drop zero slots from table?

sorted.table <- function(x, name = if (is.list(x))names(x) else deparse(substitute(x))[1]) {

digits <- factor(trunc(100*log2(1:20)%%.1), levels=0:9)
sorted.table(digits)

sorted.table(data.frame(DigitsColumn=digits))

sorted.table(digits, name="My Digits")

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
Of Sam Steingold
Sent: Wednesday, September 19, 2012 9:13 AM
To: r-help at r-project.org
Subject: Re: [R] drop zero slots from table?

Function
--8<---------------cut here---------------start------------->8---
sorted.table <- function (vec) {
  tab <- table(vec)
  tab <- tab[tab > 0]
  sort(tab, decreasing=TRUE)
}
--8<---------------cut here---------------end--------------->8---
does what I want but it prints "vec" instead of the name of its
argument:
--8<---------------cut here---------------start------------->8---

sorted.table(foo$bar)

vec
  A  B
  10 3
--8<---------------cut here---------------end--------------->8---
how do I pass all arguments of sorted.table() on to table() as is?
thanks!

* Sam Steingold <fqf at tah.bet> [2012-09-19 11:51:08 -0400]:

I find myself doing
tab <- table(...)
tab <- tab[tab > 0]
tab <- sort(tab,decreasing=TRUE)
all the time.
I am wondering if the "drop 0" (and maybe even sort?) can be effected by
some magic argument to table() which I fail to discover in the docs?
Obviously, I could use droplevels() to avoid 0 counts in the first
place, but I do not want to drop the levels in the data.

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