在 Python 中通过 strptime 解析没有前导零的小时数 [英] Parse hours without leading zeroes by strptime in Python

问题描述

假设您有这种格式的时间:

a = [..., 800.0, 830.0, 900.0, 930.0, 1000.0, 1030.0, ...]

问题是缺少几个小时的前导零.例如00:30用30表示，08:00用800.和表示00:00 表示为 2400.是否可以使用 strptime 方法将此数据解析为 time 对象?我尝试使用以下代码

hours = [time.strptime(str(int(i)), "%H%M") for i in a]

但是得到了

ValueError:未转换的数据仍然存在:0

附言我使用的是 Python 2.7.

解决方案

使用 zfill 根据需要重新添加那些零:

hours = [time.strptime(i[:-1].zfill(4), "%H%M") for i in a]

通过使用 i[:-1] 我们删除了那个讨厌的尾随点，并且 .zfill(4) 将添加足够的 0 字符向左移动到 4 位数字.

演示:

<预><代码>>>>导入时间>>>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=1(tm0_year), 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_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_min=0,=3, tm_sec=0, tm_wday=0, tm_yday=1, tm_isdst=-1)]

如果它们是浮点值，请使用 format() 函数 为您提供零填充值:

<预><代码>>>>格式(800.，'04.0f')'0800'

这样做:

hours = [time.strptime(format(i % 2400, '04.0f'), "%H%M") for i in a]

其中 % 2400 将您的值标准化为 0. 到 2399. 范围.

Suppose you have time in this format:

a = [..., 800.0, 830.0, 900.0, 930.0, 1000.0, 1030.0, ...]

The problem is that leading zeroes for hours are missing. For example 00:30 is represented by 30, 08:00 is represented by 800. and 00:00 is represented by 2400. Is it possible to parse this data to time object using strptime method? I tried using following code

hours = [time.strptime(str(int(i)), "%H%M") for i in a]

but got

ValueError: unconverted data remains: 0

P.S. I'm using Python 2.7.

解决方案

Use zfill to add those zeros back as needed:

hours = [time.strptime(i[:-1].zfill(4), "%H%M") for i in a]

By using i[:-1] we remove that pesky trailing dot, and .zfill(4) will add enough 0 characters to the left to make it to 4 digits.

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)]

If they are float values instead, use the format() function on them to give you zero-padded values:

>>> format(800., '04.0f')
'0800'

So do this:

hours = [time.strptime(format(i % 2400, '04.0f'), "%H%M") for i in a]

where % 2400 normalizes your values to the 0. to 2399. range.

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