Question about permutations (itertools)

[Vincent Davis]

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. I see ways to get a bigger list and then remove
duplicates.

list(permutations([0,1], 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)

Thanks
Vincent

 

[Peter Otten]

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

 

[Mark Dickinson]

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)]

 

[Ulrich Eckhardt]

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