Python argparse：从参数创建 timedelta 对象？

问题描述 投票：0回答：6

我正在尝试使用 

argparse
将参数转换为 
timedelta
对象。我的程序读取用户提供的字符串并将它们转换为各种 
datetime
对象以供以后使用。但我无法正确处理 
filter_length
参数。我的代码：

import datetime
import time
import argparse

def mkdate(datestring):
    return datetime.datetime.strptime(datestring, '%Y-%m-%d').date()

def mktime(timestring):
    return datetime.datetime.strptime(timestring, '%I:%M%p').time()

def mkdelta(deltatuple):
    return datetime.timedelta(deltatuple)

parser = argparse.ArgumentParser()
parser.add_argument('start_date', type=mkdate, nargs=1)
parser.add_argument('start_time', type=mktime, nargs=1, )
parser.add_argument('filter_length', type=mkdelta, nargs=1, default=datetime.timedelta(1))#default filter length is 1 day.

我运行程序，传递 

1
作为 
timedelta
值（我只希望它是一天）：

> python program.py 2012-09-16 11:00am 1

但是我收到以下错误：

>>> program.py: error: argument filter_length: invalid mkdelta value: '1'

我不明白为什么该值无效。如果我单独调用 mkdelta 函数，如下所示：

mkdelta(1)
print mkdelta(1)

返回：

datetime.timedelta(1)
1 day, 0:00:00

这正是我正在寻找的价值。有人可以帮我弄清楚如何使用 

argparse
正确进行此转换吗？

python argparse timedelta
6个回答
7
投票

注意到错误消息中 

'1'
周围的引号了吗？您将一个字符串传递给 mkdelta，而在测试代码中，您传递一个整数。

6
投票

你的函数不处理字符串参数，这是 argparse 传递给它的；拨打

int()
：

def mkdelta(deltatuple):
    return datetime.timedelta(int(deltatuple))

如果您需要支持超过天数，则必须找到一种方法来解析传入的参数到 timedelta 参数。

例如，您可以支持 

d
、
h
、
m
或 
s
后缀来表示天、小时、分钟或秒：

_units = dict(d=60*60*24, h=60*60, m=60, s=1)
def mkdelta(deltavalue):
    seconds = 0
    defaultunit = unit = _units['d']  # default to days
    value = ''
    for ch in list(str(deltavalue).strip()):
        if ch.isdigit():
            value += ch
            continue
        if ch in _units:
            unit = _units[ch]
            if value:
                seconds += unit * int(value)
                value = ''
                unit = defaultunit
            continue
        if ch in ' \t':
            # skip whitespace
            continue
        raise ValueError('Invalid time delta: %s' % deltavalue)
    if value:
        seconds = unit * int(value)
    return datetime.timedelta(seconds=seconds)

现在您的 

mkdelta
方法接受更完整的增量，甚至仍然是整数：

>>> mkdelta('1d')
datetime.timedelta(1)
>>> mkdelta('10s')
datetime.timedelta(0, 10)
>>> mkdelta('5d 10h 3m 10s')
datetime.timedelta(5, 36190)
>>> mkdelta(5)
datetime.timedelta(5)
>>> mkdelta('1')
datetime.timedelta(1)

默认单位是天。

2
投票

您可以使用自定义操作来收集所有剩余的参数并将它们解析为

timedelta
。

这将允许您编写 CLI 命令，例如 

% test.py 2012-09-16 11:00am 2 3 4 5
datetime.timedelta(2, 3, 5004)   # args.filter_length

您还可以为 

--days
、
--seconds
等提供可选参数，这样您就可以编写 CLI 命令，例如

% test.py 2012-09-16 11:00am --weeks 6 --days 0
datetime.timedelta(42)           # args.filter_length

