Question

我想以时间格式转换小数时间列表（HH，HHH）HH：MM：SS

16.       ,   16.00000381,  16.00000572,  16.00000954,
16.00001144,  16.00001335,  16.00001717,  16.00001907,
16.00002098,  16.0000248 ,  16.00002861,  16.00003052,
16.00003433,  16.00003624,  16.00003815,  16.00004196,
16.00004387,  16.00004768,  16.00004959,  16.00005341

有没有办法在python中执行此操作？

由于

Answer 1

假设值代表小时：

In [86]: import datetime as DT
In [87]: data = [16.       ,   16.00000381,  16.00000572,  16.00000954,
16.00001144,  16.00001335,  16.00001717,  16.00001907,
16.00002098,  16.0000248 ,  16.00002861,  16.00003052,
16.00003433,  16.00003624,  16.00003815,  16.00004196,
16.00004387,  16.00004768,  16.00004959,  16.00005341]
   ....:    ....:    ....:    ....: 
In [88]: map(str, [DT.timedelta(seconds=x*60*60.0) for x in data])
Out[88]: 
['16:00:00',
 '16:00:00.013716',
 '16:00:00.020592',
 '16:00:00.034344',...
 '16:00:00.157932',
 '16:00:00.171648',
 '16:00:00.178524',
 '16:00:00.192276']

Answer 2

一次性：

import datetime
x = 21.234768653
hour = int(x)
minute = int((x - int(x))*60.0)
second = int(((x - int(x))*60 - int((x - int(x))*60.0))*60.0)
(datetime.time(hour, minute, second)).strftime('%H:%M:%S')

有关时间列表：

times = [(datetime.time(int(x), int((x - int(x))*60.0), int(((x - int(x)) * 60 - int((x - int(x))*60.0))*60.0))).strftime('%H:%M:%S') for x in time_list]

Python：十进制为时间格式

2 个答案: