Hi folks,
I would like to trim the trailing spaces in my factor variables using lapply
(described in this post by Marc Schwartz:
http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22826.html) but the code is
not functioning (in this example there is only one factor with trailing
spaces):
y1 -
Thanks Marc! What would be the easiest way to coerce char-variables back to
factor-variables? Is there a way to prevent the coercion in d[] - lapply(d,
function(x) if (is.factor(x)) sub( +$, , x) else x) ?
-Lauri
2007/8/16, Marc Schwartz [EMAIL PROTECTED]:
On Thu, 2007-08-16 at 17:54 +0300,
On Thu, 2007-08-16 at 17:54 +0300, Lauri Nikkinen wrote:
Hi folks,
I would like to trim the trailing spaces in my factor variables using lapply
(described in this post by Marc Schwartz:
http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22826.html) but the code is
not functioning (in this
The easiest way might be to modify the lapply() call as follows:
d[] - lapply(d, function(x) if (is.factor(x)) factor(sub( +$, , x)) else x)
str(d)
'data.frame': 60 obs. of 3 variables:
$ x: Factor w/ 5 levels 1,2,3,4,..: 1 1 1 1 1 1 1 1 1 1 ...
$ y: num 7.01 8.33 5.48 6.51 5.61 ...
$ f:
On Thu, 16 Aug 2007, Marc Schwartz wrote:
The easiest way might be to modify the lapply() call as follows:
d[] - lapply(d, function(x) if (is.factor(x)) factor(sub( +$, , x)) else
x)
str(d)
'data.frame': 60 obs. of 3 variables:
$ x: Factor w/ 5 levels 1,2,3,4,..: 1 1 1 1 1 1 1 1 1 1
On Thu, 2007-08-16 at 17:52 +0100, Prof Brian Ripley wrote:
On Thu, 16 Aug 2007, Marc Schwartz wrote:
The easiest way might be to modify the lapply() call as follows:
d[] - lapply(d, function(x) if (is.factor(x)) factor(sub( +$, , x))
else x)
str(d)
'data.frame': 60 obs. of 3
If the problem is with the levels of the factor, why not change
them directly?
d = data.frame(a=1:5,
+ b=c('one ','two','three ','three ','two'))
d$b
[1] onetwothree three two
Levels: one three two
levels(d$b) = sub(' +$','',levels(d$b))
d$b
[1] one two three three two