% test.py 2012-09-16 11:00am --weeks 6.5 --days 0
datetime.timedelta(45, 43200)

import datetime as dt
import argparse

def mkdate(datestring):
    return dt.datetime.strptime(datestring, '%Y-%m-%d').date()

def mktime(timestring):
    return dt.datetime.strptime(timestring, '%I:%M%p').time()

class TimeDeltaAction(argparse.Action):
    def __call__(self, parser, args, values, option_string = None):
        # print '{n} {v} {o}'.format(n = args, v = values, o = option_string)
        setattr(args, self.dest, dt.timedelta(*map(float, values)))

parser = argparse.ArgumentParser()
parser.add_argument('start_date', type = mkdate)
parser.add_argument('start_time', type = mktime)
parser.add_argument('--days', type = float, default = 1)
parser.add_argument('--seconds', type = float, default = 0)
parser.add_argument('--microseconds', type = float, default = 0)
parser.add_argument('--milliseconds', type = float, default = 0)
parser.add_argument('--minutes', type = float, default = 0)
parser.add_argument('--hours', type = float, default = 0)
parser.add_argument('--weeks', type = float, default = 0)
parser.add_argument('filter_length', nargs = '*', action = TimeDeltaAction)

args = parser.parse_args()
if not args.filter_length:
    args.filter_length = dt.timedelta(
        args.days, args.seconds, args.microseconds, args.milliseconds,
        args.minutes, args.hours, args.weeks)
print(repr(args.filter_length))
0
投票

这个要点似乎可以解决您的问题：https://gist.github.com/jnothman/4057689

0
投票

以防万一有人来到这里希望向参数解析器添加 

pandas.Timedelta
（就像我刚才所做的那样），事实证明以下工作正常：

parser = argparse.ArgumentParser()
parser.add_argument('--timeout', type=pd.Timedelta)

然后：

>>> parser.parse_args(['--timeout', '24 hours'])
Namespace(timeout=Timedelta('1 days 00:00:00'))
0
投票

为了补充 Martijn 的精彩答案，这里是他发布的函数的一个变体，可以扩展到数周、数月甚至数年。它稍微复杂一点，因为它需要处理多字符时间单位，但我认为增加的复杂性值得额外的功能。它也被移植到 Python 3。在这里

import datetime


def parse_delay_into_timedelta(delta: str):
    second = 1
    minute = second * 60
    hour = minute * 60
    day = hour * 24
    week = day * 7
    month = week * 4
    year = month * 12
    units = {"s": second,
             "m": minute,
             "h": hour,
             "d": day,
             "w": week,
             "mo": month,
             "y": year}
    # We need this to handle multi-character units
    # like months
    longest = max(len(k) for k in units.keys())
    seconds = 0
    default_unit = unit = units['d']  # Defaults to days
    value = ""
    delta = delta.strip().lower()
    i = 0
    while i < len(delta):
        ch = delta[i]
        i += 1
        # isdigit will return True even for things like numbers
        # in Arabic and stuff. For my use case this is undesirable,
        # but if you care about that, just swap isnumeric() for isdigit()
        if ch.isnumeric():
            value += ch
        # Only restrict ourselves to ASCII letters. Again, this is specific
        # to my use-case. Can edit as necessary
        elif ch.isalpha() and ch.isascii():
            unit_name = ch
            j = i
            # Try to parse as many characters of the unit as possible,
            # but stop once we either reach the end of the string or
            # the length of the current (potential) unit equals the
            # length of the longest one we know
            while j < len(delta) and len(unit_name) < longest:
                if (char := delta[j]).isalpha() and char.isascii():
                    unit_name += char
                else:
                    break
                j += 1
            # If the unit was more than one character long, we
            # have to skip the next len(unit_name) - 1 characters,
            # so we don't try to parse 'mo' and then 'o' again, for
            # example. If the unit was one character long, this value
            # will just be zero
            i += len(unit_name) - 1
            try:
                unit = units[unit_name]
            except KeyError:
                raise ValueError(f"invalid unit {unit_name!r}")
            if value:
                seconds += unit * int(value)
                value = ""
                unit = default_unit
        elif ch in ' \t\r':
            # Skip useless characters
            continue
        else:
            # Unknown character
            raise ValueError(f"Invalid character {ch!r} in time delta: {delta!r}")
    if value:
        seconds = unit * int(value)

    return datetime.timedelta(seconds=seconds)

你可以这样使用它：

>>> parse_delay_into_timedelta("2y 1mo 1w 5d 6m 1h 3s")
datetime.timedelta(days=712, seconds=3963)
>>> parse_delay_into_timedelta("1s")
datetime.timedelta(seconds=1)
>>> parse_delay_into_timedelta("10s")
datetime.timedelta(seconds=10)
>>> parse_delay_into_timedelta("15h")
datetime.timedelta(seconds=54000)
最新问题
© www.soinside.com 2019 - 2024. All rights reserved.