对于第一个问题,您应该做的第一件事是使用第二个字段对列表进行排序项目获取者 https://docs.python.org/3/library/operator.html#operator.itemgetter从操作员模块:
x = [
['4', '21', '1', '14', '2008-10-24 15:42:58'],
['3', '22', '4', '2somename', '2008-10-24 15:22:03'],
['5', '21', '3', '19', '2008-10-24 15:45:45'],
['6', '21', '1', '1somename', '2008-10-24 15:45:49'],
['7', '22', '3', '2somename', '2008-10-24 15:45:51']
]
from operator import itemgetter
x.sort(key=itemgetter(1))
然后你可以使用itertools'groupby https://docs.python.org/3/library/itertools.html#itertools.groupby功能:
from itertools import groupby
y = groupby(x, itemgetter(1))
现在 y 是一个包含 (element, item iterator) 元组的迭代器。解释这些元组比显示代码更令人困惑:
for elt, items in groupby(x, itemgetter(1)):
print(elt, items)
for i in items:
print(i)
哪个打印:
21 <itertools._grouper object at 0x511a0>
['4', '21', '1', '14', '2008-10-24 15:42:58']
['5', '21', '3', '19', '2008-10-24 15:45:45']
['6', '21', '1', '1somename', '2008-10-24 15:45:49']
22 <itertools._grouper object at 0x51170>
['3', '22', '4', '2somename', '2008-10-24 15:22:03']
['7', '22', '3', '2somename', '2008-10-24 15:45:51']
对于第二部分,您应该使用此处已经提到的列表理解:
from pprint import pprint as pp
pp([y for y in x if y[3] == '2somename'])
哪个打印:
[['3', '22', '4', '2somename', '2008-10-24 15:22:03'],
['7', '22', '3', '2somename', '2008-10-24 15:45:51']]