Question

我目前有一个数据框，其中uniqueID在另一列中有多个日期。我想提取每个日期之间的小时数，但是如果下一个日期在周末之后，则忽略周末。例如，如果今天是星期五的中午12点，而下一个日期是星期二中午12点，则这两个日期之间的小时差为48小时。

这是我的数据集，具有预期的输出：

df = pd.DataFrame({"UniqueID": ["A","A","A","B","B","B","C","C"],"Date":
["2018-12-07 10:30:00","2018-12-10 14:30:00","2018-12-11 17:30:00",
"2018-12-14 09:00:00","2018-12-18 09:00:00",
"2018-12-21 11:00:00","2019-01-01 15:00:00","2019-01-07 15:00:00"],
"ExpectedOutput": ["28.0","27.0","Nan","48.0","74.0","NaN","96.0","NaN"]})

df["Date"] = df["Date"].astype(np.datetime64)

这是我到目前为止的内容，但包括周末：

df["date_diff"] = df.groupby(["UniqueID"])["Date"].apply(lambda x: x.diff() 
/ np.timedelta64(1 ,'h')).shift(-1)

谢谢！

Answer 1

想法是要删除的times的下限日期时间，并获取开始日期+一日与转移日期到hours3列之间的工作日数，numpy.busday_count，然后创建hour1和hour2列中的开始时间和结束时间（如果不是周末时间）。最后汇总所有小时数列：

df["Date"] = pd.to_datetime(df["Date"])
df = df.sort_values(['UniqueID','Date'])

df["shifted"] = df.groupby(["UniqueID"])["Date"].shift(-1)
df["hours1"] = df["Date"].dt.floor('d') 
df["hours2"] = df["shifted"].dt.floor('d') 

mask = df['shifted'].notnull()
f = lambda x: np.busday_count(x['hours1'] + pd.Timedelta(1, unit='d'), x['hours2'])
df.loc[mask, 'hours3'] = df[mask].apply(f, axis=1) * 24

mask1 = df['hours1'].dt.dayofweek < 5
hours1 = df['hours1'] + pd.Timedelta(1, unit='d') - df['Date']
df['hours1'] = np.where(mask1, hours1, np.nan) / np.timedelta64(1 ,'h')

mask1 = df['hours2'].dt.dayofweek < 5
df['hours2'] = np.where(mask1, df['shifted']-df['hours2'], np.nan) / np.timedelta64(1 ,'h')

df['date_diff'] = df['hours1'].fillna(0) + df['hours2'] + df['hours3']

print (df)
  UniqueID                Date ExpectedOutput             shifted  hours1  \
0        A 2018-12-07 10:30:00           28.0 2018-12-10 14:30:00    13.5   
1        A 2018-12-10 14:30:00           27.0 2018-12-11 17:30:00     9.5   
2        A 2018-12-11 17:30:00            Nan                 NaT     6.5   
3        B 2018-12-14 09:00:00           48.0 2018-12-18 09:00:00    15.0   
4        B 2018-12-18 09:00:00           74.0 2018-12-21 11:00:00    15.0   
5        B 2018-12-21 11:00:00            NaN                 NaT    13.0   
6        C 2019-01-01 15:00:00           96.0 2019-01-07 15:00:00     9.0   
7        C 2019-01-07 15:00:00            NaN                 NaT     9.0   

   hours2  hours3  date_diff  
0    14.5     0.0       28.0  
1    17.5     0.0       27.0  
2     NaN     NaN        NaN  
3     9.0    24.0       48.0  
4    11.0    48.0       74.0  
5     NaN     NaN        NaN  
6    15.0    72.0       96.0  
7     NaN     NaN        NaN

第一个解决方案被删除有两个原因-不准确且缓慢：

np.random.seed(2019)

dates = pd.date_range('2015-01-01','2018-01-01', freq='H')
df = pd.DataFrame({"UniqueID": np.random.choice(list('ABCDEFGHIJ'), size=100),
                   "Date": np.random.choice(dates, size=100)})
print (df)

def old(df):
    df["Date"] = pd.to_datetime(df["Date"])
    df = df.sort_values(['UniqueID','Date'])

    df["shifted"] = df.groupby(["UniqueID"])["Date"].shift(-1)

    def f(x):
        a = pd.date_range(x['Date'],  x['shifted'], freq='T')
        return ((a.dayofweek < 5).sum() / 60).round()


    mask = df['shifted'].notnull()
    df.loc[mask, 'date_diff'] = df[mask].apply(f, axis=1)  
    return df

def new(df):
    df["Date"] = pd.to_datetime(df["Date"])
    df = df.sort_values(['UniqueID','Date'])

    df["shifted"] = df.groupby(["UniqueID"])["Date"].shift(-1)
    df["hours1"] = df["Date"].dt.floor('d') 
    df["hours2"] = df["shifted"].dt.floor('d') 

    mask = df['shifted'].notnull()
    f = lambda x: np.busday_count(x['hours1'] + pd.Timedelta(1, unit='d'), x['hours2'])
    df.loc[mask, 'hours3'] = df[mask].apply(f, axis=1) * 24

    mask1 = df['hours1'].dt.dayofweek < 5
    hours1 = df['hours1'] + pd.Timedelta(1, unit='d') - df['Date']
    df['hours1'] = np.where(mask1, hours1, np.nan) / np.timedelta64(1 ,'h')

    mask1 = df['hours2'].dt.dayofweek < 5
    df['hours2'] = np.where(mask1, df['shifted'] - df['hours2'], np.nan) / np.timedelta64(1 ,'h')

    df['date_diff'] = df['hours1'].fillna(0) + df['hours2'] + df['hours3']
    return df
print (new(df))
print (old(df))

In [44]: %timeit (new(df))
22.7 ms ± 115 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [45]: %timeit (old(df))
1.01 s ± 8.03 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

日期时间的小时数差异，不包括周末

1 个答案: