问题描述
我有一个代码使用 np.busdaycount 计算不包括周末的日期差异，但我需要它在我无法获得的时间.
I have a code that calculates the date differance excluding the weekends using np.busdaycount, but i need it in the hours which i cannot able to get.
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
我需要输出时间，不包括周末.喜欢
I need the out put as hours excluding the weekend. Like
Hours
0 254
问题
inflow_date_time=2019-08-01 23:22:46End_date_time = 2019-08-05 17:43:51预计小时数 42 小时(1+24+17)
Inflow_date_time=2019-08-01 23:22:46End_date_time = 2019-08-05 17:43:51Hours expected 42 hours(1+24+17)
inflow_date_time=2019-08-03 23:22:46End_date_time = 2019-08-05 17:43:51
预计小时数 17 小时(0+0+17)
Inflow_date_time=2019-08-03 23:22:46End_date_time = 2019-08-05 17:43:51
Hours expected 17 hours(0+0+17)
inflow_date_time=2019-08-01 23:22:46End_date_time = 2019-08-05 17:43:51预计小时数 17 小时(0+0+17)
Inflow_date_time=2019-08-01 23:22:46End_date_time = 2019-08-05 17:43:51Hours expected 17 hours(0+0+17)
流入日期时间=2019-07-26 23:22:46End_date_time = 2019-08-05 17:43:51
预计小时数 138 小时(1+120+17)
Inflow_date_time=2019-07-26 23:22:46End_date_time = 2019-08-05 17:43:51
Hours expected 138 hours(1+120+17)
inflow_date_time=2019-08-05 11:22:46End_date_time = 2019-08-05 17:43:51
预计小时数 6 小时(0+0+6)
Inflow_date_time=2019-08-05 11:22:46End_date_time = 2019-08-05 17:43:51
Hours expected 6 hours(0+0+6)
请提出建议.
推荐答案
想法是按天删除times的下限日期时间，并获取开始日+一天之间的工作日数到numpy.busday_count >hours3 列 然后为开始和结束时间创建 hour1 和 hour2 列，如果不是周末时间，则按小时计算.最后将所有小时列加在一起:
Idea is floor datetimes for remove times by floor by days and get number of business days between start day + one day to hours3 column by numpy.busday_count and then create hour1 and hour2 columns for start and end hours with floor by hours if not weekends hours. Last sum all hours columns together:
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
这篇关于两天之间的差异(不包括周末)(以小时为单位)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！