我需要按日期排序列表的python列表

Question

我需要按日期对列表进行排序:

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']

从这个网站,我得到了这个代码:

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

但是,我收到此错误消息:

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 - 需要按照3d元素对列表进行排序 - 从最旧到最新

Answer 1

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('-'对于您的第一个条目).你需要打出只是日期部分strptime():

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

该str.rsplit()上的空白,分裂一次,为了提高效率:

>>> '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']

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

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