如何检查熊猫行中的连续差异

Question

I have following dataframe in pandas

   code     car_no      date         time        error_code
   123      6           2019-01-01   06:23:00    E09
   123      6           2019-01-01   06:25:00    E28
   123      6           2019-01-01   06:28:00    E09
   123      6           2019-01-01   22:00:00    E28
   123      7           2019-01-01   08:23:00    E09
   123      6           2019-01-01   09:23:00    E09
   123      7           2019-01-01   08:28:00    E28

What I want to flag, is for a specific code and car_no on the same date if E09 comes first and E28 come later with less than 2 hours difference then flag should be set. My desired dataframe is as follows

  code     car_no      date         time        error_code   flag
   123      6           2019-01-01   06:23:00    E09         1
   123      6           2019-01-01   06:25:00    E28         1 
   123      6           2019-01-01   06:28:00    E09         0
   123      6           2019-01-01   22:00:00    E28         0
   123      7           2019-01-01   08:23:00    E09         1
   123      6           2019-01-01   09:23:00    E09         0 
   123      7           2019-01-01   08:28:00    E28         0

How can I do it in pandas?

Answer 1

Writing down your conditions and do it within groupby , then we just need to assign it back

#df.time=pd.to_timedelta(df.time) 
s=df.groupby(['date','car_no']).\
      apply(lambda x : x.error_code.eq('E28')&x.error_code.shift().eq('E09')&x.time.diff().dt.seconds.lt(60*60*2))
s=(s|s.groupby(level=[0,1]).shift(-1)).reset_index(level=[0,1],drop=True)
df['flag']=s
df
Out[126]: 
   code  car_no        date     time error_code   flag
0   123       6  2019-01-01 06:23:00        E09   True
1   123       6  2019-01-01 06:25:00        E28   True
2   123       6  2019-01-01 06:28:00        E09  False
3   123       6  2019-01-01 22:00:00        E28  False
4   123       7  2019-01-01 08:23:00        E09   True
5   123       6  2019-01-01 09:23:00        E09  False
6   123       7  2019-01-01 08:28:00        E28   True