基于元素名称在python中对长列表的元素进行分组

Question

我有一个像下面这样的长列表，我分别根据 (element[0], element[3], element[2]) 对它们进行了排序。对于元素，我的意思是下划线之间的数字。list3=['20180406_145813_4_1.jpg', '20180406_145813_5_1.jpg', '20180406_145813_6_1.jpg', '20180406_175827_10_12.jpg', '20180406_175827_11_12.jpg', '20180409_190651_7_2.jpg', '20180409_190651_8_2.jpg',...]现在，我想根据 element[3] 拆分列表。我想要的结果是：[['20180406_145813_4_1.jpg', '20180406_145813_5_1.jpg', '20180406_145813_6_1.jpg'], ['20180406_175827_10_12.jpg', '20180406_175827_11_12.jpg'], ['20180409_190651_7_2.jpg', '20180409_190651_8_2.jpg'],...]我有这段代码将每个名字打印为一个列表。我不知道如何在这段代码中按 element3 分组：for imagename in list3:    element3 = imagename.split("_")[3]    for j,m in groupby(list3):        print(list(m))

Answer 1

您可以像这样使用itertools.groupby：

from itertools import groupby

list3 = ['20180406_145813_4_1.jpg',

         '20180406_145813_5_1.jpg',

         '20180406_145813_6_1.jpg',

         '20180406_175827_10_12.jpg',

         '20180406_175827_11_12.jpg',

         '20180409_190651_7_2.jpg',

         '20180409_190651_8_2.jpg']

result = [list(group) for _, group in groupby(list3, key=lambda x: x.split('_')[3])]

print(result)

输出

[['20180406_145813_4_1.jpg', '20180406_145813_5_1.jpg', '20180406_145813_6_1.jpg'], ['20180406_175827_10_12.jpg', '20180406_175827_11_12.jpg'], ['20180409_190651_7_2.jpg', '20180409_190651_8_2.jpg']]

上面的列表推导等价于以下for循环：

result = []

for _, group in groupby(list3, key=lambda x: x.split('_')[3]):

    result.append(list(group))

Answer 2

试试这个（不导入任何东西）：

list3=['20180406_145813_4_1.jpg',

       '20180406_145813_5_1.jpg',

       '20180406_145813_6_1.jpg',

       '20180406_175827_10_12.jpg',

       '20180406_175827_11_12.jpg',

       '20180409_190651_7_2.jpg',

       '20180409_190651_8_2.jpg',

       ...]

res = []

for first, second, third in zip(*[iter(list3)]*3):

    res.append([first, second, third])

只需将一个列表附加first, seconds, third到res列表中

print(res)

[['20180406_145813_4_1.jpg', '20180406_145813_5_1.jpg', '20180406_145813_6_1.jpg'], 

 ['20180406_175827_10_12.jpg', ...]]