adding NaN where regex doesn't match

Question

import pandas as pd
df= pd.DataFrame({'Date':['nothing ',
                              'This 1A1619 A124 person BL171111 the A-1-24 and ',
                              'dont Z112 but NOT 12-24-1981',
                               'nada here either',
                              'mix: 1A25629Q88 or A13B ok A1 the A16'],
                      'IDs': ['A11','B22','C33', 'D44', 'E55'],
                      })

This is a follow up and variation to pulling mixed letters and numbers. Using this code

pat = r'((?<!\S)(?:[a-zA-Z]+\d|\d+[a-zA-Z])[a-zA-Z0-9]*(?!\S))'
df['Date'].str.extractall(pat)

gives me

        0
   match    
1   0   1A1619
    1   A124
    2   BL171111
2   0   Z112
4   0   1A25629Q88
    1   A13B
    2   A1
    3   A16

I am looking to add NaN where the regex doesnt match. So I would like something this instead

        0
   match    
0   NaN
1   0   1A1619
1   A124
2   BL171111
2   0   Z112
3   NaN
4   0   1A25629Q88
    1   A13B
    2   A1
    3   A16

How would I alter my code to do so?

simple thing is loop through the list, check for individual element with regex if it matches add that to final list, if not just use NaN as default value — Code Maniac, Aug 30 '19 at 01:49

score 1 · Accepted Answer · answered Aug 30 '19 at 01:46

Given s is the return of df['Date'].str.extractall(pat), we can:

i = df.index.difference(s.index.get_level_values(0))
o = pd.DataFrame({0: np.nan}, index=[i, [0]*len(i)])
adjust = lambda s,o: pd.concat([s, o]).sort_index()

Then

>>> adjust(s,o)

                  0
  match            
0 0             NaN
1 0          1A1619
  1            A124
  2        BL171111
2 0            Z112
3 0             NaN
4 0      1A25629Q88
  1            A13B
  2              A1
  3             A16

adding NaN where regex doesn't match

1 Answers1