thanks everyone.
I think the motto should be "always specify the levels of a factor when
you create it
if you possibly can".
best wishes
Robin
On 06/29/2010 12:39 PM, Felix Andrews wrote:
Just use factor(), not levels(); you can pass a factor to factor() too.
x<- factor(c(rep("a",3
Just use factor(), not levels(); you can pass a factor to factor() too.
> x <- factor(c(rep("a",3),"b","d"), levels = letters[1:5])
> table(x)
x
a b c d e
3 1 0 1 0
Cheers,
-Felix
On 29 June 2010 20:59, Robin Hankin wrote:
> Hi
>
> suppose I have a factor 'x':
>
>> x <- as.factor(c(rep("a",3),
You could try
x<- factor(c(rep("a",3),"b","d"), levels=letters[1:4])
table(x)
# x
# a b c d
# 3 1 0 1
Hope this helps
Allan
On 29/06/10 11:59, Robin Hankin wrote:
Hi
suppose I have a factor 'x':
> x <- as.factor(c(rep("a",3),"b","d"))
> table(x)
x
a b d
3 1 1
>
>
But this is not what I wa
On Tue, 2010-06-29 at 11:59 +0100, Robin Hankin wrote:
> Hi
>
> suppose I have a factor 'x':
>
> > x <- as.factor(c(rep("a",3),"b","d"))
> > table(x)
> x
> a b d
> 3 1 1
> >
> >
>
> But this is not what I want because
> I need to include the fact that the count of "c" is zero.
>
> I can't j
Hi
suppose I have a factor 'x':
> x <- as.factor(c(rep("a",3),"b","d"))
> table(x)
x
a b d
3 1 1
>
>
But this is not what I want because
I need to include the fact that the count of "c" is zero.
I can't just change the levels of x:
> levels(x) <- c("a","b","c","d")
> table(x)
x
a b c d
3 1 1
5 matches
Mail list logo
