Hi,

table() is behaving as documented with respect to your example. local.labels
is a *character* vector with two distinct values and local.preds is a
*character* variable with one distinct value. If you were expecting your
table to divine that you wanted to include 'ah~' as a missing value in your
local.preds object, it's not going to happen under your current setup.

> local.labels <- c("ah", "ah", "ah~")
> local.preds <- c("ah", "ah", "ah")
> table(local.labels, local.preds)
            local.preds
local.labels ah
         ah   2
         ah~  1
> class(local.labels)
[1] "character"
> class(local.preds)
[1] "character"


Is this what you had in mind?

labels <- factor(local.labels)
preds <- factor(local.preds, levels = c('ah', 'ah~'))
> table(labels, preds)
      preds
labels ah ah~
   ah   2   0
   ah~  1   0

> labels
[1] ah  ah  ah~
Levels: ah ah~
> preds
[1] ah ah ah
Levels: ah ah~

There is a distinction in R between character objects and factor objects.
More generally, many [generic] functions behave differently depending on the
type of object(s) supplied as input.

HTH,
Dennis

On Wed, Jul 28, 2010 at 10:29 PM, Na'im R. Tyson <ntyson@clovermail.net>wrote:

> R-philes,
>
> I have a question about displaying counts of unused factors using the
> table() function.  I have two vectors with character data in them:
>
> local.labels("ah", "ah", "ah~")
> local.preds("ah", "ah", "ah")
>
> If I use the table function as shown below, I get an error because the
> number of levels do not match up.
>
> v.cont.table <- table(local.labels, local.preds, dnn=c("observed",
> "predicted"));
>
> Is there any way to get zero counts into the contingency table, or would I
> have to use a flat table?  Any help with sample code, or a pointer to a
> previous post, would be much appreciated.
>
> Regards,
>
> Na'im
>
> ______________________________________________
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> PLEASE do read the posting guide
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