[英]What is wrong with this Numpy/Pandas code to construct new boolean column based on the values in two other boolean columns?
[英]pandas create a boolean column based on two other columns with datetime values
我有一个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
我想基于-3 <= date2 - date1 <= 0
创建一个布尔列on_time
,如果date1
或date2
任何值为NaN
或NaT
,则使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
我想知道是否有更好的方法,更简洁有效的方法。
IIUC,您可以使用series.dt.days()
创建一个帮助器系列,并使用s.ge()
和le
进行比较:
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
## 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)
输出量
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
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