pandas groupby to nested json

I don't think think there is anything built-in to pandas to create a nested dictionary of the data. Below is some code that should work in general for a series with a MultiIndex, using a defaultdict

The nesting code iterates through each level of the MultIndex, adding layers to the dictionary until the deepest layer is assigned to the Series value.

In  [99]: from collections import defaultdict

In [100]: results = defaultdict(lambda: defaultdict(dict))

In [101]: for index, value in grouped.itertuples():
     ...:     for i, key in enumerate(index):
     ...:         if i == 0:
     ...:             nested = results[key]
     ...:         elif i == len(index) - 1:
     ...:             nested[key] = value
     ...:         else:
     ...:             nested = nested[key]

In [102]: results
Out[102]: defaultdict(<function <lambda> at 0x7ff17c76d1b8>, {2010: defaultdict(<type 'dict'>, {'govnr': {'pati mara': 500.0, 'jess rapp': 80.0}, 'mayor': {'joe smith': 100.0, 'jay gould': 12.0}})})

In [106]: print json.dumps(results, indent=4)
{
    "2010": {
        "govnr": {
            "pati mara": 500.0, 
            "jess rapp": 80.0
        }, 
        "mayor": {
            "joe smith": 100.0, 
            "jay gould": 12.0
        }
    }
}

I had a look at the solution above and figured out that it only works for 3 levels of nesting. This solution will work for any number of levels.

import json
levels = len(grouped.index.levels)
dicts = [{} for i in range(levels)]
last_index = None

for index,value in grouped.itertuples():

    if not last_index:
        last_index = index

    for (ii,(i,j)) in enumerate(zip(index, last_index)):
        if not i == j:
            ii = levels - ii -1
            dicts[:ii] =  [{} for _ in dicts[:ii]]
            break

    for i, key in enumerate(reversed(index)):
        dicts[i][key] = value
        value = dicts[i]

    last_index = index


result = json.dumps(dicts[-1])

Here is a generic recursive solution for this problem:

def df_to_dict(df):
    if df.ndim == 1:
        return df.to_dict()

    ret = {}
    for key in df.index.get_level_values(0):
        sub_df = df.xs(key)
        ret[key] = df_to_dict(sub_df)
    return ret

I'm aware this is an old question, but I came across the same issue recently. Here's my solution. I borrowed a lot of stuff from chrisb's example (Thank you!).

This has the advantage that you can pass a lambda to get the final value from whatever enumerable you want, as well as for each group.

from collections import defaultdict

def dict_from_enumerable(enumerable, final_value, *groups):
    d = defaultdict(lambda: defaultdict(dict))
    group_count = len(groups)
    for item in enumerable:
        nested = d
        item_result = final_value(item) if callable(final_value) else item.get(final_value)
        for i, group in enumerate(groups, start=1):
            group_val = str(group(item) if callable(group) else item.get(group))
            if i == group_count:
                nested[group_val] = item_result
            else:
                nested = nested[group_val]
    return d

In the question, you'd call this function like:

dict_from_enumerable(grouped.itertuples(), 'amount', 'year', 'office', 'candidate')

The first argument can be an array of data as well, not even requiring pandas.

pandas groupby to nested json

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