1
\$\begingroup\$

Here is my dataframe:

data = [['a1','b1',0], ['a2','b3',0], ['a1','b2',1], ['a1','b1',1], ['a2','b3',0]]
df = pd.DataFrame(data=data, columns = ['A','B','label'])

Except for 'label' column, each col is categorical value (string). I want to replace (numeric) values by their frequency of label = 1, e.g.:

n(a1) =  count(A == 'a1' & label = 1)/count(A == 'a1')

I used a very stupid way by iterating columns to create a dictionary, then replace df through dictionary. Is there any more simply way?

dic = {}
for col, value in df.iteritems():
    if col != 'label':
        for cat in value.unique():
            count = df[value == cat].shape[0]
            positive = df[(value == cat) & (df['label'] == 1)].shape[0]
            dic[cat] = (positive) / count
df.replace(dic, inplace=True)

My question is to make the code concise since I naively iterate cols and values. I believe pandas has a lot of convenient functions to achieve this.

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

Avoid for-loops and avoid unique() in this case. Fundamentally you're doing a grouped count, so use Pandas built-in grouping support which is vectorised. Since your numerator is filtering on label, after grouping you need to join (merge) and fillna on missing values that had no label=1.

Don't construct a dic manually. Once you have a replacement frame with a proper index based on the original values, you can just to_dict().

import pandas as pd

df = pd.DataFrame({
    'A':     ('a1', 'a2', 'a1', 'a1', 'a2'),
    'B':     ('b1', 'b3', 'b2', 'b1', 'b3'),
    'label': (   0,    0,    1,    1,    0),
})


def make_counts(col: str) -> pd.Series:
    grouped = df.groupby([col, 'label'])[col].count()
    positive = grouped.loc[:, 1].groupby(level=col).sum().rename('positive')
    count = grouped.groupby(level=col).sum().rename('count_')
    fractions = pd.merge(
        positive, count, how='right', left_index=True, right_index=True,
    )
    replacement = (fractions.positive.fillna(0) / fractions.count_).to_dict()
    return df[col].replace(replacement)


for col in ('A', 'B'):
    df[col] = make_counts(col)

print(df)
'''
          A    B  label
0  0.666667  0.5      0
1  0.000000  0.0      0
2  0.666667  1.0      1
3  0.666667  0.5      1
4  0.000000  0.0      0
'''

Assuming that "whatever you're actually doing" still uses only 0 or 1 for your labels, you should actually re-interpret this as a grouped mean rather than a grouped count:

import pandas as pd

df = pd.DataFrame({
    'A':     ('a1', 'a2', 'a1', 'a1', 'a2'),
    'B':     ('b1', 'b3', 'b2', 'b1', 'b3'),
    'label': (   0,    0,    1,    1,    0),
})


def make_counts(col: str) -> pd.Series:
    return df.groupby(col).label.transform('mean')


for col in ('A', 'B'):
    df[col] = make_counts(col)
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.