Python 如何从一个简单的字符串构造一个 timedelta 对象

How to construct a timedelta object from a simple string

提问by priestc

I'm writing a function that needs a timedelta input to be passed in as a string. The user must enter something like "32m" or "2h32m", or even "4:13" or "5hr34m56s"... Is there a library or something that has this sort of thing already implemented?

我正在编写一个需要将 timedelta 输入作为字符串传入的函数。用户必须输入诸如“32m”或“2h32m”，甚至“4:13”或“5hr34m56s”之类的东西……是否有图书馆或已经实现了此类东西的东西？

采纳答案by virhilo

For the first format(5hr34m56s), you should parse using regular expressions

对于第一种格式（5hr34m56s），您应该使用正则表达式进行解析

Here is re-based solution:

这是重新基于的解决方案：

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.iteritems():
        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)
>>> 

回答by metakermit

To me the most elegant solution, without having to resort to external libraries such as dateutilor manually parsing the input, is to use datetime'spowerful strptimestring parsing method.

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，而不必求助于诸如dateutil 之类的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)

After this you can use your timedelta object as normally, convert it to seconds to make sure we did the correct thing etc.

在此之后，您可以照常使用 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())

回答by wildwilhelm

I had a bit of time on my hands yesterday, so I developed @virhilo's answerinto a Python module, adding a few more time expression formats, including all those requested by @priestc.

昨天我有一些时间，所以我将@virhilo的答案开发到 Python 模块中，添加了更多时间表达式格式，包括@priestc请求的所有格式。

Source code is on github(MIT License) for anybody that wants it. It's also on PyPI:

源代码在 github（MIT 许可证）上，供任何需要的人使用。它也在 PyPI 上：

pip install pytimeparse

Returns the time as a number of seconds:

以秒数形式返回时间：

>>> from pytimeparse.timeparse import timeparse
>>> timeparse('32m')
1920
>>> timeparse('2h32m')
9120
>>> timeparse('4:13')
253
>>> timeparse('5hr34m56s')
20096
>>> timeparse('1.2 minutes')
72

回答by Alexey Kislitsin

If you use Python 3 then here's updated version for Hari Shankar's solution, which I used:

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

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)

回答by kztd

I wanted to input just a time and then add it to various dates so this worked for me:

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")

回答by Peter

I've modified virhilo's nice answerwith a few upgrades:

我通过一些升级修改了virhilo 的好答案：

• added a assertion that the string is a valid time string
• replace the "hr" hour-indicator with "h"
• allow for a "d" - days indicator
• allow non-integer times (e.g. 3m0.25sis 3 minutes, 0.25 seconds)

• 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)

回答by Don Kirkby

Django comes with the utility function parse_duration(). From the documentation:

Django 自带实用功能parse_duration()。从文档：

Parses a string and returns a datetime.timedelta.

Expects data in the format "DD HH:MM:SS.uuuuuu"or as specified by ISO 8601 (e.g. P4DT1H15M20Swhich is equivalent to 4 1:15:20) or PostgreSQL's day-time interval format (e.g. 3 days 04:05:06).

解析一个字符串并返回一个datetime.timedelta.

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

回答by jqmichael

Use isodatelibrary to parse ISO 8601 duration string. For example:

使用isodate库解析 ISO 8601 持续时间字符串。例如：

isodate.parse_duration('PT1H5M26S')

Also see Is there an easy way to convert ISO 8601 duration to timedelta?

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？