問題描述
我有一個代碼使用 np.busdaycount 計算不包括周末的日期差異，但我需要它在我無法獲得的時間.
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
我需要輸出時間，不包括周末.喜歡
I need the out put as hours excluding the weekend. Like
Hours
0 254
問題
inflow_date_time=2019-08-01 23:22:46End_date_time = 2019-08-05 17:43:51預計小時數 42 小時(1+24+17)
Inflow_date_time=2019-08-01 23:22:46 End_date_time = 2019-08-05 17:43:51 Hours expected 42 hours (1+24+17)
inflow_date_time=2019-08-03 23:22:46End_date_time = 2019-08-05 17:43:51
預計小時數 17 小時(0+0+17)
Inflow_date_time=2019-08-03 23:22:46
End_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預計小時數 17 小時(0+0+17)
Inflow_date_time=2019-08-01 23:22:46 End_date_time = 2019-08-05 17:43:51 Hours expected 17 hours (0+0+17)
流入日期時間=2019-07-26 23:22:46End_date_time = 2019-08-05 17:43:51
預計小時數 138 小時(1+120+17)
Inflow_date_time=2019-07-26 23:22:46
End_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
預計小時數 6 小時(0+0+6)
Inflow_date_time=2019-08-05 11:22:46
End_date_time = 2019-08-05 17:43:51
Hours expected 6 hours
(0+0+6)
請提出建議.
推薦答案
想法是按天刪除times的下限日期時間，并獲取開始日+一天之間的工作日數到hour1 和 hour2 列，如果不是周末時間，則按小時計算.最后將所有小時列加在一起:
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)
<小時>
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)
<小時>
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
這篇關于兩天之間的差異(不包括周末)(以小時為單位)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！