Permutations with repetition in Python

Question

I want to iterate over all the vertices of an n dimensional cube of size 1. I know I could do that with itertools.product as follows:

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Answer 1

For your use-case of 3d cubes, Eevee's solution is the correct one.

However for fun and to demonstrate the power of itertools, here's a linear-time solution that generalizes to higher dimensions:

from itertools import combinations

# n is the number of dimensions of the cube (3 for a 3d cube)
def generate_vertices(n):
    for number_of_ones in xrange(0, n + 1):
        for location_of_ones in combinations(xrange(0, n), number_of_ones):
            result = [0] * n
            for location in location_of_ones:
                result[location] = 1
            yield result

for vertex in generate_vertices(3):
    print vertex


# result:
# [0, 0, 0]
# [1, 0, 0]
# [0, 1, 0]
# [0, 0, 1]
# [1, 1, 0]
# [1, 0, 1]
# [0, 1, 1]
# [1, 1, 1]

Answer 2

It is not a bad idea to count depending on what you will do with the vertices because if you have to iterate over all of them doing something O(f(n)) is at least O(f(n)*2n), sorting them is O(n*2n). So it basically depends if f(n) majors n.

Aside from that here is a possible magic expression:

def magic_expression(ones, n):
    a = (0,) * (n - ones) + (1,) * ones
    previous = tuple()
    for p in itertools.permutations(a):
        if p > previous:
            previous = p
            yield p

With help from permutations with unique values.

This works because itertools.permutations yield sorted results. Note that a is initially sorted because zeros come first.

Answer 3

An (inefficient) alternative method:

>>> ['{0:03b}'.format(x) for x in range(8)]
['000', '001', '010', '011', '100', '101', '110', '111']

Or as tuples:

>>> [tuple(int(j) for j in list('{0:03b}'.format(x))) for x in range(8)]

[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (0, 1, 1),
 (1, 0, 0),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]

Sorted by number of vertices:

>>> sorted(_, key=lambda x: sum(x))

[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (1, 0, 0),
 (0, 1, 1),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]

---

Or using itertools:

>>> sorted(itertools.product((0, 1), repeat=3), key=lambda x: sum(x))

[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (1, 0, 0),
 (0, 1, 1),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]

Answer 4

If you've written more than eight lines of code to generate eight constant values, something has gone wrong.

Short of just embedding the list I want, I'd do it the dumb way:

vertices = (
    (v.count(1), v)
    for v in itertools.product((0, 1), repeat=3)
)
for count, vertex in sorted(vertices):
    print vertex

Unless you're working with 1000-hypercubes, you shouldn't have any huge performance worries.

Answer 5

Following is some code that runs faster (for medium n) and several times faster (for large n) than that of Cam or Eevee. A time comparison follows.

def cornersjc (n):   # Re: jw code
    from itertools import product
    m = (n+1)/2
    k = n-m
    # produce list g of lists of tuples on k bits
    g = [[] for i in range(k+1)]
    for j in product((0,1), repeat=k):
        g[sum(j)].append(tuple(j))
    # produce list h of lists of tuples on m bits
    if k==m:
        h = g
    else:
        h = [[] for i in range(m+1)]
        for j in product((0,1), repeat=m):
            h[sum(j)].append(tuple(j))
    # Now deliver n-tuples in proper order
    for b in range(n+1):  # Deliver tuples with b bits set
        for lb in range(max(0, b-m), min(b+1,k+1)):
            for l in g[lb]:
                for r in h[b-lb]:
                    yield l+r

The timing results shown below are from a series of %timeit calls in ipython. Each call was of a form like
%timeit [x for x in cube1s.f(n)]
with the names cornersjc, cornerscc, cornersec, cornerses in place of f (standing for my code, Cam's code, Eevee's code, and my version of Eevee's method) and a number in place of n.

n    cornersjc    cornerscc    cornersec    cornerses

5      40.3 us      45.1 us      36.4 us      25.2 us    
6      51.3 us      85.2 us      77.6 us      46.9 us    
7      87.8 us      163 us       156 us       88.4 us    
8     132 us       349 us       327 us       178 us    
9     250 us       701 us       688 us       376 us    
10    437 us      1.43 ms      1.45 ms       783 us
11    873 us      3 ms         3.26 ms      1.63 ms
12   1.87 ms      6.66 ms      8.34 ms      4.9 ms

Code for cornersjc was given above. Code for cornerscc, cornersec, and cornerses is as follows. These produce the same output as cornersjc, except that Cam's code produces a list of lists instead of a list of tuples, and within each bit-count group produces in reverse.

def cornerscc(n):   # Re: Cam's code
    from itertools import combinations
    for number_of_ones in xrange(0, n + 1):
        for location_of_ones in combinations(xrange(0, n), number_of_ones):
            result = [0] * n
            for location in location_of_ones:
                result[location] = 1
            yield result

def cornersec (n):   # Re:  Eevee's code
    from itertools import product
    vertices = ((v.count(1), v)
                for v in product((0, 1), repeat=n))
    for count, vertex in sorted(vertices):
        yield vertex

def cornerses (n):   # jw mod. of Eevee's code
    from itertools import product
    for vertex in sorted(product((0, 1), repeat=n), key=sum):
        yield vertex

Note, the last three lines of cornersjc can be replaced by

            for v in product(g[lb], h[b-lb]):
                yield v[0]+v[1]

which is cleaner but slower. Note, if yield v is used instead of yield v[0]+v[1], the code runs faster than cornersjc but (at n=5) produces pair-of-tuple results like ((1, 0), (1, 1, 0)); when yield v[0]+v[1] is used, the code runs slower than cornersjc but produces identical results, a list of tuples like (1, 0, 1, 1, 0). An example timing follows, with cornersjp being the modified cornersjc.

In [93]: for n in range(5,13):
    %timeit [x for x in cube1s.cornersjp(n)]
   ....:     
10000 loops, best of 3: 49.3 us per loop
10000 loops, best of 3: 64.9 us per loop
10000 loops, best of 3: 117 us per loop
10000 loops, best of 3: 178 us per loop
1000 loops, best of 3: 351 us per loop
1000 loops, best of 3: 606 us per loop
1000 loops, best of 3: 1.28 ms per loop
100 loops, best of 3: 2.74 ms per loop