擅长:python、mysql、java
<p>假设列表按第一个元素排序,您可以使用<a href="https://docs.python.org/2/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">^{<cd1>}</a>:</p>
<pre><code>from itertools import groupby
from operator import itemgetter
[[i[1:] for i in v] for k,v in groupby(first_list, itemgetter(0))]
#[[[45.4, 9.1]],
# [[45.5, 9.1], [45.4, 9.2], [45.4, 9.2]],
# [[45.4, 9.1], [45.4, 9.1], [45.4, 9.1]]]
</code></pre>