Python将列表元素按值分组为元组

Question

我有这个清单：

mylist = ['*', 'Drama', 'Film Noir', 'Mystery', 'Romance', '*', 'Dance', 'Drama', 'Musical', '*', 'Crime', 'Drama', '*', 'Action', 'Drama', 'Period', 'Western', '*', 'Adaptation', 'Based-on', 'Comedy', 'Romance', '*', 'Mystery', 'Thriller', '*', 'Comedy']

显然，元素“*”在列表中分隔组。如何将出现在“*”之间的元素分组？预期结果如下所示：

expected_list = [('Drama', 'Film Noir', 'Mystery', 'Romance'), ('Dance', 'Drama', 'Musical'), ('Crime', 'Drama'), ('Action', 'Drama', 'Period', 'Western'), ('Adaptation', 'Based-on', 'Comedy', 'Romance'), ('Mystery', 'Thriller'), ('Comedy')]

Answer 1

使用^{}根据条目是否为*对条目进行分组，然后使用__ne__（不相等）方法或等效的lambda表达式保留那些不是*的部分：

>>> import itertools
>>> mylist = ['*', 'Drama', 'Film Noir', 'Mystery', 'Romance', '*', 'Dance', 'Drama', 'Musical', '*', 'Crime', 'Drama', '*', 'Action', 'Drama', 'Period', 'Western', '*', 'Adaptation', 'Based-on', 'Comedy', 'Romance', '*', 'Mystery', 'Thriller', '*', 'Comedy']
>>> [tuple(g) for k, g in itertools.groupby(mylist, key="*".__ne__) if k]
[('Drama', 'Film Noir', 'Mystery', 'Romance'),
 ('Dance', 'Drama', 'Musical'),
 ('Crime', 'Drama'),
 ('Action', 'Drama', 'Period', 'Western'),
 ('Adaptation', 'Based-on', 'Comedy', 'Romance'),
 ('Mystery', 'Thriller'),
 ('Comedy',)]

Answer 2

这是一个经典的use case for ^{}：

from itertools import groupby

expected_list = [tuple(grp) for isstar, grp in groupby(mylist, key='*'.__eq__) if not isstar]

将您的输入分组为*和非*运行，将*组和tuple组放到非*组