How to get every single permutation of all substrings of a string?
algorithm, permutation, python, string, substring
Solution
This is one way of doing it with itertools.permutations:
from itertools import permutations
s = 'pet'
print [''.join(p) for i in range(1, len(s)+1) for p in permutations(s, i)]
Output:
['p', 'e', 't', 'pe', 'pt', 'ep', 'et', 'tp', 'te', 'pet', 'pte', 'ept', 'etp', 'tpe', 'tep']
Problem
I know how to get the permutations of just the plain string in python: ``` >>> from itertools import permutations >>> perms = [''.join(p) for p in permutations('stack')] >>> print perms ... ``` But how would I get the permutations of `'stac'`, `'stak'`, `'sack'`, `'stck'`, `'stc'`, `'st'`, and so forth? My desired output is: ``` >>> permutations('pet') ['pet', 'pte', 'ept', 'etp', 'tpe', 'tep', 'pe', 'ep', 'p', 'e', 't', 'pt', 'tp', 'et', 'te'] ``` What I have so far: ``` def permutate(values, size): return map(lambda p: [values[i] for i in p], permutate_positions(len(values), size)) def permutate_positions(n, size): if (n==1): return [[n]] unique = [] for p in map(lambda perm: perm[:size], [ p[:i-1] + [n-1] + p[i-1:] for p in permutate_positions(n-1, size) for i in range(1, n+1) ]): if p not in unique: unique.append(p) return unique def perm(word): all = [] for k in range(1, len(word)+1): all.append(permutate([' ']+list(word), k)) return all ``` This runs as: ``` >>> perm('pet') [[['t'], ['e'], ['p']], [['t', 'e'], ['e', 't'], ['e', 'p'], ['t', 'p'], ['p', 't'], ['p', 'e'], ['p', 'p']], [['t', 'e', 'p'], ['e', 't', 'p'], ['e', 'p', 't'], ['e', 'p', 'p'], ['t', 'p', 'e'], ['p', 't', 'e'], ['p', 'e', 't'], ['p', 'e', 'p'], ['t', 'p', 'p'], ['p', 't', 'p'], ['p', 'p', 't'], ['p', 'p', 'e']]] >>> ``` However, it has a bunch of list of lists, and with values like ['p', 'p', 't']! How do I do this? ANy help is appreciated.