Question

我正在寻找一种获取列表列表并查找元素子集的最大值的优雅方法。用一个小例子更好地解释它。给出以下输入：

data = [['1','AAA','somestuff','1/5/2018'],
        ['1','AAA','differentstuff','1/5/2018'],
        ['1','AAA','evendifferent','1/10/2018'],
        ['2','BBB','foo','1/12/2018'],
        ['2','BBB','bar','1/20/2018']]

我想返回以下列表列表：

[['1','AAA','evendifferent','1/10/2018'], 
    ['2','BBB','bar','1/20/2018']]

输出按内部列表的索引1分组，最大值基于日期（内部列表中的最后一项）。

Answer 1

您需要从字符串日期中获取日期时间-否则'1/5/2018'会比较“大”然后是'1/10/2018'，因为'5' > '1'是字符串。

您可以这样实现：

data = [['1','AAA','somestuff','1/5/2018'],
        ['1','AAA','differentstuff','1/5/2018'] ,
        ['1','AAA','evendifferent','1/10/2018'] ,
        ['2','BBB','foo','1/12/2018'] ,
        ['2','BBB','bar','1/20/2018']]


# group by AAA, BBB etc. into lists 
from collections import defaultdict

dd = defaultdict(list)
for d in data:
    dd[d[1]].append(d)

import datetime

# iterate over groups and get the maximum value of each list
for k in dd:
    # for the datetime converted da
    print(max ( dd[k], key = lambda x: datetime.datetime.strptime(x[-1],"%d/%M/%Y")))

文档：

在列表清单中寻找最大值

1 个答案: