CODEWITHSUNDEEP
python
cybertextron
cybertextron

Reputation: 10961

producing all the anagrams from a string python

I was thinking about this problem today, and I came with the following pseudocode (Python 3.2) :

def anagrams( string ):

    for c in string:
      anagram = c + anagram( string - {c} ) # remove the char from its position in the string
      print(anagram)

    return

def main():

    word = "abcd"
    anagrams( word )

    return

However, I'd like to know a pythonic way to do this operation: anagram = c + anagram( string - {c} )

How could I remove that char from the string? so for example:

"abc" -> 'a' + "bc" -> 'a' + 'b' + "c" -> 'a' + 'b' + 'c' = 'abc'
             + "cb" -> 'a' + 'c' + "b" -> 'a' + 'c' + 'b' = 'acb'
      -> 'b' + "ac" -> 'b' + 'a' + "c" -> 'b' + 'a' + 'c' = 'bac'
             + "ca" -> 'b' + 'c' + "a" -> 'b' + 'c' + 'a' = 'bca'
      -> 'c' + "ba" -> 'c' + 'b' + "a" -> 'c' + 'b' + 'a' = 'cba'
             + "ab" -> 'c' + 'a' + "b" -> 'c' + 'a' + 'b' = 'cab'

Thanks

Upvotes: 14

Views: 39265

Answers (5)

Brad Parks
Brad Parks

Reputation: 72001

Here's some of the suggestions wrapped up to make a new script, called anagram

$ anagram hit
hit
hti
iht
ith
thi
tih

anagram

#!/usr/bin/env python3
import argparse
import itertools
import sys

parser = argparse.ArgumentParser()
parser.add_argument("word", help="A word to generate anagrams for.")
args = parser.parse_args()

perms = [''.join(perm) for perm in itertools.permutations(args.word)]
for perm in perms:
    print(perm)

Upvotes: 0

deepstructure
deepstructure

Reputation: 778

Just to note, @sloth's answer gives a slightly unexpected result if the string contains more than one instance of a letter - duplicate permutations:

["".join(perm) for perm in itertools.permutations('aabc')]

Results in:

['aabc',
 'aacb',
 'abac',
 'abca',
 'acab',
 'acba',
 'aabc',
 'aacb',
 'abac',
 'abca',
 'acab',
 'acba',
 'baac',
 'baca',
 'baac',
 'baca',
 'bcaa',
 'bcaa',
 'caab',
 'caba',
 'caab',
 'caba',
 'cbaa',
 'cbaa']

If this isn't the desired result, using 'set' will eliminate the dups. Although if you want a list, you'll need to relist (also 'set' doesn't maintain order so use 'sorted'):

sorted(set(["".join(perm) for perm in itertools.permutations("aabc")]))

Results in:

['aabc',
 'aacb',
 'abac',
 'abca',
 'acab',
 'acba',
 'baac',
 'baca',
 'bcaa',
 'caab',
 'caba',
 'cbaa']

Upvotes: 5

sloth
sloth

Reputation: 101072

Why not just use itertools?

>>> import itertools
>>> ["".join(perm) for perm in itertools.permutations("abc")]
['abc', 'acb', 'bac', 'bca', 'cab', 'cba']

The documentation also contains code how the permutation is done.

Edit:

Without itertools:

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                yield perm[:i] + elements[0:1] + perm[i:]


word = "abc"
print list(all_perms(word))

Without itertools and without generators:

def all_perms(elements):
    if len(elements) <=1:
        return elements
    else:
        tmp = []
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                tmp.append(perm[:i] + elements[0:1] + perm[i:])
        return tmp

Result:

['abc', 'bac', 'bca', 'acb', 'cab', 'cba']

Upvotes: 34

Lanaru
Lanaru

Reputation: 9721

Use the itertools module.

import itertools
perms = [''.join(perm) for perm in itertools.permutations('abc')]

Upvotes: 6

mjgpy3
mjgpy3

Reputation: 8937

You can cast the word to a list, run remove and then use join to get it back together again.

word = 'abca'
letters = list(word)
letters.remove('a')  # Only deletes the first 'a'
word = ''.join(letters)

Upvotes: 0

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