From p.dalgaard@biostat.ku.dk  Sat Sep 23 15:55:58 2000
From: p.dalgaard@biostat.ku.dk (Peter Dalgaard BSA)
Date: 23 Sep 2000 16:55:58 +0200
Subject: [Rd] bug in apply with median
In-Reply-To: Larry Ammann's message of "Sat, 23 Sep 2000 08:48:49 -0500"
References: <39CCB4C0.9F104660@metronet.com>
Message-ID: <x2snqrrwch.fsf@blueberry.kubism.ku.dk>

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:

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

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

Works fine.

However,

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

Question is whether this is a bug. Apply() never promised to work on
non-arrays. When applied columnwise, it would anyway be more obvious
to use

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

-- 
   O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard@biostat.ku.dk)             FAX: (+45) 35327907
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[Rd] bug in apply with median
Peter Dalgaard BSA