擅长:python、mysql、java
<p>您可以使用<a href="https://docs.python.org/3/library/functions.html#zip" rel="nofollow noreferrer">^{<cd1>}</a>、<a href="https://docs.python.org/3/library/itertools.html#itertools.cycle" rel="nofollow noreferrer">^{<cd2>}</a>和嵌套的<a href="https://docs.python.org/3/tutorial/datastructures.html#list-comprehensions" rel="nofollow noreferrer">comprehension</a>la:</p>
<pre><code>from itertools import cycle
list_a = ['pineapple', 'banana', 'mango']
list_b = ['tropical']
[x for tpl in zip(list_a, cycle(list_b)) for x in tpl]
# ['pineapple', 'tropical', 'banana', 'tropical', 'mango', 'tropical']
</code></pre>
<p>您还可以使用<a href="https://docs.python.org/3/library/itertools.html#itertools.chain" rel="nofollow noreferrer">^{<cd3>}</a>来展平这些对,如@Matthias所建议的:</p>
<pre><code>from itertools import cycle, chain
[*chain(*zip(list_a, cycle(list_b)))] # Python >= 3.5
# or more verbose
list(chain.from_iterable(zip(list_a, cycle(list_b))))
</code></pre>