Want to create column with lists of unique values using groupby and transform

Question

Here is a sample dataset

test = pd.DataFrame({
    'a' : [1, 2, 3]*2,
    'b' : ['a', 'a', 'b', 'b', 'b', 'b',],
    'c' : [123, 456, 456, 123, 456, 123]
})

print(test)

   a  b    c
0  1  a  123
1  2  a  456
2  3  b  456
3  1  b  123
4  2  b  456
5  3  b  123

If I groupby columns 'a' and 'b' and then try to get a list of unique values ('c') in each group, I don't get the expected results using transform

# using transform
print(test.groupby([
    'a',
    'b',
]).c.transform(pd.Series.unique))

0    123
1    456
2    456
3    123
4    456
5    123

If I use unique instead, I almost get the expected output:

# almost expected output
print(test.groupby([
    'a',
    'b',
]).c.unique())

a  b
1  a         [123]
   b         [123]
2  a         [456]
   b         [456]
3  b    [456, 123]
Name: c, dtype: object

What I was hoping for was a pd.Series that looks like this using transform:

Expected Output

0         [123]
1         [456]
2    [456, 123]
3         [123]
4         [456]
5    [456, 123]
dtype: object

I know that I can use transform to get the nunique values of 'c' as a series doing this:

print(test.groupby([
    'a',
    'b',
]).c.transform(pd.Series.nunique))

0    1
1    1
2    2
3    1
4    1
5    2
Name: c, dtype: int64

Question

Why can't I do something similar with unique and transform?

Side Note

I know that I can do the groupby and unique and then reset_index and merge with the original data, but I'm hoping for a more pythonic/pandas-friendly method.

I also tried using set and transform, but that returned an error.

print(test.groupby([
    'a',
    'b',
]).c.transform(set))

TypeError: 'set' type is unordered

Answer 1

Does

test.groupby(['a','b'])['c'].transform('unique')

work for you?

Output:

0         [123]
1         [456]
2    [456, 123]
3         [123]
4         [456]
5    [456, 123]
Name: c, dtype: object