CODEWITHSUNDEEP
pythondictionaryscikit-learnmapping
yatemo
yatemo

Reputation: 11

See the mappings of a LabelEncoder

I face a problem concerning the LabelEncoder. I applied it to a data set as follows: 

data_set1 = data_set.apply(LabelEncoder().fit_transform)

... and it worked. However, now I want to get the mapping of the LabelEncoder. Therefore I used the following:

le = preprocessing.LabelEncoder()
le.fit(data_set1['column'])
le_name_mapping = dict(zip(le.classes_, le.transform(le.classes_)))
print(le_name_mapping)

I was expecting a dictionary that would look like the following:

{apple: 0, banana: 1, kiwi: 2}

and so on... Instead the output was the following:

{0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9}

Do you guys have any idea why and how to fix it?

Upvotes: 1

Views: 739

Answers (1)

Wajsbrot
Wajsbrot

Reputation: 341

I think this simple piece of code: 

data = ['apple', 'banana', 'kiwi', 'apple']
le = LabelEncoder()
le.fit(data)
le.classes_

outputs what you want: array(['apple', 'banana', 'kiwi'], dtype='<U6'). The first item corresponds to label 0, the second one is labelled as 1, etc.

If you want the corresponding dictionary you can get it with labels_dict = {index: value for index, value in enumerate(le.classes_)}, such that labels_dict is {0: 'apple', 1: 'banana', 2: 'kiwi'}.

Upvotes: 4

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