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"with" and "by" and NA:

3 messages · Aldi Kraja, Jorge Ivan Velez, Duncan Murdoch

Aldi Kraja
# 
Hi,

I have a data.frame with many variables for which I am performing the 
mean by subgroup, for a pair of variables at a time, where one of them 
for each pair defines the subgroup. The subgroups in the x$cm1 are 0, 1 
and 2.
x
ph1 cm1
0.2345 2
1.2222 1
2.0033 0
0.0000 2
1.0033 1
0.2345 0
1.2222 2
2.0033 0
0.0000 1
1.0033 2

 > meanbygroup <- as.vector(with(x, by(x$ph1, x$cm1, mean)))
 > meanbygroup
if the ph1 has no missing values the above statements work fine:
[1] 1.4137000 0.7418333 0.6150000

In the moment that I introduce in the ph1 a missing value in the ph1 as NA
x
ph1 cm1
0.2345 2
NA      1
1.2222 1
.............

the above transforms into
[1] 1.4137000 NA 0.6150000

Question: is there a way I can protect this calculations from the NA 
values in the ph1 (some kind of: na.rm=T)?

TIA,

Aldi


--
Duncan Murdoch
#
On 25/03/2009 7:36 PM, Aldi Kraja wrote:
You don't need with() here, as you are explicitly extracting the vectors 
from x.
You could use with(), and extract the vectors from a subset of x:

with(x[!is.na(x$ph1),], by(ph1, cm1, mean))

This is untested.  If you had provided sample data in a usable format I 
would have tried it, but you didn't, and I'm too lazy to create my own.

Duncan Murdoch