Question

我正在使用python 3，我正在尝试找到一种方法来获取列表的所有排列，同时执行一些约束。

例如，我有一个列表L=[1, 2, 3, 4, 5, 6, 7]

我想找到所有的排列。但是，我的约束是：

1应始终在2之前。
3应该在4之前，而5应该在5之前。
最后，6应该在7之前。

当然，我可以生成所有排列并忽略那些不遵循这些约束的排列，但是我想这不会有效。

Answer 1

此方法使用简单的过滤器过滤排列。

import itertools

groups = [(1,2),(3,4,5),(6,7)]
groupdxs = [i for i, group in enumerate(groups) for j in range(len(group))]
old_combo = []
for dx_combo in itertools.permutations(groupdxs):
    if dx_combo <= old_combo: # as simple filter
        continue
    old_combo = dx_combo
    iters = [iter(group) for group in groups]
    print [next(iters[i]) for i in dx_combo]

我们在这里做的是找到permutations of a multiset。（在这种情况下，多重集合是groupdxs。）这是一个paper，详细说明了O（1）算法。

Answer 2

def partial_permutations(*groups):
    groups = list(filter(None, groups)) # remove empties.
    # Since we iterate over 'groups' twice, we need to
    # make an explicit copy for 3.x for this approach to work.
    if not groups:
        yield []
        return
    for group in groups:
        for pp in partial_permutations(*(
             g[1:] if g == group else g
             for g in groups
        )):
            yield [group[0]] + pp

Answer 3

一种方法是采用其中一种交换算法，当您要将元素交换到最终位置时，请检查它是否按正确顺序排列。下面的代码就是这样。

但首先让我展示一下它的用法：

L = [1, 2, 3, 4, 5, 6, 7]
constraints = [[1, 2], [3, 4, 5], [6, 7]]

A = list(p[:] for p in constrained_permutations(L, constraints)) # copy the permutation if you want to keep it
print(len(A))
print(["".join(map(str, p)) for p in A[:50]])

和输出：

210
['1234567', '1234657', '1234675', '1236457', '1236475', '1236745', '1263457', '1263475', '1263745', '1267345', '1324567', '1324657', '1324675', '1326457', '1326475', '1326745', '1342567', '1342657', '1342675', '1345267', '1345627', '1345672', '1346527', '1346572', '1346257', '1346275', '1346725', '1346752', '1364527', '1364572', '1364257', '1364275', '1364725', '1364752', '1362457', '1362475', '1362745', '1367245', '1367425', '1367452', '1634527', '1634572', '1634257', '1634275', '1634725', '1634752', '1632457', '1632475', '1632745', '1637245']

但现在代码：

def _permute(L, nexts, numbers, begin, end):
    if end == begin + 1:
        yield L
    else:
        for i in range(begin, end):
            c = L[i]
            if nexts[c][0] == numbers[c]:
                nexts[c][0] += 1
                L[begin], L[i] = L[i], L[begin]
                for p in _permute(L, nexts, numbers, begin + 1, end):
                    yield p
                L[begin], L[i] = L[i], L[begin]
                nexts[c][0] -= 1


def constrained_permutations(L, constraints):
    # warning: assumes that L has unique, hashable elements
    # constraints is a list of constraints, where each constraint is a list of elements which should appear in the permatation in that order
    # warning: constraints may not overlap!
    nexts = dict((a, [0]) for a in L)
    numbers = dict.fromkeys(L, 0) # number of each element in its constraint
    for constraint in constraints:
        for i, pos in enumerate(constraint):
            nexts[pos] = nexts[constraint[0]]
            numbers[pos] = i

    for p in _permute(L, nexts, numbers, 0, len(L)):
        yield p

具有约束的Python排列

3 个答案: