Python不使用if语句重复代码，还有其他方法吗？

def add_timeframe(time, timeframe): if 'H' in timeframe: ┆ try: ┆ ┆ period = int(re.sub('\D', '', timeframe)) ┆ ┆ return convert_datetime(time + datetime.timedelta(hours=period)) ┆ except ValueError: ┆ ┆ return convert_datetime(time + datetime.timedelta(hours=1)) if 'D' in timeframe: ┆ try: ┆ ┆ period = int(re.sub('\D', '', timeframe)) ┆ ┆ return convert_datetime(time + datetime.timedelta(days=period)) ┆ except ValueError: ┆ ┆ return convert_datetime(time + datetime.timedelta(days=1)) if 'W' in timeframe: ┆ try: ┆ ┆ period = int(re.sub('\D', '', timeframe)) ┆ ┆ return convert_datetime(time + datetime.timedelta(weeks=period)) ┆ except ValueError: ┆ ┆ return convert_datetime(time + datetime.timedelta(weeks=period)) if 'M' in timeframe: ┆ try: ┆ ┆ period = int(re.sub('\D', '', timeframe)) ┆ ┆ return convert_datetime(time + datetime.timedelta(days=365/12*period)) ┆ except ValueError: ┆ ┆ return convert_datetime(time + datetime.timedelta(days=365/12))

2条回答

网友

1楼 · 编辑于 2024-04-26 04:19:50

我通常用字典来避免很多“如果”。我将每个条件映射到字典并执行。这是我的第一张照片：

我创建了一个用于添加月份的函数，因为timedelta没有它。然后使用re将数字和字母作为元组。所以“4M”应该是（'4'，'M'）。然后我将M映射为月加法，因此4*（add months函数），W表示add weeks等

import calendar
import datetime
import re


# add month hack
def add_month(num_months, date=None):
    '''Add N months'''

    assert num_months > 0, 'Positive N only'

    if date is None:
        date = datetime.datetime.now()

    for num in range(num_months):
        month_days = calendar.monthrange(date.year, date.month)[1]
        dt = date + datetime.timedelta(days=month_days)

        if dt.day != date.day:
             dt.replace(day=1) - datetime.timedelta(days=1)
        else:
            dt

        date = dt
    return dt



def delta(data, pattern):
    # dict instead of lots of ifs

    time_convert = {'M': lambda x : add_month(x),
                   'W': lambda x :datetime.timedelta(weeks=x),
                   'D': lambda x: datetime.timedelta(days=x),
                   'H': lambda x: datetime.timedelta(hours=x),
                          }


    _ = [re.match(pattern, item).groups() for item in data]

    return [time_convert.get(letter)(int(number))for number, letter in _] 

# test 1
data = ['1M', '7D', '4M', '2H']
pattern = '(\d+)(\w+)'

s = delta(data, pattern)
print(s)

如果我们期待不干净的数据，我们需要创建一个数据准备功能，将确保我们的数据是在一个格式，我们希望digitsletter（s）。因为如果我们只收到字母数据，我们的代码就会失败。如果我们只有一个字母，我们可以添加1，而不是try catch。这将是我的第二步：

import calendar
import datetime
import re


# add month hack
def add_month(num_months, date=None):
    '''Add N months'''

    assert num_months > 0, 'Positive N only'

    if date is None:
        date = datetime.datetime.now()

    for num in range(num_months):
        month_days = calendar.monthrange(date.year, date.month)[1]
        dt = date + datetime.timedelta(days=month_days)

        if dt.day != date.day:
             dt.replace(day=1) - datetime.timedelta(days=1)
        else:
            dt

        date = dt
    return dt


def data_prep(data, check_patter='\d+'):
    '''Our data preparation happens here'''

    _ = [bool(re.search(check_patter,item)) for item in data]

    for index, truth in enumerate(_):
        if not truth:
            data[index] = '1'+data[index]

    return data




def delta(data, pattern):

    time_convert = {'M': lambda x : add_month(x),
                'W': lambda x :datetime.timedelta(weeks=x),
                'D': lambda x: datetime.timedelta(days=x),
                'H': lambda x: datetime.timedelta(hours=x),
                       }


    # clean data. if M  > 1M
    data = data_prep(data)
    _ = [re.match(pattern, item).groups() for item in data]

    return [time_convert.get(letter)(int(number))for number, letter in _] 


data = ['1M', '7D', '4M', '2H','H']

pattern = '(\d+)(\w+)'


s = delta(data, pattern)

print(s)

在数据准备功能中，您将处理所有可能的不干净：）

网友

2楼 · 编辑于 2024-04-26 04:19:50

您可以使用re在一次go中提取数字部分和所有句点字符。。你知道吗

>>> import re
>>> inp = ["1M", "7D", "6M", "2H", "M"]
>>> [re.findall('(\d)?(M|D|H)', x) for x in inp]
[[('1', 'M')], [('7', 'D')], [('6', 'M')], [('2', 'H')], [('', 'M')]]

>>> extracted = [re.findall('(\d)?(M|D|H)', x) for x in inp]
>>> [(int(x[0][0] or '1'), x[0][1]) for x in extracted if x] # Filter out invalids.
[(1, 'M'), (7, 'D'), (6, 'M'), (2, 'H'), (1, 'M')]

然后您可以使用convert_datetime(..)和您在原始代码中正在做的其他东西。你知道吗

PS：我会执行更多的错误检查上面的代码只是建议一个稍微更python的方式做同样的事情。你知道吗

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