熊猫根据其他两个具有日期时间值的列创建一个布尔列

Question

I have the a df ,

date1        date2
2019-05-31   2019-06-01
NaT          NaN
2018-07-01   2018-08-01
NaT          2019-06-03
2019-01-01   NaN

I want to create a boolean column on_time based on -3 <= date2 - date1 <= 0 , if any values in date1 or date2 is NaN or NaT , make on_time = False ;

    a = df['date1'].isna()
    b = df['date2'].isna()

    df['on_time'] = (a | b)

    m = (-3 <= (df.loc[~a&~b, 'date1'] - df.loc[~a&~b, 'date2']).dt.days) & \
        ((df.loc[~a&~b, 'date1'] - df.loc[~a&~b, 'date2']).dt.days <= 0)

    df['on_time'] = m

I am wondering if there is a better way to do it, more concise and efficient way.

Answer 1

IIUC, you can create a helper series with series.dt.days() and compare using s.ge() and le :

s=(df.date2-df.date1).dt.days
df=df.assign(on_time=s.ge(-3)&s.le(0))

---

       date1      date2  on_time
0 2019-05-31 2019-06-01    False
1        NaT        NaT    False
2 2018-07-01 2018-08-01    False
3        NaT 2019-06-03    False
4 2019-01-01        NaT    False

Answer 2

## if the dates are of type str
df['date1'] = pd.to_datetime(df['date1'])
df['date2'] = pd.to_datetime(df['date2'])


(df['date2'] - df['date1']).apply(lambda x: True if -3<= x.days <=0   else False)

Output

       date1      date2  on_time
0 2019-05-31 2019-06-01    False
1        NaT        NaT    False
2 2018-07-01 2018-08-01    False
3        NaT 2019-06-03    False
4 2019-01-01        NaT    False