Is it possible to apply sklearn.preprocessing.LabelEncoder() on a 2D list?

Question

Say I have a list as given below:

l = [
       ['PER', 'O', 'O', 'GEO'],
       ['ORG', 'O', 'O', 'O'],
       ['O', 'O', 'O', 'GEO'],
       ['O', 'O', 'PER', 'O']
    ]

I want to encode the 2D list with LabelEncoder().

It should look something like:

l = [
       [1, 0, 0, 2],
       [3, 0, 0, 0],
       [0, 0, 0, 2],
       [0, 0, 1, 0]
    ]

Is it possible? If not, is there any workaround?

Thanks in advance!

What is your expected output? The question is unclear. – G. Anderson Apr 20 '21 at 18:43 — G. Anderson, Apr 20 '21 at 18:43
I just updated the question. Thanks for pointing out. – thenocturnalguy Apr 20 '21 at 18:46 — thenocturnalguy, Apr 20 '21 at 18:46

score 1 · Accepted Answer · answered Apr 20 '21 at 18:54

You can flatten the list, fit the encoder with all the potential values and then use the encoder to transform each sublist, as shown below:

from sklearn.preprocessing import LabelEncoder

l = [
       ['PER', 'O', 'O', 'GEO'],
       ['ORG', 'O', 'O', 'O'],
       ['O', 'O', 'O', 'GEO'],
       ['O', 'O', 'PER', 'O']
    ]

flattened_l = [e for sublist in l for e in sublist]

# flattened_l is ['PER', 'O', 'O', 'GEO', 'ORG', 'O', 'O', 'O', 'O', 'O', 'O', 'GEO', 'O', 'O', 'PER', 'O']

le = LabelEncoder().fit(flattened_l)

# See the mapping generated by the encoder:
list(enumerate(le.classes_))
# [(0, 'GEO'), (1, 'O'), (2, 'ORG'), (3, 'PER')]

# And, finally, transform each sublist:
res = [list(le.transform(sublist)) for sublist in l]
res

# Getting the result you want:
# [[3, 1, 1, 0], [2, 1, 1, 1], [1, 1, 1, 0], [1, 1, 3, 1]]

Is it possible to apply sklearn.preprocessing.LabelEncoder() on a 2D list?

1 Answers1