Don Smythe
Don Smythe

Reputation: 9814

pandas convert text feature to numeric value

I can convert all text features in a pandas dataframe by casting to 'category' using the df.astype() method as below. However I find category hard to work with (eg for plotting data) and would prefer to create a new column of integers

#convert all objects to categories
object_types = dataset.select_dtypes(include=['O'])
for col in object_types:
    dataset['{0}_category'.format(col)] = dataset[col].astype('category')

I can convert the text to integers using this hack:

#convert all objects to int values
object_types = dataset.select_dtypes(include=['O'])

new_cols = {}
for col in object_types:
    data_set = set(dataset[col].tolist())
    data_indexed = {}
    for i, item in enumerate(data_set):
        data_indexed[item] = i
    new_list = []
    for item in dataset[col].tolist():
        new_list.append(data_indexed[item])
    new_cols[col]=new_list

for key, val in new_cols.items():
    dataset['{0}_int_value'.format(key)] = val

But is there a better (or existing) way to do the same?

Upvotes: 11

Views: 16716

Answers (2)

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210982

I would use factorize method, which is designed for this particular task:

In [90]: x
Out[90]:
    A  B
9   c  z
10  c  z
4   b  x
5   b  y
1   a  w
7   b  z

In [91]: x.apply(lambda col: pd.factorize(col, sort=True)[0])
Out[91]:
    A  B
9   2  3
10  2  3
4   1  1
5   1  2
1   0  0
7   1  3

or:

In [92]: x.apply(lambda col: pd.factorize(col)[0])
Out[92]:
    A  B
9   0  0
10  0  0
4   1  1
5   1  2
1   2  3
7   1  0

Upvotes: 15

piRSquared
piRSquared

Reputation: 294576

consider df

df = pd.DataFrame(dict(A=list('aaaabbbbcccc'),
                       B=list('wwxxxyyzzzzz')))

df

enter image description here

you can convert to integers like this

def intify(s):
    u = np.unique(s)
    i = np.arange(len(u))
    return s.map(dict(zip(u, i)))

or shorter version

def intify(s):
    u = np.unique(s)
    return s.map({k: i for i, k in enumerate(u)})

df.apply(intify)

Or in a single line

df.apply(lambda s: s.map({k:i for i,k in enumerate(s.unique())}))

enter image description here

Upvotes: 4

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