我有一个代码，可以使用np.busdaycount计算除周末以外的日期差异，但是我需要在我无法获得的小时数内。
import datetime import numpy as np df.Inflow_date_time= [pandas.Timestamp('2019-07-22 21:11:26')] df.End_date_time= [pandas.Timestamp('2019-08-02 11:44:47')] df['Day'] = ([np.busday_count(b,a) for a, b in zip(df['End_date_time'].values.astype('datetime64[D]'),df['Inflow_date_time'].values.astype('datetime64[D]'))]) Day 0 9
除周末外，我的工作时间为小时。喜欢
Hours 0 254
问题
Inflow_date_time = 2019-08-01 23:22:46 End_date_time = 2019-08-05 17:43:51预期小时数42小时（1 + 24 + 17）
Inflow_date_time = 2019-08-03 23:22:46 End_date_time = 2019-08-05 17:43:51 预计工时17小时（0 + 0 + 17）
Inflow_date_time = 2019-08-01 23:22:46 End_date_time = 2019-08-05 17:43:51预期的小时数17小时（0 + 0 + 17）
Inflow_date_time = 2019-07-26 23:22:46 End_date_time = 2019-08-05 17:43:51 预期小时138小时（1 + 120 + 17）
Inflow_date_time = 2019-08-05 11:22:46 End_date_time = 2019-08-05 17:43:51 预期小时数6小时（0 + 0 + 6）
请提出建议。
想法是按日期逐层删除times的楼层日期时间，并获取开始日期+一天之间的工作日数，然后按hours3列（如果不是周末时间）numpy.busday_count创建hour1和hour2按楼层按小时数的开始和结束时间列。最后汇总所有小时数列：
times
hours3
numpy.busday_count
hour1
hour2
df = pd.DataFrame(columns=['Inflow_date_time','End_date_time', 'need']) df.Inflow_date_time= [pd.Timestamp('2019-08-01 23:22:46'), pd.Timestamp('2019-08-03 23:22:46'), pd.Timestamp('2019-08-01 23:22:46'), pd.Timestamp('2019-07-26 23:22:46'), pd.Timestamp('2019-08-05 11:22:46')] df.End_date_time= [pd.Timestamp('2019-08-05 17:43:51')] * 5 df.need = [42,17,41,138,6] #print (df)
df["hours1"] = df["Inflow_date_time"].dt.ceil('d') df["hours2"] = df["End_date_time"].dt.floor('d') one_day_mask = df["Inflow_date_time"].dt.floor('d') == df["hours2"] df['hours3'] = [np.busday_count(b,a)*24 for a, b in zip(df['hours2'].dt.strftime('%Y-%m-%d'), df['hours1'].dt.strftime('%Y-%m-%d'))] mask1 = df['hours1'].dt.dayofweek < 5 hours1 = df['hours1'] - df['Inflow_date_time'].dt.floor('H') df['hours1'] = np.where(mask1, hours1, np.nan) / np.timedelta64(1 ,'h') mask2 = df['hours2'].dt.dayofweek < 5 df['hours2'] = (np.where(mask2, df['End_date_time'].dt.floor('H')-df['hours2'], np.nan) / np.timedelta64(1 ,'h')) df['date_diff'] = df['hours1'].fillna(0) + df['hours2'].fillna(0) + df['hours3'] one_day = (df['End_date_time'].dt.floor('H') - df['Inflow_date_time'].dt.floor('H')) / np.timedelta64(1 ,'h') df["date_diff"] = df["date_diff"].mask(one_day_mask, one_day)
print (df) Inflow_date_time End_date_time need hours1 hours2 hours3 \ 0 2019-08-01 23:22:46 2019-08-05 17:43:51 42 1.0 17.0 24 1 2019-08-03 23:22:46 2019-08-05 17:43:51 17 NaN 17.0 0 2 2019-08-01 23:22:46 2019-08-05 17:43:51 41 1.0 17.0 24 3 2019-07-26 23:22:46 2019-08-05 17:43:51 138 NaN 17.0 120 4 2019-08-05 11:22:46 2019-08-05 17:43:51 6 13.0 17.0 -24 date_diff 0 42.0 1 17.0 2 42.0 3 137.0 4 6.0
