128

我正在编写一个需要将字符串解析为timedelta. 用户必须输入类似"32m"or "2h32m",甚至"4:13"or "5hr34m56s"... 是否有库或已经实现了这种东西的东西?

4

11 回答 11

128

对我来说,最优雅的解决方案是使用datetime强大的字符串解析方法,而无需求助于dateutil等外部库或手动解析输入。strptime

from datetime import datetime, timedelta
# we specify the input and the format...
t = datetime.strptime("05:20:25","%H:%M:%S")
# ...and use datetime's hour, min and sec properties to build a timedelta
delta = timedelta(hours=t.hour, minutes=t.minute, seconds=t.second)

在此之后,您可以像往常一样使用您的 timedelta 对象,将其转换为秒以确保我们做了正确的事情等。

print(delta)
assert(5*60*60+20*60+25 == delta.total_seconds())
于 2012-09-10T13:20:58.973 回答
96

昨天我有一点时间,所以我将@virhilo答案开发成一个Python 模块,添加了更多时间表达式格式,包括@priestc要求的所有格式。

任何想要它的人都可以在 github(MIT 许可证)上找到源代码。它也在 PyPI 上:

pip install pytimeparse

以秒数返回时间:

>>> from pytimeparse.timeparse import timeparse
>>> timeparse('32m')
1920
>>> timeparse('2h32m')
9120
>>> timeparse('4:13')
253
>>> timeparse('5hr34m56s')
20096
>>> timeparse('1.2 minutes')
72
于 2014-02-01T12:26:20.367 回答
90

对于第一种格式(5hr34m56s),您应该使用正则表达式进行解析

这是基于重新的解决方案:

import re
from datetime import timedelta


regex = re.compile(r'((?P<hours>\d+?)hr)?((?P<minutes>\d+?)m)?((?P<seconds>\d+?)s)?')


def parse_time(time_str):
    parts = regex.match(time_str)
    if not parts:
        return
    parts = parts.groupdict()
    time_params = {}
    for name, param in parts.items():
        if param:
            time_params[name] = int(param)
    return timedelta(**time_params)


>>> from parse_time import parse_time
>>> parse_time('12hr')
datetime.timedelta(0, 43200)
>>> parse_time('12hr5m10s')
datetime.timedelta(0, 43510)
>>> parse_time('12hr10s')
datetime.timedelta(0, 43210)
>>> parse_time('10s')
datetime.timedelta(0, 10)
>>> 
于 2011-01-07T17:06:25.463 回答
15

我只想输入一个时间,然后将其添加到不同的日期,所以这对我有用:

from datetime import datetime as dtt

time_only = dtt.strptime('15:30', "%H:%M") - dtt.strptime("00:00", "%H:%M")
于 2017-03-01T22:27:47.893 回答
12

我通过一些升级修改了 virhilo 的好答案:

  • 添加了字符串是有效时间字符串的断言
  • 用“h”替换“hr”小时指示器
  • 允许使用 "d" - 天数指标
  • 允许非整数时间(例如3m0.25s3 分 0.25 秒)

.

import re
from datetime import timedelta


regex = re.compile(r'^((?P<days>[\.\d]+?)d)?((?P<hours>[\.\d]+?)h)?((?P<minutes>[\.\d]+?)m)?((?P<seconds>[\.\d]+?)s)?$')


def parse_time(time_str):
    """
    Parse a time string e.g. (2h13m) into a timedelta object.

    Modified from virhilo's answer at https://stackoverflow.com/a/4628148/851699

    :param time_str: A string identifying a duration.  (eg. 2h13m)
    :return datetime.timedelta: A datetime.timedelta object
    """
    parts = regex.match(time_str)
    assert parts is not None, "Could not parse any time information from '{}'.  Examples of valid strings: '8h', '2d8h5m20s', '2m4s'".format(time_str)
    time_params = {name: float(param) for name, param in parts.groupdict().items() if param}
    return timedelta(**time_params)
于 2018-08-19T10:55:29.907 回答
7

Django 带有实用功能parse_duration()。从文档中:

解析一个字符串并返回一个datetime.timedelta.

