How to flatten a pandas dataframe with some column

问题:

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     name     columnA                               columnB
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"}]
...

As you can see, not all the rows have the same number of elements in the json strings for a column.

What I need to do is keep the normal columns like id and name as it is and flatten the json columns like so:

id    name   columnA.dist   columnA.time   columnB.pos.1st   columnB.pos.2nd   columnB.pos.3rd     columnB.pos.total
1     John   600            0:12.10        500               300               200                 1000 
2     Mark   600            NaN            500               300               Nan                 800 

I have tried using json_normalize like so:

from pandas.io.json import json_normalize
json_normalize(df)

But there seems to be some problems with keyerror. What is the correct way of doing this?

回答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.

回答2:

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)

回答3:

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