Question

相对python /编程新手在这里。我正在使用itertools permutations方法来缩小产生的结果，使用“if”语句来创建我需要的列表。

在下面的代码中，您将看到我想从排列中返回一个缩小的列表，其中逗号的索引与列表的逗号的索引匹配。但是，此代码仅在我手动将索引放入“if”语句时才有效。由于我不会提前知道我想要列出的列表中有多少逗号，我该如何为此编码呢？

from itertools import permutations

prp1 = ["UP", ",", "DOWN", "LEFT", ",", "RIGHT"]

# grab indices for commas in 'prp1' list
comma_indices = []
for index, p in enumerate(prp1):
    if p == ",":
        commas.append(index)

perms = permutations(prp1)
lst = []
for perm in perms:
    # Presently I have manually inserted the indices of commas,
    # but how to do this non-manual for unknown amount of commas here?
    if perm[1] == ',' and perm[4] == ',':
        lst.append(perm)

for l in lst:
    print l

我确实考虑过做这样的事情：

for perm in perms:
    for comma in commas:
        if...........

..但是我一次只能访问1个逗号。

任何帮助非常感谢，达伦

编辑：

需要置换的更多示例列表：

prp2 = ['down', ',', 'up', 'left', 'down']
prp3 = ['down', 'down', ',', 'up', ',', 'left', ',', 'right', ',', 'left']

Answer 1

您已经知道如何在没有硬编码的情况下找到逗号所在的索引：

comma_indices = []
for index, p in enumerate(prp1):
    if p == ",":
        commas.append(index)

因此，您可以对perm应用相同的方法，看看它是否匹配，例如

observed_comma_indices = [index for index, p in enumerate(perm) if p == ","]
if comma_indices == observed_comma_indices:
    lst.append(perm)

但也有其他方法。例如，您不能生成所有排列，而只保留您想要的排列，而只能在第一时间创建所需的排列：

from itertools import permutations

def permute_some(seq, indices_to_permute):
    for perm in permutations(indices_to_permute):
        newseq = seq[:]
        for i, p in zip(indices_to_permute, perm):
            newseq[i] = seq[p]
        yield newseq

给出了

>>> seq = ["A", "B", ",", "C", ","]
>>> perm_ii = [i for i,x in enumerate(seq) if x != ","]
>>> for p in permute_some(seq, perm_ii):
...     print p
...     
['A', 'B', ',', 'C', ',']
['A', 'C', ',', 'B', ',']
['B', 'A', ',', 'C', ',']
['B', 'C', ',', 'A', ',']
['C', 'A', ',', 'B', ',']
['C', 'B', ',', 'A', ',']

Python - 使用“if”语句缩小itertools排列结果。如何重复元素？

1 个答案: