Python: create all derivatives of string by replacing its symbols

Question

I've got some dict like that

d = {'a' : ['1', '+', '='], 'b' : ['2'], c : ['3']}

My goal is to create a list of all possible derivatives from initial word by replacing its elements like that:

word = 'abcde'
rez = ['abcde', 
       '1bcde', '+bcde', '=bcde', 
       'a2cde', '12cde', '+2cde', '=2cde', 
       'ab3de', '1b3de', '+b3de', '=b3de', 'a23de', '123de', '+23de', '=23de']

Order of words in rez is not important.

I feel that there should be some easy way with itertools , but I can't see it. Any nice and beautiful solution would be very welcome.

Answer 1

Sure, itertools is always the answer for these kind of problems. There may be better alternatives but the first that comes to my mind is using itertools.product :

from itertools import product

[''.join(chars) for chars in product(*[[x] + d.get(x, []) for x in word])]

Output

['abcde', 'ab3de', 'a2cde', 'a23de', '1bcde', '1b3de', '12cde', '123de',
 '+bcde', '+b3de', '+2cde', '+23de', '=bcde', '=b3de', '=2cde', '=23de']

Answer 2

Here is one approach. Since you want to allow non-replacement (eg, leaving "a" as "a"), it is better to include the original character in the list of replacement values, so that the dict has, eg, 'a': ['a', '1', '+', '='] . This can be done with:

for k in d:
    d[k].append(k)

Then:

subs = [[(k, v) for v in vs] for k, vs in d.iteritems()]  
rez = []
for comb in itertools.product(*subs):
    baseword = word
    for before, after in comb:
        baseword = baseword.replace(before, after)
    rez.append(baseword)

The result:

['123de', '1b3de', '12cde', '1bcde', '+23de', '+b3de', '+2cde', '+bcde', '=23de', '=b3de', '=2cde', '=bcde', 'a23de', 'ab3de', 'a2cde', 'abcde']

Answer 3

Just for fun. No import , no def , no indentation.

d = {'a' : ['1', '+', '='], 'b' : ['2'], 'c' : ['3']}
for key in d: d[key].append(key)
print reduce(lambda items, f: sum(map(f, items), []), [lambda msg, c=c: [msg.replace(c, r) for r in d[c]] for c in d.keys()], ["abcde"])

Result:

[
    '123de', '1b3de', '12cde', '1bcde', 
    '+23de', '+b3de', '+2cde', '+bcde', 
    '=23de', '=b3de', '=2cde', '=bcde', 
    'a23de', 'ab3de', 'a2cde', 'abcde'
]