How can I convert a Pandas DataFrame to a three level nested dictionary?

Question

How can I convert a Pandas DataFrame to a three level nested dictionary using column names?

The columns are not first three columns and I want it to group by column artist then group by column album, and I need it to be case insensitive, preferably without using defaultdict.

This is a minimal reproducible example:

from collections import defaultdict                                               
from itertools import product                                                     
from pandas import DataFrame                                                      
tree = defaultdict(lambda: defaultdict(dict))                                     
columns = {'a': str(), 'b': str(), 'c': str(), 'd': int(), 'e': int(), 'f': int()}
df = DataFrame(columns, index=[])                                                 
for i, j, k in product('abcd', repeat=3):                                         
    tree[i][j][k] = list(map('abcd'.index, (i, j, k)))                            
    df.loc[len(df)] = [i, j, k, *list(map('abcd'.index, (i, j, k)))]

How can I get a nested dictionary similar to tree from df?

I am really sorry I can provide any actual examples because they wouldn't be minimal.

I tried to use .groupby() but I only ever saw it being used with one column and I really don't know what to do with the pandas.core.groupby.generic.DataFrameGroupBy object it returns, I just started using it today.

Currently I can do this:

tree1 = dict()                                                                                  
for index, row in df.iterrows():                                                                
    if not tree1.get(row['a'].lower()):                                                         
        tree1[row['a'].lower()] = dict()                                                        
    if not tree1[row['a'].lower()].get(row['b'].lower()):                                       
        tree1[row['a'].lower()][row['b'].lower()] = dict()                                      
    tree1[row['a'].lower()][row['b'].lower()][row['c'].lower()] = [row['d'], row['e'], row['f']]

I actually implemented case insensitive str and dict but for the sake of brevity (they are very long) I wouldn't use it here.

But according to this answer https://stackoverflow.com/a/55557758/16383578 such method is bad, what is a better way?

Timus · Accepted Answer · 2021-07-28 19:53:33Z

I would probably do it like this:

cols = ['a', 'b', 'c']
for col in cols:
    df[col] = df[col].str.casefold()
tree = {}
for (a, b, c), values in (df.set_index(cols).T.to_dict(orient='list')
                            .items()):
    tree.setdefault(a, {}).setdefault(b, {})[c] = values

or

...
for (a, b, c), values in (df.set_index(cols).apply(list, axis=1)
                            .to_dict()).items():
    tree.setdefault(a, {}).setdefault(b, {})[c] = values

This produces the same result (when the first part that casefolds is included)

def to_dict(df):
    return df.set_index(df.columns[0]).iloc[:, 0].to_dict()

df['values'] = df[['d', 'e', 'f']].apply(list, axis=1)
df = df[['a', 'b', 'c', 'values']]
tree = (df.set_index(['a', 'b'])
          .groupby(['a', 'b']).apply(to_dict)
          .reset_index('b')
          .groupby('a').apply(to_dict)
          .to_dict())

but I think it's a bit too convoluted.

Results:

{'a': {'a': {'a': [0, 0, 0], 'b': [0, 0, 1], 'c': [0, 0, 2], 'd': [0, 0, 3]},
       'b': {'a': [0, 1, 0], 'b': [0, 1, 1], 'c': [0, 1, 2], 'd': [0, 1, 3]},
       'c': {'a': [0, 2, 0], 'b': [0, 2, 1], 'c': [0, 2, 2], 'd': [0, 2, 3]},
       'd': {'a': [0, 3, 0], 'b': [0, 3, 1], 'c': [0, 3, 2], 'd': [0, 3, 3]}},
 'b': {'a': {'a': [1, 0, 0], 'b': [1, 0, 1], 'c': [1, 0, 2], 'd': [1, 0, 3]},
       'b': {'a': [1, 1, 0], 'b': [1, 1, 1], 'c': [1, 1, 2], 'd': [1, 1, 3]},
       'c': {'a': [1, 2, 0], 'b': [1, 2, 1], 'c': [1, 2, 2], 'd': [1, 2, 3]},
       'd': {'a': [1, 3, 0], 'b': [1, 3, 1], 'c': [1, 3, 2], 'd': [1, 3, 3]}},
 'c': {'a': {'a': [2, 0, 0], 'b': [2, 0, 1], 'c': [2, 0, 2], 'd': [2, 0, 3]},
       'b': {'a': [2, 1, 0], 'b': [2, 1, 1], 'c': [2, 1, 2], 'd': [2, 1, 3]},
       'c': {'a': [2, 2, 0], 'b': [2, 2, 1], 'c': [2, 2, 2], 'd': [2, 2, 3]},
       'd': {'a': [2, 3, 0], 'b': [2, 3, 1], 'c': [2, 3, 2], 'd': [2, 3, 3]}},
 'd': {'a': {'a': [3, 0, 0], 'b': [3, 0, 1], 'c': [3, 0, 2], 'd': [3, 0, 3]},
       'b': {'a': [3, 1, 0], 'b': [3, 1, 1], 'c': [3, 1, 2], 'd': [3, 1, 3]},
       'c': {'a': [3, 2, 0], 'b': [3, 2, 1], 'c': [3, 2, 2], 'd': [3, 2, 3]},
       'd': {'a': [3, 3, 0], 'b': [3, 3, 1], 'c': [3, 3, 2], 'd': [3, 3, 3]}}}

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