擅长:python、mysql、java
<p>您可以使用列表理解来获得预期的排列:</p>
<pre><code>>>> [i for i in perms if i[0] not in {3,4}]
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1)]
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<p>注意,使用<code>set</code>容器检查会员资格。以及如果你有更多的过滤器编号,效率会更高!你知道吗</p>
<p>如果不希望元组以<code>3,1</code>和<code>4,1</code>开头,只需执行以下操作:</p>
<pre><code>>>> perms = it.permutations(my_list)
>>> [i for i in perms if i[:2] !=[4,1] and i[:2] !=[4,1]]
[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]
>>>
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