擅长:python、mysql、java
<p>您可以尝试<a href="https://docs.python.org/3/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">^{<cd1>}</a>:</p>
<pre><code>>>> from itertools import groupby
>>> x = [4, 4, 1, 6, 6, 6, 1, 1, 1, 1]
>>> new_list = [list(group) for _, group in groupby(x)]
>>> new_list
[[4, 4], [1], [6, 6, 6], [1, 1, 1, 1]]
>>>
</code></pre>
<p>另一种方法是:</p>
<pre><code>>>> master_list, new_list = [], []
>>> for elem in x:
... if not new_list:
... new_list.append(elem)
... elif elem == new_list[-1]:
... new_list.append(elem)
... else:
... master_list.append(new_list)
... new_list = [elem]
>>> master_list.append(new_list)
>>> master_list
[[4, 4], [1], [6, 6, 6], [1, 1, 1, 1]]
</code></pre>