从 Pandas DataFrame 创建嵌套的 JSON，并将分组的行作为最深嵌套中的属性

Question

I'm looking for a solution to build a nested dict / JSON with the last three columns "name" , "color" , "amount" as attributes inside a "products" list. The values from the cat1-cat3 columns should be the keys.

The provided DataFrame looks like this:

import pandas as pd

df = pd.DataFrame({
    'cat1': ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'C'],
    'cat2': ['BB', 'BB', 'BC', 'BB', 'BB', 'BB', 'BC', 'BC'],
    'cat3': ['CC', 'CC', 'CD', 'CD', 'CD', 'CC', 'CD', 'CE'],
    'name': ['P1', 'P2', 'P3', 'P1', 'P4', 'P1', 'P3','P6'],
    'color': ['red', 'blue', 'green', 'green', 'yellow', 'red', 'blue', 'blue']
    'amount': [132, 51, 12, 421, 55, 11, 123, 312]
})

This would be the desired output:

{
   "A":{
      "BB":{
         "CC":{
            "products":[
               {
                  "name":"P1",
                  "color":"red",
                  "amount":132
               },
               {
                  "name":"P2",
                  "color":"blue",
                  "amount":51
               }
            ]
         }
      },
      "BC":{
         "CD":{
            "products":[
               {
                  "name":"P3",
                  "color":"green",
                  "amount":12
               }
            ]
         }
      }
   },
   "B":{
      "BB":{
         "CD":{
            "products":[
               {
                  "name":"P1",
                  "color":"green",
                  "amount":421
               },
               {
                  "name":"P4",
                  "color":"yellow",
                  "amount":55
               }
            ]
         }
      }
   },
   "C":{
      "BB":{
         "CC":{
            "products":[
               {
                  "name":"P1",
                  "color":"red",
                  "amount":11
               }
            ]
         }
      },
      "BC":{
         "CD":{
            "products":[
               {
                  "name":"P3",
                  "color":"blue",
                  "amount":123
               }
            ]
         },
         "CE":{
            "products":[
               {
                  "name":"P6",
                  "color":"blue",
                  "amount":312
               }
            ]
         }
      }
   }
}

@BEN_YO provided a recursive solution for this problem without the inner products part.

So I'm actually looking for an adaption of this method with an inner list:

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.iloc[:,1:]) for k,g in grouped}
     return d
     
recur_dictify(df)

Answer 1

If another way is fine , you can try the below, it is a little dirty though (you can try optimizing it)

cols = ['name','color','amount']
u = df[df.columns.difference(cols)].join(df[cols].agg(dict,1).rename('d'))
v = (u.groupby(['cat1','cat2','cat3'])['d'].agg(list).reset_index("cat3"))

v = v.groupby(v.index).apply(lambda x: dict(zip(x['cat3'],x['d'])))
v.index = pd.MultiIndex.from_tuples(v.index,names=['cat1','cat2'])
d = v.unstack(0).to_dict()

---

print(d)
{'A': {'BB': {'CC': [{'amount': 132, 'color': 'red', 'name': 'P1'},
                     {'amount': 51, 'color': 'blue', 'name': 'P2'}]},
       'BC': {'CD': [{'amount': 12, 'color': 'green', 'name': 'P3'}]}},
 'B': {'BB': {'CD': [{'amount': 421, 'color': 'green', 'name': 'P1'},
                     {'amount': 55, 'color': 'yellow', 'name': 'P4'}]},
       'BC': nan},
 'C': {'BB': {'CC': [{'amount': 11, 'color': 'red', 'name': 'P1'}]},
       'BC': {'CD': [{'amount': 123, 'color': 'blue', 'name': 'P3'}],
              'CE': [{'amount': 312, 'color': 'blue', 'name': 'P6'}]}}}

Answer 2

We can groupby on cat1 , cat2 and cat3 and recursively build the dictionary based on the grouped categories:

def set_val(d, k, v):
    if len(k) == 1:
        d[k[0]] = v
    else:
        d[k[0]] = set_val(d.get(k[0], {}), k[1:], v)
    return d


dct = {}
for k, g in df.groupby(['cat1', 'cat2', 'cat3']):
    set_val(dct, k, {'products': g[['name', 'color', 'amount']].to_dict('r')})

---

print(dct)

{'A': {'BB': {'CC': {'products': [{'amount': 132, 'color': 'red', 'name': 'P1'},
                                  {'amount': 51, 'color': 'blue', 'name': 'P2'}]}},
       'BC': {'CD': {'products': [{'amount': 12, 'color': 'green', 'name': 'P3'}]}}},
 'B': {'BB': {'CD': {'products': [{'amount': 421, 'color': 'green', 'name': 'P1'},
                                  {'amount': 55, 'color': 'yellow', 'name': 'P4'}]}}},
 'C': {'BB': {'CC': {'products': [{'amount': 11, 'color': 'red', 'name': 'P1'}]}},
       'BC': {'CD': {'products': [{'amount': 123, 'color': 'blue', 'name': 'P3'}]},
              'CE': {'products': [{'amount': 312, 'color': 'blue', 'name': 'P6'}]}}}}

Answer 3

This is a generic method adapted from Shubham Sharma's great Solution

def gen_nested_dict(dataframe, group, inner_key, inner_dict):
    def set_val(d, k2, v):
        if len(k2) == 1:
            d[k2[0]] = v
        else:
            d[k2[0]] = set_val(d.get(k2[0], {}), k2[1:], v)
        return d

    dct = {}
    for k, g in dataframe.groupby(group):
        set_val(dct, k, {inner_key: g[inner_dict].to_dict('records')})

    return dct

 mydct = gen_nested_dict(df, ['cat1', 'cat2', 'cat3'], 'products', ['name', 'color', 'amount'])