擅长:python、mysql、java
<p>您可以使用来自<code>itertools</code>的<a href="https://docs.python.org/3/library/itertools.html#itertools.repeat" rel="nofollow noreferrer">^{<cd1>}</a>并将其与您的列表压缩。你知道吗</p>
<pre><code>>>> from itertools import repeat
>>> mb = [3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61]
>>> zip(mb, repeat(1))
[(3, 1), (5, 1), (7, 1), (11, 1), (13, 1), (17, 1), (19, 1), (23, 1), (29, 1), (31, 1), (37, 1), (41, 1), (43, 1), (47, 1), (53, 1), (59, 1), (61, 1)]
</code></pre>
<p>或者你可以使用这样的列表:</p>
<pre><code>>>> [(x, 1) for x in mb]
[(3, 1), (5, 1), (7, 1), (11, 1), (13, 1), (17, 1), (19, 1), (23, 1), (29, 1), (31, 1), (37, 1), (41, 1), (43, 1), (47, 1), (53, 1), (59, 1), (61, 1)]
</code></pre>
<hr/>
<p><strong>到您的解决方案:</strong>在解决方案中,您在第一次循环迭代后返回结果。所以它还没有正确的值。试着把回路移到你的回路之外。你知道吗</p>