Re: Question about permutations (itertools)
list(combinations_with_replacement('01',3))
('0', '0', '0')
('0', '0', '1')
('0', '1', '1')
('1', '1', '1')
Is it possible to get combinations_with_replacement to return numbers
rather than strings? (see above)
list(combinations_with_replacement(range(0,2), 3))
[(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)]
Hopefully that'll give you some ideas.
Cheers,
Xav
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Re: Question about permutations (itertools)
Vincent Davis wrote:
As a note I am doing this in py3.
I am looking for the most efficient (speed) way to produce an an
iterator to of permutations.
One of the problem I am having it that neither combinations nor
permutations does not exactly what I want directly.
For example If I want all possible ordered lists of 0,1 of length 3
(0,0,0)
(0,0,1)
(0,1,1)
(1,1,1)
(1,0,1)
(1,1,0)
(1,0,0)
I don't see a way to get this directly from the itertools. But maybe I
am missing something.
for t in itertools.product([0, 1], repeat=3):
... print(t)
...
(0, 0, 0)
(0, 0, 1)
(0, 1, 0) # Seems you missed one
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)
I see ways to get a bigger list and then remove
duplicates.
list(permutations([0,1], 3))
[]
list(combinations_with_replacement('01',3))
('0', '0', '0')
('0', '0', '1')
('0', '1', '1')
('1', '1', '1')
Is it possible to get combinations_with_replacement to return numbers
rather than strings? (see above)
for t in itertools.combinations_with_replacement([0, 1], 3):
... print(t)
...
(0, 0, 0)
(0, 0, 1)
(0, 1, 1)
(1, 1, 1)
Peter
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Re: Question about permutations (itertools)
On Mon, May 31, 2010 at 8:17 AM, Xavier Ho cont...@xavierho.com wrote:
list(combinations_with_replacement('01',3))
('0', '0', '0')
('0', '0', '1')
('0', '1', '1')
('1', '1', '1')
Is it possible to get combinations_with_replacement to return numbers
rather than strings? (see above)
list(combinations_with_replacement(range(0,2), 3))
[(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)]
Thanks, for some reason I didn't combinations_with_replacement took a
list as an argument.
Hopefully that'll give you some ideas.
Cheers,
Xav
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Re: Question about permutations (itertools)
On May 31, 3:04 pm, Vincent Davis vinc...@vincentdavis.net wrote: For example If I want all possible ordered lists of 0,1 of length 3 (0,0,0) (0,0,1) (0,1,1) (1,1,1) (1,0,1) (1,1,0) (1,0,0) I don't see a way to get this directly from the itertools. But maybe I am missing something. In this case, you're missing itertools.product: list(itertools.product([0, 1], repeat=3)) [(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)] -- Mark -- http://mail.python.org/mailman/listinfo/python-list
Re: Question about permutations (itertools)
Vincent Davis wrote: I am looking for the most efficient (speed) way to produce an an iterator to of permutations. One of the problem I am having it that neither combinations nor permutations does not exactly what I want directly. For example If I want all possible ordered lists of 0,1 of length 3 (0,0,0) (0,0,1) (0,1,1) (1,1,1) (1,0,1) (1,1,0) (1,0,0) I don't see a way to get this directly from the itertools. But maybe I am missing something. I see ways to get a bigger list and then remove duplicates. You have three digits where each digit can have two values (binary digits, a.k.a. bits), so the number of combinations is 2*2*2 = 8. Even if the possible values where unevenly distributed, you could calculate the number of combinations by multiplying. Then, there are two different approaches: 1. count with an integer and then dissect into digits # Note: Using // for integer division in Python3! digit0 = n % base0 digit1 = (n // base0) % base1 digit2 = (n // base0 // base1) % base2 2. simulate digits and detect overflow Here you simply count up the ones and if they overflow, you reset them to zero and count up the tens. What I don't really understand is what you mean with ordered lists. Uli -- Sator Laser GmbH Geschäftsführer: Thorsten Föcking, Amtsgericht Hamburg HR B62 932 -- http://mail.python.org/mailman/listinfo/python-list
