pandas group by remove outliers

Question

I want to remove outliers based on percentile 99 values by group wise.

 import pandas as pd
 df = pd.DataFrame({\'Group\': [\'A\',\'A\',\'A\',\'B\',\'B\',\'         


        

Answer 1

I don't think you want to use quantile, as you'll exclude your lower values:

import pandas as pd
df = pd.DataFrame({'Group': ['A','A','A','B','B','B','B'], 'count': [1.1,11.2,1.1,3.3,3.40,3.3,100.0]})
print(pd.DataFrame(df.groupby('Group').quantile(.01)['count']))

output:

       count
Group       
A        1.1
B        3.3

Those aren't outliers, right? So you wouldn't want to exclude them.

You could try setting left and right limits by using standard deviations from the median maybe? This is a bit verbose, but it gives you the right answer:

left = pd.DataFrame(df.groupby('Group').median() - pd.DataFrame(df.groupby('Group').std()))
right = pd.DataFrame(df.groupby('Group').median() + pd.DataFrame(df.groupby('Group').std()))

left.columns = ['left']
right.columns = ['right']

df = df.merge(left, left_on='Group', right_index=True)
df = df.merge(right, left_on='Group', right_index=True)

df = df[(df['count'] > df['left']) & (df['count'] < df['right'])]
df = df.drop(['left', 'right'], axis=1)
print(df)

output:

  Group  count
0     A    1.1
2     A    1.1
3     B    3.3
4     B    3.4
5     B    3.3

Answer 2

Here is my solution:

def is_outlier(s):
    lower_limit = s.mean() - (s.std() * 3)
    upper_limit = s.mean() + (s.std() * 3)
    return ~s.between(lower_limit, upper_limit)

df = df[~df.groupby('Group')['count'].apply(is_outlier)]

You can write your own is_outlier function