Pandas : Add arrays as values of column

Question

I have a Pandas DataFrame df that stores some numeric values :

print(df)

       value 
0          0
1          2
2          4
3          5
4          8

And I have a function that converts a numerical value to a one-hot vector

print(to_categorical(0))
[1 0 0 0 0 0 0 0 0 0]

print(to_categorical(5))
[0 0 0 0 0 5 0 0 0 0]

etc...

So, I can call my function over my columns of numeric value :

print(to_categorical(df['value'))

[[1 0 0 0 0 0 0 0 0 0]
 [0 0 1 0 0 0 0 0 0 0]
 [0 0 0 0 1 0 0 0 0 0]
 [0 0 0 0 0 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 1 0]]

And now I want to store my results as a new column. Here is what I expect from my example :

df['one-hot'] = to_categorical(df['value')
print(df)

        value                    one-hot
0          0       [1 0 0 0 0 0 0 0 0 0]
1          2       [0 0 1 0 0 0 0 0 0 0]
2          4       [0 0 0 0 1 0 0 0 0 0]
3          5       [0 0 0 0 0 1 0 0 0 0]
4          8       [0 0 0 0 0 0 0 0 1 0]

But this give me an error since pandas tries to flatten my array into multiple colums. How can I do that ?

Possible duplicate of How do I get a DataFrame Index / Series column as an array or list? — Georgy
– Georgy, Commented Mar 28, 2019 at 11:08

jezrael · Accepted Answer · 2019-03-28 09:50:44Z

3

First I think working with lists in pandas is not good idea, but is is possible by convert to lists:

df['one-hot'] = to_categorical(df['value').tolist()

answered Mar 28, 2019 at 9:50

jezrael

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3 Comments

Nakeuh Over a year ago

In my case, I want to have a structure that stores a mapping between my values (few thousands of unique values), and the corresponding one-hot vectors (so vectors of few thousands of values) What do you think would be a better approach ?

jezrael Over a year ago

@Nakeuh - Better is create new DataFrame - df1 = pd.DataFrame(to_categorical(df['value'), index=df.index)

Nakeuh Over a year ago

Hm I see. I suppose that the 'all in one DataFrame' is less computing efficient ? I think I will stay with the non-efficient way for now (I prefer having only one object and I am not really concerned about efficiency in my use case), but I keep in mind your suggestion for the future. Thanks !

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