擅长:python、mysql、java
<p>我会这样想:</p>
<pre><code>def find_element(nested_lst, what):
for idx, sublst in enumerate(nested_lst):
try:
idx2 = sublst.index(what)
return (idx, idx2)
except ValueError:
pass
</code></pre>
<p>应该有用。你知道吗</p>
<p>示例:</p>
<pre><code>>>> def find_element(nested_lst, what):
... for idx, sublst in enumerate(nested_lst):
... try:
... idx2 = sublst.index(what)
... return (idx, idx2)
... except ValueError:
... pass
...
>>> birds = ['duck', 'chicken', 'goose']
>>> cats = ['tiger', 'lion']
>>> humans = ['human']
>>> find_element([birds, cats, humans], 'human')
(2, 0)
>>> find_element([birds, cats, humans], 'gator') # returns None if not found.
>>> find_element([birds, cats, humans], 'tiger')
(1, 0)
</code></pre>
<p>值得注意的是,平均而言,<code>list.index</code>是一个O(N)操作,这意味着列表不是测试成员资格的最有效的数据结构。如果您的实际数据支持它,那么可以考虑改用<code>set</code>。你知道吗</p>