"DD HH:MM:SS.uuuuuu"期望采用 ISO 8601 或 PostgreSQL 的日时间间隔格式(例如 )指定的格式或P4DT1H15M20S指定4 1:15:20的数据3 days 04:05:06

于 2019-01-03T00:28:48.540 回答
5

如果你想使用 : 作为分隔符,我使用这个函数:

import re
from datetime import timedelta

def timedelta_parse(value):
    """
    convert input string to timedelta
    """
    value = re.sub(r"[^0-9:]", "", value)
    if not value:
        return

    return timedelta(**{key:float(val)
                        for val, key in zip(value.split(":")[::-1], 
                                            ("seconds", "minutes", "hours", "days"))
               })

例子:

In [4]: timedelta_pars("1:0:0:1")
Out[4]: datetime.timedelta(days=1, seconds=1)

In [5]: timedelta_pars("123.5")
Out[5]: datetime.timedelta(seconds=123, microseconds=500000)

In [6]: timedelta_pars("1:6:34:9.983")
Out[6]: datetime.timedelta(days=1, seconds=23649, microseconds=983000)

In [8]: timedelta_pars("23:45:00")
Out[8]: datetime.timedelta(seconds=85500)
于 2021-03-26T00:40:14.677 回答
3

如果 Pandas 已经在您的依赖项中,它会做得很好:

>>> import pandas as pd
>>> pd.Timedelta('5hr34m56s')
Timedelta('0 days 05:34:56')

>>> pd.Timedelta('2h32m')
Timedelta('0 days 02:32:00')

>>> pd.Timedelta('5hr34m56s')
Timedelta('0 days 05:34:56')

>>> # It is pretty forgiving:
>>> pd.Timedelta('2 days 24:30:00 10 sec')
Timedelta('3 days 00:30:10')

datetime.timedelta如果您喜欢该类型,请转换为:

>>> pd.Timedelta('1 days').to_pytimedelta()
datetime.timedelta(1)

不幸的是,这不起作用:

>>> pd.Timedelta('4:13')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "pandas\_libs\tslibs\timedeltas.pyx", line 1217, in 
pandas._libs.tslibs.timedeltas.Timedelta.__new__
  File "pandas\_libs\tslibs\timedeltas.pyx", line 454, in 
pandas._libs.tslibs.timedeltas.parse_timedelta_string
ValueError: expected hh:mm:ss format

Pandas 实际上有相当广泛的日期和时间工具,尽管这不是它的主要用途。

要安装熊猫:

# If you use pip
pip install pandas

# If you use conda
conda install pandas
于 2021-08-06T19:24:40.817 回答
3

使用isodate库解析 ISO 8601 持续时间字符串。例如:

isodate.parse_duration('PT1H5M26S')

另请参阅是否有一种简单的方法可以将 ISO 8601 持续时间转换为 timedelta?

于 2019-12-17T20:06:26.677 回答
2

如果您使用 Python 3,那么这里是 Hari Shankar 解决方案的更新版本,我使用的是:

from datetime import timedelta
import re

regex = re.compile(r'(?P<hours>\d+?)/'
                   r'(?P<minutes>\d+?)/'
                   r'(?P<seconds>\d+?)$')

def parse_time(time_str):
    parts = regex.match(time_str)
    if not parts:
        return
    parts = parts.groupdict()
    print(parts)
    time_params = {}
    for name, param in parts.items():
        if param:
            time_params[name] = int(param)
    return timedelta(**time_params)
于 2016-09-01T10:57:02.200 回答
0

考虑尝试tempora.parse_timedelta

$ pip-run 'tempora>=4.1.1'
Collecting tempora>=4.1.1
  Downloading tempora-4.1.1-py3-none-any.whl (15 kB)
Collecting jaraco.functools>=1.20
  Using cached jaraco.functools-3.3.0-py3-none-any.whl (6.8 kB)
Collecting pytz
  Using cached pytz-2021.1-py2.py3-none-any.whl (510 kB)
Collecting more-itertools
  Using cached more_itertools-8.8.0-py3-none-any.whl (48 kB)
Installing collected packages: more-itertools, pytz, jaraco.functools, tempora
Successfully installed jaraco.functools-3.3.0 more-itertools-8.8.0 pytz-2021.1 tempora-4.1.1
Python 3.9.2 (v3.9.2:1a79785e3e, Feb 19 2021, 09:06:10) 
[Clang 6.0 (clang-600.0.57)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from tempora import parse_timedelta
>>> parse_timedelta("32m")
datetime.timedelta(seconds=1920)
>>> parse_timedelta("2h32m")
datetime.timedelta(seconds=9120)
>>> parse_timedelta("4:13")
datetime.timedelta(seconds=15180)
>>> parse_timedelta("5hr34m56s")
datetime.timedelta(seconds=20096)
于 2021-06-20T22:58:31.757 回答