pandas aggregate items as list and filter based on legth

Question

Let me show with an example, I have this dataframe:

I want to end up with to this dataframe: (so I group by "column_1" and "last_column" and I aggregate by "column_2" to get the items as a list)

If you notice, when column_1 = ‘yes’ it doesn’t appear that row, SINCE THE LENGTH OF THE RESULT IS 1.

I’m able to filter and aggregate as a list separately, but not both together…

df.groupby(
['column_1', 'last_column']
)['column_2'].agg(list).filter(lambda x : len(x)<2)

I’m getting the following error:

Dataframe:

import pandas as pd

data = {'column_1': ['no', 'no', 'no', 'no', 'yes', 'yes', 'yes', 'yes'], 
    'column_2': ['spain', 'france', 'italy', 'germany', 'spain', 'france', 'italy', 'germany'], 
    "last_column": ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B']}

df = pd.DataFrame.from_dict(data)

Asked By: Tonino Fernandez

||

Source

Answer 1

How about this?:

df2 = df.groupby(
['column_1', 'last_column'], as_index=False
)['column_2'].agg(list)
df2['len'] = df2['column_2'].apply(lambda x : len(x))
df2 = df2[df2['len']<=2]
df2

Answered By: gtomer

Answer 2

You can try:

df_out = df.groupby(["column_1", "last_column"])["column_2"].agg(list)
df_out = df_out.groupby(level=0).filter(lambda y: len(y) > 1)

print(df_out.reset_index())

Prints:

  column_1 last_column          column_2
0       no           A  [italy, germany]
1       no           B   [spain, france]

Dataframe used:

  column_1 column_2 last_column
0       no    spain           B
1       no   france           B
2       no    italy           A
3       no  germany           A
4      yes    spain           B
5      yes   france           B
6      yes    italy           B
7      yes  germany           B

Answered By: Andrej Kesely

pandas aggregate items as list and filter based on legth

Question:

Answers: