On May 26, 2010, at 4:05 AM, Ivan Calandra wrote:
What about (not tested because no sample data, and I'm definitely
unsure
of my lapply() solution):
lapply(data[2:13], FUN=as.character)
lapply(data[2:13], FUN=as.numeric)
?
That would not preserve any of the output since no assignments were
made. Consider instead:
data[ , -1] <- lapply(data[, -1] , function(x)
as.numeric(as.character(x)) )
Would be safer to make a copy of "data" and work on it, and of course,
there is always:
require(fortunes)
fortune("dog")
--
David.
data[2:13] should correspond to the 12 Month columns.
There might be a way to combine both lines but I don't think lapply
would work with 2 function names.
When I can't find a vectorized solution, or one with *apply, I use a
for
loop. I find it very easy to understand and it's fast enough for my
"small" datasets
HTH
Ivan
Le 5/26/2010 09:57, Mohan L a écrit :
On Wed, May 26, 2010 at 1:13 PM, Ivan Calandra
<ivan.calan...@uni-hamburg.de <mailto:ivan.calan...@uni-hamburg.de>>
wrote:
Hi,
What about:
as.numeric(as.character(data$State))
?
What I what is, I want to convert all the column excluding State to
factor to numeric in the data frame. So that I will send this data
frame to another function to total .
Any help.
Thanks & Rg
Mohan L
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**********
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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