Anagrams

From Rosetta Code
Revision as of 22:22, 25 September 2008 by rosettacode>DanBron (→‎{{header|J}}: shorter, faster (possibly wrong results in general case when /:&> requires fill?))
Task
Anagrams
You are encouraged to solve this task according to the task description, using any language you may know.

Two or more words can be composed of the same characters, but in a different order. Using the word list at http://www.puzzlers.org/pub/wordlists/unixdict.txt, find the sets of words that share the same characters that contain the most words in them.

J

   (#~a:~:{:"1)(]/.~/:&>)<;._2 toJ fread jpath'~temp\unixdict.txt'
+-----+-----+-----+-----+-----+
|abel |able |bale |bela |elba |
+-----+-----+-----+-----+-----+
|alger|glare|lager|large|regal|
+-----+-----+-----+-----+-----+
|angel|angle|galen|glean|lange|
+-----+-----+-----+-----+-----+
|caret|carte|cater|crate|trace|
+-----+-----+-----+-----+-----+
|elan |lane |lean |lena |neal |
+-----+-----+-----+-----+-----+
|evil |levi |live |veil |vile |
+-----+-----+-----+-----+-----+

Python

Python 2.5 shell input (IDLE) <python>>>> import urllib >>> from collections import defaultdict >>> words = urllib.urlopen('http://www.puzzlers.org/pub/wordlists/unixdict.txt').read().split() >>> len(words) 25104 >>> anagram = defaultdict(list) # map sorted chars to anagrams >>> for word in words: anagram[str(sorted(word))].append( word )


>>> count, max_anagrams = max((len(ana), ana) for ana in anagram.itervalues()) >>> for ana in anagram.itervalues(): if len(ana) >= count: print ana


['caret', 'carte', 'cater', 'crate', 'trace'] ['alger', 'glare', 'lager', 'large', 'regal'] ['evil', 'levi', 'live', 'veil', 'vile'] ['angel', 'angle', 'galen', 'glean', 'lange'] ['elan', 'lane', 'lean', 'lena', 'neal'] ['abel', 'able', 'bale', 'bela', 'elba'] >>> count 5 >>></python>