Change values in pandas dataframe according to value_counts()

Question

I have following pandas dataframe :

import pandas as pd 
from pandas import Series, DataFrame

data = DataFrame({\'Qu1\': [\'apple\', \'potato\', \'cheese\',         


        

Answer 1

You could:

value_counts = df.apply(lambda x: x.value_counts())

         Qu1  Qu2  Qu3
apple    1.0  3.0  1.0
banana   2.0  4.0  NaN
cheese   3.0  NaN  3.0
egg      1.0  NaN  1.0
potato   2.0  NaN  3.0
sausage  NaN  2.0  1.0

Then build a dictionary that will contain the replacements for each column:

import cycle
replacements = {}
for col, s in value_counts.items():
    if s[s<2].any():
        replacements[col] = dict(zip(s[s < 2].index.tolist(), cycle(['other'])))

replacements
{'Qu1': {'egg': 'other', 'apple': 'other'}, 'Qu3': {'egg': 'other', 'apple': 'other', 'sausage': 'other'}}

Use the dictionary to replace the values:

df.replace(replacements)

      Qu1      Qu2     Qu3
0   other  sausage   other
1  potato   banana  potato
2  cheese    apple   other
3  banana    apple  cheese
4  cheese    apple  cheese
5  banana  sausage  potato
6  cheese   banana  cheese
7  potato   banana  potato
8   other   banana   other

or wrap the loop in a dictionary comprehension:

from itertools import cycle

df.replace({col: dict(zip(s[s < 2].index.tolist(), cycle(['other']))) for col, s in value_counts.items() if s[s < 2].any()})

However, this is not only more cumbersome but also slower than using .where. Testing with 3,000 columns:

df = pd.concat([df for i in range(1000)], axis=1)

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 9 entries, 0 to 8
Columns: 3000 entries, Qu1 to Qu3
dtypes: object(3000)

Using .replace():

%%timeit
value_counts = df.apply(lambda x: x.value_counts())
df.replace({col: dict(zip(s[s < 2].index.tolist(), cycle(['other']))) for col, s in value_counts.items() if s[s < 2].any()})

1 loop, best of 3: 4.97 s per loop

vs .where():

%%timeit
df.where(df.apply(lambda x: x.map(x.value_counts()))>=2, "other")

1 loop, best of 3: 2.01 s per loop

Answer 2

I would create a dataframe of same shape where the corresponding entry is the value count:

data.apply(lambda x: x.map(x.value_counts()))
Out[229]: 
   Qu1  Qu2  Qu3
0    1    2    1
1    2    4    3
2    3    3    1
3    2    3    3
4    3    3    3
5    2    2    3
6    3    4    3
7    2    4    3
8    1    4    1

And, use the results in df.where to return "other" where the corresponding entry is smaller than 2:

data.where(data.apply(lambda x: x.map(x.value_counts()))>=2, "other")

      Qu1      Qu2     Qu3
0   other  sausage   other
1  potato   banana  potato
2  cheese    apple   other
3  banana    apple  cheese
4  cheese    apple  cheese
5  banana  sausage  potato
6  cheese   banana  cheese
7  potato   banana  potato
8   other   banana   other