probably basic, but couldn't find it in any other question. I tried:
print ["".join(seq) for seq in itertools.permutations("00011")]
but got lots of duplications, seems like itertools doesn't understand all zeroes and all ones are the same...
what am I missing?
EDIT:
oops. Thanks to Gareth I've found out this question is a dup of: permutations with unique values . Not closing it as I think my phrasing of the question is clearer.
set("".join(seq) for seq in itertools.permutations("00011"))
list(itertools.combinations(range(5), 2))
returns a list of 10 positions where the two ones can be within the five-digits (others are zero):
[(0, 1),
(0, 2),
(0, 3),
(0, 4),
(1, 2),
(1, 3),
(1, 4),
(2, 3),
(2, 4),
(3, 4)]
For your case with 2 ones and 13 zeros, use this:
list(itertools.combinations(range(5), 2))
which returns a list of 105 positions. And it is much faster than your original solution.
Now the function:
def combiner(zeros=3, ones=2):
for indices in itertools.combinations(range(zeros+ones), ones):
item = ['0'] * (zeros+ones)
for index in indices:
item[index] = '1'
yield ''.join(item)
print list(combiner(3, 2))
['11000',
'01100',
'01010',
'01001',
'00101',
'00110',
'10001',
'10010',
'00011',
'10100']
and this needs 14.4µs.
list(combiner(13, 2))
returning 105 elements needs 134µs.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.