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Recipe 66463: permutation

A small but efficient recursive function for printing the permutation of characters in a given string.

Python 
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#!/usr/bin/env python
__version__ = "1.1"
"""permute.py-- A small but efficient recursive function for printing the
permutation of the characters of the specified string.
"""
import sys

def printList(alist, blist=[]):
    if not len(alist): print ''.join(blist)
    for i in range(len(alist)):
        blist.append(alist.pop(i))
        printList(alist, blist)
        alist.insert(i, blist.pop())

if __name__ == '__main__':
    k='love'
    if len(sys.argv)>1: k = sys.argv[1]
    printList(list(k))

Discussion

Done initially as a programming excercise this script uses stacks to build the permutations of the characters in the specified string. This does not check for duplicate characters in the specified string but could be modified easily to do so too. Even though the algorithm above uses text/strings as an example, this can easily be used to work with lists of any sort.

I also plan to submit one that does combinations.

Comments

  1. 1. At 10:49 a.m. on 13 sep 2001, Robin Houston said:

    A faster way. If you're generating the permutations of a long list, speed matters a lot. The code below is about four times faster than the recipe presented above.

    def permute(alist, level=0):
    index, copy, printing = level, list(alist), level+1 == len(alist)

    while 1:
        if printing:
            print ''.join(copy)
        else:
            permute(copy, 1+level);

        if index != 0:
            copy[index-1], copy[index] = copy[index], copy[index-1]

        index -= 1
        if index < 0:
            break
  2. 2. At 4:52 p.m. on 9 nov 2004, Shalabh Chaturvedi said:

    A different way using generators. The logic behind generating permutations is a little more visible in the following (IMO):

    def getPermutations(a):
   if len(a)==1:
      yield a
   else:
      for i in range(len(a)):
         this = a[i]
         rest = a[:i] + a[i+1:]
         for p in getPermutations(rest):
            yield this + p
  3. 3. At 2:45 p.m. on 5 may 2007, Eyal Lotem said:

    mutable default argument. Its best to avoid those by using:

    def ...(...blist=None...):
  if blist == None:
    blist = []
  ...

    Or otherwise, this function can only run once in parallel and is not reentrant and not safe (when using threads, or if callbacks eventually re-call it).

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