group by key with pandas series and export to_dict() - pandas

Question

I have a dictionary that looks like this:
d = {1:0, 2:0, 3:1, 4:0, 5:2, 6:1, 7:2, 8:0}
And I want to group by .keys() such as I get:
pandas_ordered = { 0:[1,2,4,8], 1:[3,6], 2:[5,7] }
But with this command for
pd.Series(list(d.values())).groupby(list(partition.keys())).to_dict()
Bellow is an example:
# Example:
import pandas as pd
d = {1:0, 2:0, 3:1, 4:0, 5:2, 6:1, 7:2, 8:0}
def pandas_groupby(dictionary):
values = list(dictionary.values())
keys = list(dictionary.keys())
return pd.Series(values).groupby(keys).to_dict()
pandas_groupby(d)
The above code produces the error:
AttributeError: Cannot access callable attribute 'to_dict' of
'SeriesGroupBy' objects, try using the 'apply' method
Any ideas on how to do this?

Answer 1

You dict is already given by the groups in your groupby
d = {1:0, 2:0, 3:1, 4:0, 5:2, 6:1, 7:2, 8:0}
s = pd.Series(d)
s.groupby(s).groups
{0: Int64Index([1, 2, 4, 8], dtype='int64'),
1: Int64Index([3, 6], dtype='int64'),
2: Int64Index([5, 7], dtype='int64')}
But of course, can always agg and customize
s.groupby(s).agg(lambda x: tuple(x.index)).to_dict()
{0: (1, 2, 4, 8), 1: (3, 6), 2: (5, 7)}