我需要按日期对列表的python列表进行排序 [英] i need to sort a python list of lists by date

问题描述

我需要按日期对列表列表进行排序:

I need to sort a list of lists by date:

mylist=['ML-M01 Qual Report 07/31/13', 'ML36220010 Complete Qual Testing 07/24/13', 'ML36220045 Final FRB 07/13/13', 'ML9822D2600 Brief to PM 08/5/13']

从此站点，我获得了以下代码:

from this site, I got this code:

sorted(mylist, key=lambda x: datetime.datetime.strptime(x[2], '%m/%d/%y'))

但是，我收到此错误消息:

however, I get this error message:

sorted(mylist,key=lambda x: datetime.datetime.strptime(x[2], '%m/%d/%y'))
Traceback (most recent call last):
  File "<pyshell#25>", line 1, in <module>
    sorted(mylist,key=lambda x: datetime.datetime.strptime(x[2], '%m/%d/%y'))
  File "<pyshell#25>", line 1, in <lambda>
  sorted(mylist,key=lambda x: datetime.datetime.strptime(x[2], '%m/%d/%y'))
  File "C:\Python25\Lib\_strptime.py", line 330, in strptime
    (data_string, format))
ValueError: time data did not match format:  data=-  fmt=%m/%d/%y

需要帮助-需要按3d元素对列表进行排序-最旧到最新

need help - need to sort the list by the 3d element - oldest to newest

推荐答案

列表中的每个条目都是字符串，而不是列表，因此 x [2] 是单个字符('-'作为您的第一个条目).您需要将日期部分 just 拆分为 strptime():

Each entry in your list is a string, not a list, so x[2] is a single character ('-' for your first entry). You need to split out just the date portion to strptime():

sorted(mylist, key=lambda x: datetime.datetime.strptime(x.rsplit(None, 1)[-1], '%m/%d/%y'))

str.rsplit()在空格上一次分割，以提高效率:

The str.rsplit() splits on whitespace, once, for efficiency:

>>> 'ML-M01 Qual Report 07/31/13'.rsplit(None, 1)
['ML-M01 Qual Report', '07/31/13']
>>> 'ML-M01 Qual Report 07/31/13'.rsplit(None, 1)[-1]
'07/31/13'

结果:

>>> sorted(mylist, key=lambda x: datetime.datetime.strptime(x.rsplit(None, 1)[-1], '%m/%d/%y'))
['ML36220045 Final FRB 07/13/13', 'ML36220010 Complete Qual Testing 07/24/13', 'ML-M01 Qual Report 07/31/13', 'ML9822D2600 Brief to PM 08/5/13']

请注意， sorted()返回一个新的排序列表，原始列表不受影响.如果要对 mylist 进行排序，请调用 list.sort()方法:

Note that sorted() returns a new sorted list, the original list is not affected. If you want to sort mylist in place, call the list.sort() method on it:

mylist.sort(key=lambda x: datetime.datetime.strptime(x.rsplit(None, 1)[-1], '%m/%d/%y'))

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