问题
How can a 1:1 stratified sampling be performed in python?
Assume the Pandas Dataframe df to be heavily imbalanced. It contains a binary group and multiple columns of categorical sub groups.
df = pd.DataFrame({'id':[1,2,3,4,5], 'group':[0,1,0,1,0], 'sub_category_1':[1,2,2,1,1], 'sub_category_2':[1,2,2,1,1], 'value':[1,2,3,1,2]})
display(df)
display(df[df.group == 1])
display(df[df.group == 0])
df.group.value_counts()
For each member of the main group==1 I need to find a single match of group==0 with.
A StratifiedShuffleSplit from scikit-learn will only return a random portion of data, not a 1:1 match.
回答1:
If I understood correctly you could use np.random.permutation:
import numpy as np
import pandas as pd
np.random.seed(42)
df = pd.DataFrame({'id': [1, 2, 3, 4, 5], 'group': [0, 1, 0, 1, 0], 'sub_category_1': [1, 2, 2, 1, 1],
'sub_category_2': [1, 2, 2, 1, 1], 'value': [1, 2, 3, 1, 2]})
# create new column with an identifier for a combination of categories
columns = ['sub_category_1', 'sub_category_2']
labels = df.loc[:, columns].apply(lambda x: ''.join(map(str, x.values)), axis=1)
values, keys = pd.factorize(labels)
df['label'] = labels.map(dict(zip(keys, values)))
# build distribution of sub-categories combinations
distribution = df[df.group == 1].label.value_counts().to_dict()
# select from group 0 only those rows that are in the same sub-categories combinations
mask = (df.group == 0) & (df.label.isin(distribution))
# do random sampling
selected = np.ravel([np.random.permutation(group.index)[:distribution[name]] for name, group in df.loc[mask].groupby(['label'])])
# display result
result = df.drop('label', axis=1).iloc[selected]
print(result)
Output
group id sub_category_1 sub_category_2 value
4 0 5 1 1 2
2 0 3 2 2 3
Note that this solution assumes the size of the each possible sub_category combination of group 1 is less than the size of the corresponding sub-group in group 0. A more robust version involves using np.random.choice with replacement:
selected = np.ravel([np.random.choice(group.index, distribution[name], replace=True) for name, group in df.loc[mask].groupby(['label'])])
The version with choice does not have the same assumption as the one with permutation, although it requires at least one element for each sub-category combination.
来源：https://stackoverflow.com/questions/54653317/python-11-stratified-sampling-per-each-group