"with" and "by" and NA:
On 25/03/2009 7:36 PM, Aldi Kraja wrote:
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$cm1, mean)))
You don't need with() here, as you are explicitly extracting the vectors from x.
> meanbygroup
if the ph1 has no missing values the above statements work fine: [1] 1.4137000 0.7418333 0.6150000 In the moment that I introduce in the ph1 a missing value in the ph1 as NA x ph1 cm1 0.2345 2 NA 1 1.2222 1 ............. the above transforms into [1] 1.4137000 NA 0.6150000 Question: is there a way I can protect this calculations from the NA values in the ph1 (some kind of: na.rm=T)?
You could use with(), and extract the vectors from a subset of x: with(x[!is.na(x$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