我有一个日期列表和一个时间列表。当小时属于不同的第二天时(因此,当一个时间小于前一个时间时),我如何创建一个以键为日期、以时间为值、以下一个日期为键的字典?这两个列表当前都是字符串。
我的日期列表:
['2019/07/27',
'2019/07/29',]我的次数列表:
['06:55:40',
'06:55:41',
'08:48:33',
'08:48:33',
'09:02:54',
'09:02:54',
'09:02:54',
'10:02:19',
'10:02:20',
'07:34:52',
'07:34:52',
'07:35:03',
'09:22:19',
'09:22:19',
'09:22:19',
'09:22:23',
'11:17:24',
'11:17:27',
'13:24:57',]预期输出:
{'2019/07/27': ['06:55:40',
'06:55:41',
'08:48:33',
'08:48:33',
'09:02:54',
'09:02:54',
'09:02:54',
'10:02:19',
'10:02:20',],
'2019/07/29': ['07:34:52',
'07:34:52',
'07:35:03',
'09:22:19',
'09:22:19',
'09:22:19',
'09:22:23',
'11:17:24',
'11:17:27',
'13:24:57',]}发布于 2019-08-10 00:40:53
我的解决方案如下。我把时间列表切开了。
i = 1
j = 0
length = len(times)
while i < length:
if times[i-1] > times[i] or i is length-1: # i is the index of the start time on the next day
dict[dates[j]] = times[:i]
times = times[i:]
j += 1
i = 1
length = len(times)
else:
i += 1https://stackoverflow.com/questions/57432663
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