使用此日期格式反向排序列表的最佳方法是什么？

Question

我有一个包含数百万条记录的数据库,如下所示:

[("Something", "10/08/2017", "something else", "something more", "etc"),
("Something", "16/08/2017", "something else", "something more", "etc"),
("Something", "14/07/2017", "something else", "something more", "etc"),
("Something", "14/08/2017", "something else", "something more", "etc"),
("Something", "15/07/2017", "something else", "something more", "etc"),
("Something", "14/08/2017", "something else", "something more", "etc")]

这是DD/MM/YYYY我需要它按第二个值反转排序.

第一次尝试:
我试过这个,但它只按白天而不是按月分类.

def sort_by(my_list, index):
    my_list.sort(key=lambda x: x[index], reverse=True)
    return my_list

结果的图片(忽略颜色,它来自我的GUI):https://image.prntscr.com/image/ZeIgzxNaQ2OD4VX0Yztnhw.png
正如你所看到的那样,它只在白天进行排序,而不是按月进行排序,所以它保持不变每当一天变化时,从一个月跳到另一个月.那是一团糟.

第二次尝试:

def sort_by_date(my_list, index):
    for record, data in enumerate(my_list):
        split = data[index].split("/")
        altered_record = my_list[record]
        altered_record[index] = split[2]+"-"+split[1]+"-"+split[0]
    my_list.sort(key=lambda x: x[index], reverse=True)
    return my_list

到那时我意识到它不起作用,因为元组不允许项目分配,所以我必须将整个数据转换为列表而不是元组列表.即使我这样做,我怀疑它是否有效,并且需要花费大量时间来完成这项工作.

所以我的问题是:
你如何按DD/MM/YYYY对元组列表进行排序？

根据来自@ juanpa.arrivillaga的评论,所需的输出是按日期排序的反向排序列表:

31/12/2017
30/12/2017
29/12/2017
28/12/2017
(And the next days of that month)
30/11/2017
29/11/2017
28/11/2017
27/11/2017
(And so go on)

编辑:如果可能,我想将日期保留为字符串,而不是更改日期类型.

Answer 1

您可以使用Python的datetime模块:

from pprint import pprint
from datetime import datetime

database = [("Something", "10/08/2017", "something else", "something more", "etc"),
            ("Something", "16/08/2017", "something else", "something more", "etc"),
            ("Something", "14/07/2017", "something else", "something more", "etc"),
            ("Something", "14/08/2017", "something else", "something more", "etc"),
            ("Something", "15/07/2017", "something else", "something more", "etc"),
            ("Something", "14/08/2017", "something else", "something more", "etc")]

pprint(sorted(database, key=lambda x: datetime.strptime(x[1], "%d/%m/%Y"), reverse=True))

哪个有输出:

[('Something', '16/08/2017', 'something else', 'something more', 'etc'),
 ('Something', '14/08/2017', 'something else', 'something more', 'etc'),
 ('Something', '14/08/2017', 'something else', 'something more', 'etc'),
 ('Something', '10/08/2017', 'something else', 'something more', 'etc'),
 ('Something', '15/07/2017', 'something else', 'something more', 'etc'),
 ('Something', '14/07/2017', 'something else', 'something more', 'etc')]