如何计算两个时间戳之间的小时差并排除周末

Question

我有一个这样的数据框：

     Folder1                   Folder2                 
0   2021-11-22 12:00:00      2021-11-24 10:00:00
1   2021-11-23 10:30:00      2021-11-25 18:30:00    
2   2021-11-12 10:30:00      2021-11-15 18:30:00    
3   2021-11-23 10:00:00            NaN         

         

使用此代码：

def strfdelta(td: pd.Timestamp):
    seconds = td.total_seconds()
    hours = int(seconds // 3600)
    minutes = int((seconds % 3600) // 60)
    seconds = int(seconds % 60)
    return f"{hours:02}:{minutes:02}:{seconds:02}"
            
df["Folder1"] = pd.to_datetime(df["Folder1"])
df["Folder2"] = pd.to_datetime(df["Folder2"])

bm1 = df["Folder1"].notna() & df["Folder2"].notna()
bm2 = df["Folder1"].notna() & df["Folder2"].isna()

df["Time1"] = (df.loc[bm1, "Folder2"] - df.loc[bm1, "Folder1"]).apply(strfdelta)
df["Time2"] = (datetime.now() - df.loc[bm2, "Folder1"]).apply(strfdelta)

我有这个 df：

     Folder1                   Folder2                           Time1     Time2
0   2021-11-22 12:00:00      2021-11-24 10:00:00                46:00:00    NaN
1   2021-11-23 10:30:00      2021-11-25 18:30:00                56:00:00    NaN
2   2021-11-12 10:30:00      2021-11-15 18:30:00                80:00:00    NaN
3   2021-11-23 10:00:00            NaN                             NaN     03:00:00

基本上，这就是我想要的，但是，在计算 Folder1 和 Folder2 的时间戳之间的差异时，如何排除周末时间？我应该改变什么才能拥有这样的df：

     Folder1                   Folder2                           Time1     Time2
0   2021-11-22 12:00:00      2021-11-24 10:00:00                46:00:00    NaN
1   2021-11-23 10:30:00      2021-11-25 18:30:00                56:00:00    NaN
2   2021-11-12 10:30:00      2021-11-15 18:30:00                32:00:00    NaN
3   2021-11-23 10:00:00            NaN                            NaN     03:00:00

因此，在索引 2 的行中，13.11 和 14.11 是周末，因此，在时间 1 中，差异应该是 32 而不是 80

Answer 1

我认为您可以像这样利用pandas.date_range 函数与pandas.tseries.offsets.CustomBusinessHour 结合使用：

# import pandas and numpy
import pandas as pd
import numpy as np

# construct dataframe
df = pd.DataFrame()
df["Folder1"] = pd.to_datetime(
    pd.Series(
        [
            "2021-11-22 12:00:00",
            "2021-11-23 10:30:00",
            "2021-11-12 10:30:00",
            "2021-11-23 10:00:00",
        ]
    )
)
df["Folder2"] = pd.to_datetime(
    pd.Series(
        [
            "2021-11-24 10:00:00", 
            "2021-11-25 18:30:00", 
            "2021-11-15 18:30:00", 
            np.NaN
        ]
    )
)

# define custom business hours
cbh = pd.tseries.offsets.CustomBusinessHour(start="0:00", end="23:59")

# actual calculation
df["Time1"] = df[~(df["Folder1"].isnull() | df["Folder2"].isnull())].apply(
    lambda row: len(
        pd.date_range(
            start=row["Folder1"], 
            end=row["Folder2"], 
            freq=cbh)),
    axis=1,
)

df.head()

这对我来说是：

print(df.head())
              Folder1             Folder2  Time1
0 2021-11-22 12:00:00 2021-11-24 10:00:00   46.0
1 2021-11-23 10:30:00 2021-11-25 18:30:00   56.0
2 2021-11-12 10:30:00 2021-11-15 18:30:00   32.0
3 2021-11-23 10:00:00                 NaT    NaN

作为奖励，您还可以使用它更有效地进行 Time2 计算：

df["Time2"] = df[df["Folder2"].isnull()].apply(
    lambda row: len(
        pd.date_range(
            start=row["Folder1"],
            end=datetime.datetime.now(),
            freq=cbh)),
    axis=1,
)

这对我来说产生了（欧洲中部时间 14:45）：

print(df.head())
              Folder1             Folder2  Time1  Time2
0 2021-11-22 12:00:00 2021-11-24 10:00:00   46.0    NaN
1 2021-11-23 10:30:00 2021-11-25 18:30:00   56.0    NaN
2 2021-11-12 10:30:00 2021-11-15 18:30:00   32.0    NaN
3 2021-11-23 10:00:00                 NaT    NaN    5.0

Answer 2

df = pd.DataFrame({'Folder1': ['2021-11-22 12:00:00', '2021-11-23 10:30:00', '2021-11-12 10:30:00', '2021-11-23 10:00:00'],
                   'Folder2': ['2021-11-24 10:00:00', '2021-11-25 18:30:00', '2021-11-15 18:30:00', None]})
df[['Folder1','Folder2']] = df[['Folder1','Folder2']].astype('datetime64')

def strfdelta(t1, t2):
    hd = pd.date_range(t1, t2, freq='W-SAT').append(pd.date_range(t1, t2, freq='W-SUN'))
    sec = (t2-t1).total_seconds() - len(hd)*24*3600
    return f"{int(sec//3600):02d}:{int((sec%3600)//60):02d}:{int(sec%60):02d}"

now = pd.to_datetime('now')
df['Time1'] = df.fillna(now).apply(lambda x: strfdelta(x['Folder1'], x['Folder2']), axis=1)
print(df) 

打印：

              Folder1             Folder2     Time1
0 2021-11-22 12:00:00 2021-11-24 10:00:00  46:00:00
1 2021-11-23 10:30:00 2021-11-25 18:30:00  56:00:00
2 2021-11-12 10:30:00 2021-11-15 18:30:00  32:00:00
3 2021-11-23 10:00:00                 NaT  20:58:26

Answer 3

df['Folder1']=pd.to_datetime(df['Folder1'])
df['Folder2']=pd.to_datetime(df['Folder2']).fillna(df['Folder1'])

df['missing']=df.apply(lambda x: pd.date_range(start=x['Folder1'], end=x['Folder2'], freq='D'), axis=1)#Create column with missing date periods



df=(df.assign(time=np.where((df['missing'].apply(lambda x: x.strftime('%w'))).map(set).astype(str).str.contains('0|6'),#Where missing periods have a Saturday or Sunday
                            
                            (df['Folder2']-df['Folder1']).astype('timedelta64[h]')-48,# When above condition is met, subtract two 48 hours from the two days columns difference
                            (df['Folder2']-df['Folder1']).astype('timedelta64[h]'))#When condition not met substract just the two date columns)
             ).drop(columns=['missing']) )             
print(df)



Folder1             Folder2  time
0 2021-11-22 12:00:00 2021-11-24 10:00:00  46.0
1 2021-11-23 10:30:00 2021-11-25 18:30:00  56.0
2 2021-11-12 10:30:00 2021-11-15 18:30:00  32.0
3 2021-11-23 10:00:00 2021-11-23 10:00:00   0.0