在python中生成置换[closed]

def perm(m, n):
  if m == 1:  # base case
    return [(n,)]
  perms = []
  for s in range(1, n):  # combine possible start values: 1 through n-1 ...
    for p in perm(m-1, n-s):  # ... with all appropriate smaller perms 
      perms.append((s,) + p)
  return perms

>>> perm(1, 5)
[(5,)]
>>> perm(2, 5)
[(1, 4), (2, 3), (3, 2), (4, 1)]
>>> perm(3, 5)
[(1, 1, 3), (1, 2, 2), (1, 3, 1), (2, 1, 2), (2, 2, 1), (3, 1, 1)]

考虑一个列表 l = [1,2,3,4,5]

itertools.permutations

p = itertools.permutations(list(range(1,6)),2)

my_elems = [el for el in p if sum(el) == 5]

输出

[(1, 4), (2, 3), (3, 2), (4, 1)]

看看你给出的第二个例子,我想你想要的是 product ,不是 permutations

p = itertools.product(list(range(1,6)),repeat=3)

my_elems = [el for el in p if sum(el) == 5]

#[(1, 1, 3), (1, 2, 2), (1, 3, 1), (2, 1, 2), (2, 2, 1), (3, 1, 1)]

这对第一个案子也有效。