Use zfill
根据需要添加这些零:
hours = [time.strptime(i[:-1].zfill(4), "%H%M") for i in a]
通过使用i[:-1]
我们删除那个讨厌的尾随点,并且.zfill(4)
将添加足够的0
向左移动 4 个字符,使其成为 4 位数字。
Demo:
>>> import time
>>> a = ['800.', '830.', '900.', '30.']
>>> [time.strptime(i[:-1].zfill(4), "%H%M") for i in a]
[time.struct_time(tm_year=1900, tm_mon=1, tm_mday=1, tm_hour=8, tm_min=0, tm_sec=0, tm_wday=0, tm_yday=1, tm_isdst=-1), time.struct_time(tm_year=1900, tm_mon=1, tm_mday=1, tm_hour=8, tm_min=30, tm_sec=0, tm_wday=0, tm_yday=1, tm_isdst=-1), time.struct_time(tm_year=1900, tm_mon=1, tm_mday=1, tm_hour=9, tm_min=0, tm_sec=0, tm_wday=0, tm_yday=1, tm_isdst=-1), time.struct_time(tm_year=1900, tm_mon=1, tm_mday=1, tm_hour=0, tm_min=30, tm_sec=0, tm_wday=0, tm_yday=1, tm_isdst=-1)]
如果它们是浮点值,请使用format()功能对它们进行零填充值:
>>> format(800., '04.0f')
'0800'
所以这样做:
hours = [time.strptime(format(i % 2400, '04.0f'), "%H%M") for i in a]
where % 2400
将您的值标准化为 0. 到 2399. 范围。