Group values of a chosen column into a list when creating a dictionary from a pandas data frame with a non-unique index

user_13506329 0
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I have a dataframe that looks like his

_____________________
|col1 | col2 | col3 |
---------------------
| a   | b    | c    |
| d   | b    | c    |
| e   | f    | g    |
| h   | f    | j    |
---------------------

I want to get a dictionary structure that looks as follows

{
    b : { col1: [a,d], col2: b, col3: c},
    f : { col1: [e, h], col2: f, col3: [g, j]}
}

I have seen this answer. But it seems like overkill for what I want to do as it converts every value of the key inside the nested dictionary into a list. I would only like to convert col1 into a list when creating the dictionary. Is this possible?

pandas dataframe python-3.x python
1 Answers

Use custom lambda function for return unique values in list if there is multiple them else scalar in lambda function:

d = (df.set_index('col2', drop=False)
       .groupby(level=0)
       .agg(lambda x: list(set(x)) if len(set(x)) > 1 else list(set(x))[0])
       .to_dict('index'))
print (d)
{'b': {'col1': ['d', 'a'], 'col2': 'b', 'col3': 'c'}, 
 'f': {'col1': ['h', 'e'], 'col2': 'f', 'col3': ['j', 'g']}}

If order is important use dict.fromkeys for remove duplicates:

d = (df.set_index('col2', drop=False)
       .groupby(level=0)
       .agg(lambda x: list(dict.fromkeys(x)) if len(set(x)) > 1 else list(set(x))[0])
       .to_dict('index'))
print (d)
{'b': {'col1': ['a', 'd'], 'col2': 'b', 'col3': 'c'},
 'f': {'col1': ['e', 'h'], 'col2': 'f', 'col3': ['g', 'j']}}
user_29010020
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