Generating unique permutations

Question

I'm trying to generate unique permutations of length two from a string but I'm getting repeated values. What am I doing wrong? Here's the code:

a = 'abba'
from itertools import permutations
x = []
x = [y for y in list(permutations(a,2)) if y not in x]
'''
output was this:
[('a', 'b'), ('a', 'b'), ('a', 'a'), ('b', 'a'), ('b', 'b'), ('b', 'a'), ('b', 'a'), ('b', 'b'),('b', 'a'), ('a', 'a'), ('a', 'b'), ('a', 'b')]
'''

jonrsharpe · Accepted Answer · 2014-05-17T12:15:00.533

1

The list comprehension builds a list, then assigns it to x, so x == [] the whole time it's running. Every time the list comprehension checks y not in x, x is still an empty list, so of course y is never in it.

If order isn't crucial, you could use a set instead:

x = set(permutations(a, 2))

Otherwise, unroll the list comprehension:

x = []
for y in permutations(a, 2):
    if y not in x:
        x.append(y)

edited May 17 '14 at 12:15

answered May 17 '14 at 12:07

jonrsharpe

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I'm sorry. Could you explain plz? :3 – May 17 '14 at 12:09
Which part don't you understand? – jonrsharpe May 17 '14 at 12:10
I understand that it first builds a list and then assigns it to x but what does it has to do with those duplicates? :| – May 17 '14 at 12:11
Because every time the list comprehension checks `y not in x`, *`x` is still an empty list*. – jonrsharpe May 17 '14 at 12:12

Generating unique permutations

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