How to flatten a pandas dataframe with some columns as json?

Question

I have a dataframe df that loads data from a database. Most of the columns are json strings while some are even list of jsons. For example:

id           


        

Answer 1

Here's a solution using json_normalize() again by using a custom function to get the data in the correct format understood by json_normalize function.

import ast
from pandas.io.json import json_normalize

def only_dict(d):
    '''
    Convert json string representation of dictionary to a python dict
    '''
    return ast.literal_eval(d)

def list_of_dicts(ld):
    '''
    Create a mapping of the tuples formed after 
    converting json strings of list to a python list   
    '''
    return dict([(list(d.values())[1], list(d.values())[0]) for d in ast.literal_eval(ld)])

A = json_normalize(df['columnA'].apply(only_dict).tolist()).add_prefix('columnA.')
B = json_normalize(df['columnB'].apply(list_of_dicts).tolist()).add_prefix('columnB.pos.') 

Finally, join the DFs on the common index to get:

df[['id', 'name']].join([A, B])

---

EDIT:- As per the comment by @MartijnPieters, the recommended way of decoding the json strings would be to use json.loads() which is much faster when compared to using ast.literal_eval() if you know that the data source is JSON.

Answer 2

The quickest seems to be:

json_struct = json.loads(df.to_json(orient="records"))    
df_flat = pd.io.json.json_normalize(json_struct) #use pd.io.json

Answer 3

TL;DR Copy-paste the following function and use it like this: flatten_nested_json_df(df)

This is the most general function I could come up with:

def flatten_nested_json_df(df):

    df = df.reset_index()

    print(f"original shape: {df.shape}")
    print(f"original columns: {df.columns}")


    # search for columns to explode/flatten
    s = (df.applymap(type) == list).all()
    list_columns = s[s].index.tolist()

    s = (df.applymap(type) == dict).all()
    dict_columns = s[s].index.tolist()

    print(f"lists: {list_columns}, dicts: {dict_columns}")
    while len(list_columns) > 0 or len(dict_columns) > 0:
        new_columns = []

        for col in dict_columns:
            print(f"flattening: {col}")
            # explode dictionaries horizontally, adding new columns
            horiz_exploded = pd.json_normalize(df[col]).add_prefix(f'{col}.')
            horiz_exploded.index = df.index
            df = pd.concat([df, horiz_exploded], axis=1).drop(columns=[col])
            new_columns.extend(horiz_exploded.columns) # inplace

        for col in list_columns:
            print(f"exploding: {col}")
            # explode lists vertically, adding new columns
            df = df.drop(columns=[col]).join(df[col].explode().to_frame())
            new_columns.append(col)

        # check if there are still dict o list fields to flatten
        s = (df[new_columns].applymap(type) == list).all()
        list_columns = s[s].index.tolist()

        s = (df[new_columns].applymap(type) == dict).all()
        dict_columns = s[s].index.tolist()

        print(f"lists: {list_columns}, dicts: {dict_columns}")

    print(f"final shape: {df.shape}")
    print(f"final columns: {df.columns}")
    return df

It takes a dataframe that may have nested lists and/or dicts in its columns, and recursively explodes/flattens those columns.

It uses pandas' pd.json_normalize to explode the dictionaries (creating new columns), and pandas' explode to explode the lists (creating new rows).

Simple to use:

# Test
df = pd.DataFrame(
    columns=['id','name','columnA','columnB'],
    data=[
        [1,'John',{"dist": "600", "time": "0:12.10"},[{"pos": "1st", "value": "500"},{"pos": "2nd", "value": "300"},{"pos": "3rd", "value": "200"}, {"pos": "total", "value": "1000"}]],
        [2,'Mike',{"dist": "600"},[{"pos": "1st", "value": "500"},{"pos": "2nd", "value": "300"},{"pos": "total", "value": "800"}]]
    ])

flatten_nested_json_df(df)

It's not the most efficient thing on earth, and it has the side effect of resetting your dataframe's index, but it gets the job done. Feel free to tweak it.

Answer 4

create a custom function to flatten columnB then use pd.concat

def flatten(js):
    return pd.DataFrame(js).set_index('pos').squeeze()

pd.concat([df.drop(['columnA', 'columnB'], axis=1),
           df.columnA.apply(pd.Series),
           df.columnB.apply(flatten)], axis=1)