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?
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.