied.

Christoph,

After a quick look at the code for median, I'm amazed that it gives a
correct result for any data frame.

median() isn't really intended for use with data frames; there's no
data.frame method. The correct and safe approach is to use
sapply(df, median)

This was recently discussed on the R-devel list:
http://r.789695.n4.nabble.com/median-and-data-frames-td3478921.html

Sarah

On Thu, May 26, 2011 at 5:22 PM, Christoph J?ckel
<christoph.jaeckel at wi.tum.de> wrote:
> Hi together,
>
> below is a small example which produces outcome I do not understand,
> namely that the median function works fine on a data.frame without
> negative numbers, but doesn't work on a data.frame with one negative
> number. I'm sure there is a reasonable explanation for that or better,
> that I'm doing something wrong and someone could guide me how to solve
> it. I tried googling it, but couldn't find a solution:
>
>> #Set up data frame
>> df <- data.frame(V1=c(1,2,3,4),V2=c(2,3,4,5))
>> #Both work fine
>> mean(df)
> ?V1 ?V2
> 2.5 3.5
>> median(df)
> [1] 2.5 3.5
>>
>> #Now, I just make one number negative in the data.frame
>> df <- data.frame(V1=c(1,2,3,-4),V2=c(2,3,4,5))
>> mean(df)#Works fine
> ?V1 ?V2
> 0.5 3.5
>> median(df)#Why do I ?get that error?
> [1] ?NA 0.5
> Warnmeldung:
> In mean.default(X[[1L]], ...) :
> ?argument is not numeric or logical: returning NA
>> #It works fine on both columns seperately
>> median(df$V1)
> [1] 1.5
>> median(df$V2)
> [1] 3.5
>>
>> sessionInfo()
> R version 2.13.0 (2011-04-13)
> Platform: x86_64-pc-mingw32/x64 (64-bit)
>
> locale:
> [1] C
>
> attached base packages:
> [1] graphics ?grDevices datasets ?stats ? ? utils ? ? methods ? base
>
> other attached packages:
> [1] urca_1.2-5 ? ? ?zoo_1.6-5 ? ? ? svSocket_0.9-51
>
> loaded via a namespace (and not attached):
> [1] grid_2.13.0 ? ? lattice_0.19-23 svMisc_0.9-61 ? tcltk_2.13.0
> tools_2.13.0
>
>
> Thanks Christoph
> --
>
-- 
Sarah Goslee
http://www.functionaldiversity.org

Hi Christoph,

Use:

sapply(df, median)

median() does not have methods for a data frame (read: it is never
meant to be used directly on data frames so do not expect sensical
results).

Cheers,

Josh

On Thu, May 26, 2011 at 2:22 PM, Christoph J?ckel
<christoph.jaeckel at wi.tum.de> wrote:
> Hi together,
>
> below is a small example which produces outcome I do not understand,
> namely that the median function works fine on a data.frame without
> negative numbers, but doesn't work on a data.frame with one negative
> number. I'm sure there is a reasonable explanation for that or better,
> that I'm doing something wrong and someone could guide me how to solve
> it. I tried googling it, but couldn't find a solution:
>
>> #Set up data frame
>> df <- data.frame(V1=c(1,2,3,4),V2=c(2,3,4,5))
>> #Both work fine
>> mean(df)
> ?V1 ?V2
> 2.5 3.5
>> median(df)
> [1] 2.5 3.5
>>
>> #Now, I just make one number negative in the data.frame
>> df <- data.frame(V1=c(1,2,3,-4),V2=c(2,3,4,5))
>> mean(df)#Works fine
> ?V1 ?V2
> 0.5 3.5
>> median(df)#Why do I ?get that error?
> [1] ?NA 0.5
> Warnmeldung:
> In mean.default(X[[1L]], ...) :
> ?argument is not numeric or logical: returning NA
>> #It works fine on both columns seperately
>> median(df$V1)
> [1] 1.5
>> median(df$V2)
> [1] 3.5
>>
>> sessionInfo()
> R version 2.13.0 (2011-04-13)
> Platform: x86_64-pc-mingw32/x64 (64-bit)
>
> locale:
> [1] C
>
> attached base packages:
> [1] graphics ?grDevices datasets ?stats ? ? utils ? ? methods ? base
>
> other attached packages:
> [1] urca_1.2-5 ? ? ?zoo_1.6-5 ? ? ? svSocket_0.9-51
>
> loaded via a namespace (and not attached):
> [1] grid_2.13.0 ? ? lattice_0.19-23 svMisc_0.9-61 ? tcltk_2.13.0
> tools_2.13.0
>
>
> Thanks Christoph
> --
>
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
>
> Christoph J?ckel (Dipl.-Kfm.)
>
>
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
>
> Research Assistant
>
> Chair for Financial Management and Capital Markets | Lehrstuhl f?r
> Finanzmanagement und Kapitalm?rkte
>
> TUM School of Management | Technische Universit?t M?nchen
>
> Arcisstr. 21 | D-80333 M?nchen | Germany
>
> ______________________________________________
> R-help at 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.
>


-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

On May 26, 2011, at 4:22 PM, Christoph J?ckel wrote:
> Hi together,
> 
> below is a small example which produces outcome I do not understand,
> namely that the median function works fine on a data.frame without
> negative numbers, but doesn't work on a data.frame with one negative
> number. I'm sure there is a reasonable explanation for that or better,
> that I'm doing something wrong and someone could guide me how to solve
> it. I tried googling it, but couldn't find a solution:
> 
>> #Set up data frame
>> df <- data.frame(V1=c(1,2,3,4),V2=c(2,3,4,5))
>> #Both work fine
>> mean(df)
> V1  V2
> 2.5 3.5
>> median(df)
> [1] 2.5 3.5
>> 
>> #Now, I just make one number negative in the data.frame
>> df <- data.frame(V1=c(1,2,3,-4),V2=c(2,3,4,5))
>> mean(df)#Works fine
> V1  V2
> 0.5 3.5
>> median(df)#Why do I  get that error?
> [1]  NA 0.5
> Warnmeldung:
> In mean.default(X[[1L]], ...) :
>  argument is not numeric or logical: returning NA
>> #It works fine on both columns seperately
>> median(df$V1)
> [1] 1.5
>> median(df$V2)
> [1] 3.5

This was actually just discussed late last month. See the thread here:

  https://stat.ethz.ch/pipermail/r-devel/2011-April/060731.html

The bottom line is that median does not have a 'method' for data frames,
whereas mean does.

HTH,

Marc Schwartz