Question

我有一个如下的数据集，每个ID可以在任何给定的时间和持续时间内签入和签出

    Product product = _uow.Products.SelectProduct(id);

    if (product == null)
    {
        return RedirectToPage("Productslist");
    }
    else
    {
        return View("ProductPage", productVm);
    }

从此计算出的“签入分钟数”需要分为下表所示的小时时段，以便即使在整个签入结帐时，我也可以按小时和天数按每个ID的小时数来计算累计总数天。

帮助表示感谢：）

            ID  checkin_datetime    checkout_datetime
            4   04-01-2019 13:07    04-01-2019 13:09
            4   04-01-2019 13:09    04-01-2019 13:12
            4   04-01-2019 14:06    04-01-2019 14:07
            4   04-01-2019 14:55    04-01-2019 15:06
            22  04-01-2019 20:23    04-01-2019 21:32
            22  04-01-2019 21:38    04-01-2019 21:42
            25  04-01-2019 23:22    04-02-2019 00:23
            29  04-02-2019 01:00    04-02-2019 06:15

创建数据框的代码：

            ID  checkin_datetime    checkout_datetime   day         HR  Minutes
            4   04-01-2019 13:07    04-01-2019 13:09    04-01-2019  13  2
            4   04-01-2019 13:09    04-01-2019 13:12    04-01-2019  13  3
            4   04-01-2019 14:06    04-01-2019 14:07    04-01-2019  14  1
            4   04-01-2019 14:55    04-01-2019 15:06    04-01-2019  14  5
            4   04-01-2019 14:55    04-01-2019 15:06    04-01-2019  15  6
            22  04-01-2019 20:23    04-01-2019 21:32    04-01-2019  20  27
            22  04-01-2019 20:23    04-01-2019 21:32    04-01-2019  21  32
            22  04-01-2019 21:38    04-01-2019 21:42    04-01-2019  21  4
            25  04-01-2019 23:22    04-02-2019 00:23    04-01-2019  23  28
            25  04-01-2019 23:22    04-02-2019 00:23    04-02-2019  0   23
            29  04-02-2019 01:00    04-02-2019 06:15    04-02-2019  1   60
            29  04-02-2019 01:00    04-02-2019 06:15    04-02-2019  2   60
            29  04-02-2019 01:00    04-02-2019 06:15    04-02-2019  3   60
            29  04-02-2019 01:00    04-02-2019 06:15    04-02-2019  4   60
            29  04-02-2019 01:00    04-02-2019 06:15    04-02-2019  5   60
            29  04-02-2019 01:00    04-02-2019 06:15    04-02-2019  6   15

Answer 1

非常简单：
-在此期间，您只需从签入中减去签出即可（datetime可以做到）。
-要在几分钟内得到它-用一分钟的timedelta除以一分钟（我将使用内置的pandas）。
-要从datetime中获取小时，请调用.hour，然后类似地.date()来获取日期（第一个是属性，第二个是方法-注意括号）。

df['Hour'] = df['checkin_datetime'].apply(lambda x: x.hour)
df['Date'] = df['checkin_datetime'].apply(lambda x: x.date())
df['duration'] = df['checkout_datetime']-df['checkin_datetime']
df['duration_in_minutes'] = (df['checkout_datetime']-df['checkin_datetime'])/pd.Timedelta(minutes=1)

[编辑]：我有一个解决方案，可以将持续时间分成几个小时，但这不是最优雅的方法。

df2 = pd.DataFrame(
index=pd.DatetimeIndex(
    start=df['checkin_datetime'].min(),
    end=df['checkout_datetime'].max(),freq='1T'),
    columns = ['is_checked_in','ID'], data=0)

for index, row in df.iterrows():
    df2['is_checked_in'][row['checkin_datetime']:row['checkout_datetime']] = 1
    df2['ID'][row['checkin_datetime']:row['checkout_datetime']] = row['ID']

df3 = df2.resample('1H').aggregate({'is_checked_in': sum,'ID':max})
df3['Hour'] = df3.index.to_series().apply(lambda x: x.hour)

Answer 2

import pandas as pd

data={'ID':[4,4,4,4,22,22,25,29],
  'checkin_datetime':['04-01-2019 13:07','04-01-2019 13:09','04-01-2019 14:06','04-01-2019 14:55','04-01-2019 20:23'
  ,'04-01-2019 21:38','04-01-2019 23:22','04-02-2019 01:00'],
  'checkout_datetime':['04-01-2019 13:09','04-01-2019 13:12','04-01-2019 14:07','04-01-2019 15:06','04-01-2019 21:32'
                       ,'04-01-2019 21:42','04-02-2019 00:23'
                       ,'04-02-2019 06:15']
}

df = pd.DataFrame(data,columns= ['ID', 'checkin_datetime','checkout_datetime'])

df['checkout_datetime'] = pd.to_datetime(df['checkout_datetime'])
df['checkin_datetime'] = pd.to_datetime(df['checkin_datetime'])
df['Hour'] = df['checkin_datetime'].apply(lambda x: x.hour)
df['Date'] = df['checkin_datetime'].apply(lambda x: x.date())
df['duration'] = df['checkout_datetime']-df['checkin_datetime']
df['duration_in_minutes'] = (df['checkout_datetime']-df['checkin_datetime'])/pd.Timedelta(minutes=1)
with pd.option_context('display.max_rows', None, 'display.max_columns', None):  # more options can be specified also
    print(df)

我认为Itamar Muskhkin先前给出的答案是绝对正确的。

Python将时间间隔分为小时时段

2 个答案: