LabelEncoding in Pandas on a column with list of strings across rows

Question

I would like to LabelEncode a column in pandas where each row contains a list of strings. Since a similar string/text carries a same meaning across rows, encoding should respect that, and ideally encode it with a unique number. Imagine:

import pandas as pd

df =pd.DataFrame({
                  'A':[['OK', 'NG', 'Repair', 'Peace'],['Sky', 'NG', 'Fixed', 'Conflict'],['Crossed', 'OK', 'Engine', 'Peace'],['OK', 'Beats', 'RPi', 'Country']]
                  })

# df
                              A
0       [OK, NG, Repair, Peace]
1    [Sky, NG, Fixed, Conflict]
2  [Crossed, OK, Engine, Peace]
3     [OK, Beats, RPi, Country]

when I do the the followings:

le = LabelEncoder()
df['LabelEncodedA'] = df['A'].apply(le.fit_transform)

it returns:

                              A LabelEncodedA
0       [OK, NG, Repair, Peace]  [1, 0, 3, 2]
1    [Sky, NG, Fixed, Conflict]  [1, 3, 2, 0]
2  [Crossed, OK, Engine, Peace]  [0, 2, 1, 3]
3     [OK, Beats, RPi, Country]  [2, 0, 3, 1]

Which is not the intended result. Here each row is LabelEncoded in isolation. And a string e.g. 'OK' in the first row is not encoded to as the one in third or fourth row. Ideally I would like to have them encoded globally across rows. Perhaps one way may be to create a corpus out of that column, and using Tokenization or LabelEncoding obtain a mapping to encode manually the lists? How to convert then in pandas column containing list of strings to a corpus text? Or are there any better approaches?

Expected result (hypothetical):

                              A LabelEncodedA
0       [OK, NG, Repair, Peace]  [0, 1, 2, 3]
1    [Sky, NG, Fixed, Conflict]  [4, 1, 5, 6]
2  [Crossed, OK, Engine, Peace]  [7, 0, 8, 9]
3     [OK, Beats, RPi, Country]  [0, 10, 11, 12]

Answer 1

One approach would be to explode the column, then factorize to encode the column as categorical variable, then group the encoded column and aggregate using list

a = df['A'].explode()
a[:] = a.factorize()[0]
df['Encoded'] = a.groupby(level=0).agg(list)

Result

                              A         Encoded
0       [OK, NG, Repair, Peace]    [0, 1, 2, 3]
1    [Sky, NG, Fixed, Conflict]    [4, 1, 5, 6]
2  [Crossed, OK, Engine, Peace]    [7, 0, 8, 3]
3     [OK, Beats, RPi, Country]  [0, 9, 10, 11]