Anagrams: Difference between revisions
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{{task|Text processing}} |
{{task|Text processing}} |
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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. |
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. |
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=={{header|C++}}== |
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<cpp>#include <iostream> |
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#include <fstream> |
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#include <string> |
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#include <map> |
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#include <vector> |
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#include <algorithm> |
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#include <iterator> |
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int main() { |
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std::ifstream in("unixdict.txt"); |
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std::map<std::string, std::vector<std::string> > anagrams; |
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std::string word; |
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size_t count = 0; |
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while (std::getline(in, word)) { |
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std::string key = word; |
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std::sort(key.begin(), key.end()); |
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// note: the [] operator automatically inserts a new value if key does not exist |
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anagrams[key].push_back(word); |
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count = std::max(count, anagrams[key].size()); |
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} |
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in.close(); |
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for (std::map<std::string, std::vector<std::string> >::iterator it = anagrams.begin(); |
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it != anagrams.end(); |
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it++) |
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if (it->second.size() >= count) { |
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std::copy(it->second.begin(), it->second.end(), |
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std::ostream_iterator<std::string>(std::cout, ", ")); |
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std::cout << std::endl; |
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} |
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return 0; |
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}</cpp> |
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Output: |
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abel, able, bale, bela, elba, |
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caret, carte, cater, crate, trace, |
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angel, angle, galen, glean, lange, |
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alger, glare, lager, large, regal, |
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elan, lane, lean, lena, neal, |
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evil, levi, live, veil, vile, |
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== {{header|D}} == |
== {{header|D}} == |
Revision as of 01:42, 15 October 2008
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.
C++
<cpp>#include <iostream>
- include <fstream>
- include <string>
- include <map>
- include <vector>
- include <algorithm>
- include <iterator>
int main() {
std::ifstream in("unixdict.txt"); std::map<std::string, std::vector<std::string> > anagrams;
std::string word; size_t count = 0; while (std::getline(in, word)) { std::string key = word; std::sort(key.begin(), key.end()); // note: the [] operator automatically inserts a new value if key does not exist anagrams[key].push_back(word); count = std::max(count, anagrams[key].size()); }
in.close();
for (std::map<std::string, std::vector<std::string> >::iterator it = anagrams.begin(); it != anagrams.end(); it++) if (it->second.size() >= count) { std::copy(it->second.begin(), it->second.end(), std::ostream_iterator<std::string>(std::cout, ", ")); std::cout << std::endl; } return 0;
}</cpp> Output:
abel, able, bale, bela, elba, caret, carte, cater, crate, trace, angel, angle, galen, glean, lange, alger, glare, lager, large, regal, elan, lane, lean, lena, neal, evil, levi, live, veil, vile,
D
D 1, using Phobos (to download the word list you need the Tango Std Lib). <d> import std.stdio, std.stream;
void main() {
string[][string] anags; foreach (string w; new BufferedFile("unixdict.txt")) anags[w.dup.sort] ~= w.dup; int lmax; foreach (a; anags) lmax = lmax < a.length ? a.length : lmax; foreach (a; anags) if (a.length == lmax) writefln(a);
} </d>
Haskell
import Data.List groupon f x y = f x == f y main = do f <- readFile "./../Puzzels/Rosetta/unixdict.txt" let words = lines f wix = groupBy (groupon fst) . sort $ zip (map sort words) words mxl = maximum $ map length wix mapM_ (print . map snd) . filter ((==mxl).length) $ wix
Sample output:
*Main> main ["abel","able","bale","bela","elba"] ["caret","carte","cater","crate","trace"] ["angel","angle","galen","glean","lange"] ["alger","glare","lager","large","regal"] ["elan","lane","lean","lena","neal"] ["evil","levi","live","veil","vile"]
J
(#~a:~:{:"1)(]/.~/:~&>)<;.2]1!:1<'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 | +-----+-----+-----+-----+-----+
Java
<java>import java.net.*; import java.io.*; import java.util.*;
public class WordsOfEqChars {
public static void main(String[] args) throws IOException { URL url = new URL("http://www.puzzlers.org/pub/wordlists/unixdict.txt"); InputStreamReader isr = new InputStreamReader(url.openStream()); BufferedReader reader = new BufferedReader(isr);
Map<String, Collection<String>> anagrams = new HashMap<String, Collection<String>>(); String word; int count = 0; while ((word = reader.readLine()) != null) { char[] chars = word.toCharArray(); Arrays.sort(chars); String key = new String(chars); if (!anagrams.containsKey(key)) anagrams.put(key, new ArrayList<String>()); anagrams.get(key).add(word); count = Math.max(count, anagrams.get(key).size()); }
reader.close();
for (Collection<String> ana : anagrams.values()) if (ana.size() >= count) System.out.println(ana); }
}</java> Output:
[angel, angle, galen, glean, lange] [elan, lane, lean, lena, neal] [alger, glare, lager, large, regal] [abel, able, bale, bela, elba] [evil, levi, live, veil, vile] [caret, carte, cater, crate, trace]
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>