如何实现受限组合迭代器？

Question

I want something equivalent to the following code. The following code generate poker possible hand patterns.

from itertools import combinations
m = 13
n = 4
x = range(m) * n
y = 5
pattern = set()
for i in combinations(x, y):
    pattern.add(tuple(sorted(i)))

I tried to use itertools.combinations_with_replacement . Not surprisingly there are combinations like (0, 0, 0, 0, 0) or (1, 1, 1, 1, 1). But there are no 5 Queens and 5 Kings in a cards. So I don't want to take 5 same things. How do I implement restricted combinations iterator.

from itertools import combinations_with_replacement
m = 13
n = 4
x = range(m)
y = 5
pattern = []
for i in combinations_with_replacement(x, y):
    pattern.append(i)

I want like the following code.

Pseudo code:

m = 13
n = 4
x = range(m)
y = 5
pattern = []
for i in combinations_with_replacement_restricted(x, y, max=n):
    pattern.append(i)

PS Because I'm English learner, please modify my grammar mistakes.

Answer 1

After reading the docs: https://docs.python.org/2/library/itertools.html?highlight=combinations#itertools.combinations_with_replacement

I found that a built-in solution for your target don't exists.

So if you want to make this, I think the two solution may be suitable:

[1]. Make all your cards distinct:

from itertools import combinations
m = 13
n = 4
x = range(m * n)
y = 5
pattern = set()
# combinations from range(0, 52)
for i in combinations(x, y):
    # so p/n maps to range(0, 13)
    pattern.add((p / n for p in sorted(i)))

[2]. Exclude all invalid results:

from itertools import combinations
m = 13
n = 4
x = range(m) * n
y = 5
pattern = set()
for i in combinations(x, y):
    if min(i) == max(i):
        continue
    pattern.add(tuple(sorted(i)))

Answer 2

You could do this :

import itertools
# each card is a pair (value,  color)
cards = [ (value, color) for value in xrange(13) for color in xrange(4) ] 
# all 5 cards combinations within 52 cards
for hand in itertools.combinations( cards, 5 ) :
  print hand