cartesian product in pandas

In recent versions of Pandas (>= 1.2) this is built into merge so you can do:

from pandas import DataFrame
df1 = DataFrame({'col1':[1,2],'col2':[3,4]})
df2 = DataFrame({'col3':[5,6]})    

df1.merge(df2, how='cross')

This is equivalent to the previous pandas < 1.2 answer but is easier to read.

For pandas < 1.2:

If you have a key that is repeated for each row, then you can produce a cartesian product using merge (like you would in SQL).

from pandas import DataFrame, merge
df1 = DataFrame({'key':[1,1], 'col1':[1,2],'col2':[3,4]})
df2 = DataFrame({'key':[1,1], 'col3':[5,6]})

merge(df1, df2,on='key')[['col1', 'col2', 'col3']]

Output:

   col1  col2  col3
0     1     3     5
1     1     3     6
2     2     4     5
3     2     4     6

See here for the documentation: http://pandas.pydata.org/pandas-docs/stable/merging.html

Use pd.MultiIndex.from_product as an index in an otherwise empty dataframe, then reset its index, and you're done.

a = [1, 2, 3]
b = ["a", "b", "c"]

index = pd.MultiIndex.from_product([a, b], names = ["a", "b"])

pd.DataFrame(index = index).reset_index()

out:

   a  b
0  1  a
1  1  b
2  1  c
3  2  a
4  2  b
5  2  c
6  3  a
7  3  b
8  3  c

Minimal code needed for this one. Create a common 'key' to cartesian merge the two:

df1['key'] = 0
df2['key'] = 0

df_cartesian = df1.merge(df2, how='outer')

This won't win a code golf competition, and borrows from the previous answers - but clearly shows how the key is added, and how the join works. This creates 2 new data frames from lists, then adds the key to do the cartesian product on.

My use case was that I needed a list of all store IDs on for each week in my list. So, I created a list of all the weeks I wanted to have, then a list of all the store IDs I wanted to map them against.

The merge I chose left, but would be semantically the same as inner in this setup. You can see this in the documentation on merging, which states it does a Cartesian product if key combination appears more than once in both tables - which is what we set up.

days = pd.DataFrame({'date':list_of_days})
stores = pd.DataFrame({'store_id':list_of_stores})
stores['key'] = 0
days['key'] = 0
days_and_stores = days.merge(stores, how='left', on = 'key')
days_and_stores.drop('key',1, inplace=True)