在Python中对嵌套列表进行排序和分组

Question

我有以下数据结构(列表清单)

[
 ['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']
]

我希望能够

1. 使用函数对列表重新排序,以便我可以按列表中的每个项目进行分组.例如,我希望能够按第二列分组(这样所有21个都在一起)

2. 使用函数仅显示每个内部列表中的某些值.例如,我想将此列表缩小为仅包含'2somename'的第4个字段值

所以列表看起来像这样

[
     ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
     ['7', '22', '3', '2somename', '2008-10-24 15:45:51']
]

Answer 1

对于第一个问题,您应该做的第一件事是使用来自运算符模块的itemgetter按第二个字段对列表进行排序:

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的groupby函数:

from itertools import groupby
y = groupby(x, itemgetter(1))

现在y是一个包含元组(元素,项迭代器)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)

哪个印刷品:

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,您应该使用已在此处提到的列表推导:

from pprint import pprint as pp
pp([y for y in x if y[3] == '2somename'])

哪个印刷品:

[['3', '22', '4', '2somename', '2008-10-24 15:22:03'],
 ['7', '22', '3', '2somename', '2008-10-24 15:45:51']]

Answer 2

如果你把它分配给var"a"......

python 2.x:

#1:

a.sort(lambda x,y: cmp(x[1], y[1]))

#2:

filter(lambda x: x[3]=="2somename", a)

python 3:

#1:

a.sort(key=lambda x: x[1])