# Re: [Rd] median and data frames

From: Martin Maechler <maechler_at_stat.math.ethz.ch>
Date: Fri, 29 Apr 2011 16:25:09 +0200

```    > On Wed, Apr 27, 2011 at 12:44 PM, Patrick Burns
> <pburns_at_pburns.seanet.com> wrote:
```

>> Here are some data frames:
>>
>> df3.2 <- data.frame(1:3, 7:9)
>> df4.2 <- data.frame(1:4, 7:10)
>> df3.3 <- data.frame(1:3, 7:9, 10:12)
>> df4.3 <- data.frame(1:4, 7:10, 10:13)
>> df3.4 <- data.frame(1:3, 7:9, 10:12, 15:17)
>> df4.4 <- data.frame(1:4, 7:10, 10:13, 15:18)
>>
>> Now here are some commands and their answers:
```    >>> median(df4.4)
```

>> [1]  8.5 11.5
```    >>> median(df3.2[c(1,2,3),])
```

>> [1] 2 8
```    >>> median(df3.2[c(1,3,2),])
```

>> [1]  2 NA
>> Warning message:
>> In mean.default(X[[2L]], ...) :
>>  argument is not numeric or logical: returning NA
>>
>>
>>
>> The sessionInfo is below, but it looks
>> to me like the present behavior started
>> in 2.10.0.
>>
>> Sometimes it gets the right answer.  I'd
>> be grateful to hear how it does that -- I
>> can't figure it out.
>>

> Hello, Pat.

> Nice poetry there! I think I have an actual answer, as opposed to the     > usual crap I spew.

> I would agree if you said median.data.frame ought to be written to     > work columnwise, similar to mean.data.frame.

> apply and sapply always give the correct answer

>> apply(df3.3, 2, median)
> X1.3 X7.9 X10.12
> 2 8 11

[...........]

exactly

> mean.data.frame is now implemented as

> mean.data.frame <- function(x, ...) sapply(x, mean, ...)

exactly.

My personal oppinion is that mean.data.frame() should never have been written.
People should know, or learn, to use apply functions for such a task.

The unfortunate fact that mean.data.frame() exists makes people think that median.data.frame() should too, and then

```  var.data.frame()
sd.data.frame()
min.data.frame()
max.data.frame()
```

...
...

all just in order to *not* to have to know sapply() ????

No, rather not.

My vote is for deprecating mean.data.frame().

Martin

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