On Tue, Sep 14, 2010 at 12:44 AM, David Winsemius
<dwinsem...@comcast.net> wrote:
> The second argument to mean is trim. I am not sure what mean(1, 3) is
> supposed to do but what it return is 1.
>
Thanks for the info. On this particular point I find the documentation
confusing. In ?mapply :
'‘mapply’ applies
‘FUN’ to the first elements of each ... argument, '
mapply(FUN, ..., MoreArgs = NULL, [..]
' ...: arguments to vectorize over (list or vector).'
In my understanding this suggests that '...' can take several comma
separated objects, so that in
mapply(mean, tree[[1]]$node$values, tree[[2]]$node$values)
the second object should not be treated as a 'MoreArgs' argument. But
I'm probably wrong.
> If you wanted 2,3,4 ..., 11 then you
> would perhaps do:
>
> mean( mapply(c, tree[[1]]$node$values, tree[[2]]$node$values) )
>
I think the original poster was more interested in finding the mean()
by rows. Instead of
> mean( mapply(c, tree[[1]]$node$values, tree[[2]]$node$values) )
[1] 6.5
he probably looks for
> apply( mapply(c, tree[[1]]$node$values, tree[[2]]$node$values), 2, mean )
[1] 2 3 4 5 6 7 8 9 10 11
although I'm positive there is a neater way to do this. For example,
apply( data.frame(tree[[1]]$node$values, tree[[2]]$node$values), 1, mean )
Assembling your data in a data.frame prior to using an *pply function
would eliminate the need to write them all by
hand. Regards
Liviu
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