# [Rd] bug in apply with median

Martin Maechler Martin Maechler <maechler@stat.math.ethz.ch>
Tue, 26 Sep 2000 19:51:02 +0200

```>>>>> "PD" == Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> writes:
(Sep 23):

PD> Larry Ammann <ammann@metronet.com> writes:
>> I have found a problem in R version 1.1.1 when using apply with the
>> median function.
>> The problem can be illustrated with the following data matrix:
>>
>> X1  X2  X3
>> 1     2     3
>> 4    5     6
>> 7    8    NA
>>
>> Enter this data matrix as X and then try
>> apply(X,2,median,na.rm=T)
>> The problem here is that the median function returns a named scalar if
>> the number of
>> observations is odd, but returns an unnamed scalar if the number of
>> observations is
>> even. This confuses the apply function in this case at:

PD> for ( i in 1:1000 ) cat("Data frames are not matrices!\n")

==============================
>> X<-matrix(c(1:8,NA),3,3)
>> X
PD> [,1] [,2] [,3]
PD> [1,]    1    4    7
PD> [2,]    2    5    8
PD> [3,]    3    6   NA
>> apply(X,2,median,na.rm=T)
PD> [1] 2.0 5.0 7.5

PD> Works fine.

PD> However,

>> Z<-data.frame(X)
>> apply(Z,2,median,na.rm=T)
PD> Error in names(x) == ans.names : comparison (1) is possible only for vector types

PD> Question is whether this is a bug. Apply() never promised to work on
PD> non-arrays.

I think we should
either   make it promiss and do this
or	   make it produce a warning    when used with data.frames

The above problem is too common for non-experts to stumble on..

PD> When applied columnwise, it would anyway be more obvious to use

>> sapply(Z,median,na.rm=T)
PD> X1  X2  X3
PD> 2.0 5.0 7.5

This could be what
apply(<dataframe>, 2, * )
should do
whereas we could have
apply(dfr, 1, *)  === apply(data.matrix(dfr), 1, *)
when dfr is a data.frame.

Opinions ?
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```