将 Dataframe 转换为 Python 中的嵌套字典

Question

I'm looking for a way to convert a dataframe into a dictionary, very similar to what has been asked here:

Convert pandas DataFrame to a nested dict

Assuming a sample data frame

name    v1  v2  v3
0        A  A1  A11 1
1        A  A2  A12 2
2        B  B1  B12 3
3        C  C1  C11 4
4        A  A2  A21 6
5        A  A2  A21 8

The number of columns may differ and so does the column names.

I'm looking to generate:

{
    'A' : { 
        'A1' : { 'A11' : 1 },
        'A2' : { 'A12' : 2 , 'A21' : 6 , 'A21' : 8 },
        'B1' : {}, 
        'C1' : {}
    }, 

    'B' : { 
        'A1' : {},
        'A2' : {},
        'B1' : { 'B12' : 3}, 
        'C1' : {}
    },

    'C' : { 
        'A1' : {},
        'A2' : {},
        'B1' : {} ,
        'C1' : { 'C11' : 4}
    }

}

The method suggested elsewhere is via recursion:

def recur_dictify(frame):
    if len(frame.columns) == 1:
        if frame.values.size == 1: return frame.values[0][0]
        return frame.values.squeeze()
    grouped = frame.groupby(frame.columns[0])
    d = {k: recur_dictify(g.ix[:,1:]) for k,g in grouped}
    return d

Which gives:

>>> pprint.pprint(recur_dictify(df))
{'A': {'A1': {'A11': 1}, 'A2': {'A12': 2, 'A21': [6,8]}},
 'B': {'B1': {'B12': 3}},
 'C': {'C1': {'C11': 4}}}

But doesn't replicate the empty dict nest at level v2, and groups the repetition of A2 -A21 into array[6,8]. I've looked at Convert a Pandas DataFrame to a dictionary , no luck so far.

Answer 1

I assume that:

• the index has no name
• column name has values A, B, C, D
• etc.

and df contains the output of your recur_dictify above:

ky = frame.v1.unique() # I assume it's ['A1','B1','C1']

for k in df:
    for l in ky:
        if l not in df[k]:
            df[k][l] = {}

You original dataframe is strange though. The B2 entry does not appear anywhere in your result.