How to take a multi-indexed DataFrame to a nested dictionary structure?

Question

How can you make a dataframe that has a multi-index and make it into a nice nested dictionary?

Here's what I've tried so far, and it's close, however, the keys are tuples. Looking to break those out into more dictionary keys.

What I've Tried:

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
    'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
    'Type':['100','4','7','101','100','100','4','7'],
    'time':[np.linspace(0,10,2) for i in range(8)]}

nn = pd.DataFrame(that)
nn = nn.set_index(['Food','Color','Type'])
vv = {}
for idx in nn.index:
   vv[idx] = nn.loc[idx]

vv
Out[1]: 
{('Apple', 'Red', '100'): time    [0.0, 10.0]
 Name: (Apple, Red, 100), dtype: object,
 ('Apple', 'Green', '4'): time    [0.0, 10.0]
 Name: (Apple, Green, 4), dtype: object,
 ('Apple', 'Yellow', '7'): time    [0.0, 10.0]
 Name: (Apple, Yellow, 7), dtype: object,
 ('Apple', 'Red', '101'): time    [0.0, 10.0]
 Name: (Apple, Red, 101), dtype: object,
 ('Banana', 'Red', '100'): time    [0.0, 10.0]
 Name: (Banana, Red, 100), dtype: object,
 ('Banana', 'Green', '100'): time    [0.0, 10.0]
 Name: (Banana, Green, 100), dtype: object,
 ('Orange', 'Green', '4'): time    [0.0, 10.0]
 Name: (Orange, Green, 4), dtype: object,
 ('Orange', 'Yellow', '7'): time    [0.0, 10.0]
 Name: (Orange, Yellow, 7), dtype: object}

What I want the output to look like.

vv = {'Apple':{'Red':{'100':[0,10],'101':[0,10]},
               'Green':{'4':[0,10]},
               'Yellow':{'7':[0,10]}},
      'Banana':{'Red':{'100':[0,10]},
                'Green':{'100':[0,10]}}
      'Orange':{'Green':{'4':[0,10]},
                'Yellow':{'7':[0,10]}}}

Edit: Changed the range back to 8... was a typo, and changed number of points in linspace to just be 2 points for simplicity to reflect the example.

Edit 2: Looking for a general way to do this. In particular, a colleague of mine has written a treeView model in pyqt that accepts a nested dictionary for the tree. I just want to be able to take the dataframes that I have created to be quickly transformed into the format needed.

For those curious on how to do this in general, here you go. Nice little function I wrote. Works more for what I need.

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
        'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
        'Type':['100','4','7','101','100','100','4','7'],
        'time':[np.linspace(0,10,2) for i in range(8)]}

x = pd.DataFrame(that)

def NestedDict_fromDF(iDF,keyorder,values):
    if not isinstance(keyorder,list):
        keyorder = [keyorder]
    if not isinstance(values,list):
        values = [values]
    for i in reversed(range(len(keyorder))):
        if keyorder[i] not in iDF:
            keyorder.pop(i)
    for i in reversed(range(len(values))):
        if values[i] not in iDF:
            values.pop(i)
    rdict = {}
    if keyorder:
        ndf = iDF.set_index(keyorder)
        def makeDict(basedict,group):
            for k,g in group:
                basedict[k] = {}
                try:
                    makeDict(basedict[k], g.droplevel(0).groupby(level=0))
                except:
                    if values:
                        basedict[k] = g[values].reset_index(drop=True)
                    else:
                        basedict[k] = []
            return basedict
        rdict = makeDict({}, ndf.groupby(level=0))
    return rdict

yy = NestedDict_fromDF(x,['Food','Color','Type','Integer'],['time'])


{'Apple': {'Green': {'4':DataFrame},
           'Red': {'100':DataFrame,
                   '101':DataFrame},
           'Yellow': {'7':DataFrame}},
 'Banana': {'Green': {'100':DataFrame},
            'Red': {'100':DataFrame}},
 'Orange': {'Green': {'4':DataFrame},
            'Yellow': {'7':DataFrame}}}

the dataframe nn is not reproducible , it throws a ValueError: arrays must all be same length error — anky
– anky, Commented Feb 8, 2020 at 7:22
Change the range(2) to range(8), and for the purpose of example changing linspace to [0.0, 0.10] would be better. — Sayandip Dutta
– Sayandip Dutta, Commented Feb 8, 2020 at 7:23

Sayandip Dutta · Accepted Answer · 2020-02-08 08:51:59Z

1

It grew too complex, too fast:

from pprint import pprint
import pandas as pd

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
    'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
    'Type':['100','4','7','101','100','100','4','7'],
    'time':[np.linspace(0,10,2) for i in range(8)]}

nn = pd.DataFrame(that)
df = nn.groupby(['Food', 'Color', 'Type']).agg(list)
d = {}
new_df = df.groupby(level=[0,1]).apply(lambda df:df.xs(df.name).to_dict()).to_dict() #[1]
for (food, color), v in new_df.items():
    if not food in d:
        d[food] = {color: {Type: time[0].tolist() for Type, time in v['time'].items()}}
    else:
        d[food][color] = {Type: time[0].tolist() for Type, time in v['time'].items()}
pprint(d)

Output:

{'Apple': {'Green': {'4': [0.0, 10.0]},
           'Red': {'100': [0.0, 10.0], '101': [0.0, 10.0]},
           'Yellow': {'7': [0.0, 10.0]}},
 'Banana': {'Green': {'100': [0.0, 10.0]}, 'Red': {'100': [0.0, 10.0]}},
 'Orange': {'Green': {'4': [0.0, 10.0]}, 'Yellow': {'7': [0.0, 10.0]}}}

[1] taken from: DataFrame with MultiIndex to dict

Huh! got it finally!

that = {'Food':['Apple','Apple','Apple','Apple','Banana','Banana','Orange','Orange'],
    'Color':['Red','Green','Yellow','Red','Red','Green','Green','Yellow'],
    'Type':['100','4','7','101','100','100','4','7'],
    'time':[np.linspace(0,10,2) for i in range(8)]}

nn = pd.DataFrame(that)
nn = nn.set_index(['Food','Color','Type'])
group = nn.groupby(level=0)
d = {k: g.droplevel(0).groupby(level=0)
       .apply(lambda df:df.xs(df.name)['time']
       .apply(lambda x:x.tolist()).to_dict())
       .to_dict() for k,g in group}
pprint(d)

{'Apple': {'Green': {'4': [0.0, 10.0]},
           'Red': {'100': [0.0, 10.0], '101': [0.0, 10.0]},
           'Yellow': {'7': [0.0, 10.0]}},
 'Banana': {'Green': {'100': [0.0, 10.0]}, 'Red': {'100': [0.0, 10.0]}},
 'Orange': {'Green': {'4': [0.0, 10.0]}, 'Yellow': {'7': [0.0, 10.0]}}}

edited Feb 8, 2020 at 8:51

answered Feb 8, 2020 at 7:49

Sayandip Dutta

15.9k4 gold badges27 silver badges57 bronze badges

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3 Comments

Carl Over a year ago

I have seen that solution in your [1]. I like your answer, but in actuality I'm looking for a general way to do this. This example was just a boiled down version of what I need.

Sayandip Dutta Over a year ago

@Carl I think I got it, please see if it works for you.

Carl Over a year ago

@Syandip Dutta Made a function that will do this in general in the last edit, just in case you were curious.

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