find n most frequent words in a file

Question

I want to find, say, 10 most common word in a text file. Firstly, solution should be optimized for keystrokes (in other words - my time). Secondly, for the performance. Here is what I have so far to get top 10:

cat test.txt | tr -c '[:alnum:]' '[\n*]' | uniq -c | sort -nr | head  -10
  6 k
  2 g
  2 e
  2 a
  1 r
  1 k22
  1 k
  1 f
  1 eeeeeeeeeeeeeeeeeeeee
  1 d

I could make a java, python etc. program where I store (word, numberOfOccurences) in a dictionary and sort the value or I could use MapReduce, but I optimize for keystrokes.

Are there any false positives? Is there a better way?

Answer 1

That's pretty much the most common way of finding "N most common things", except you're missing a sort, and you've got a gratuitious cat:

tr -c '[:alnum:]' '[\n*]' < test.txt | sort | uniq -c | sort -nr | head  -10

If you don't put in a sort before the uniq -c you'll probably get a lot of false singleton words. uniq only does unique runs of lines, not overall uniquness.

EDIT: I forgot a trick, "stop words". If you're looking at English text (sorry, monolingual North American here), words like "of", "and", "the" almost always take the top two or three places. You probably want to eliminate them. The GNU Groff distribution has a file named "eign" in it which contains a pretty decent list of stop words. My Arch distro has /usr/share/groff/current/eign, but I think I've also seen /usr/share/dict/eign or /usr/dict/eign in old Unixes.

You can use stop words like this:

tr -c '[:alnum:]' '[\n*]' < test.txt |
fgrep -v -f /usr/share/groff/current/eign |
sort | uniq -c | sort -nr | head  -10

My guess is that most human languages need similar "stop words" removed from meaningful word frequency counts, but I don't know where to suggest getting other languages stop words lists.