Splitting dictionary/list inside a Pandas Column into Separate Columns

Question

I have data saved in a postgreSQL database. I am querying this data using Python2.7 and turning it into a Pandas DataFrame. However, the last column of this dat

Answer 1

I've concatenated those steps in a method, you have to pass only the dataframe and the column which contains the dict to expand:

def expand_dataframe(dw: pd.DataFrame, column_to_expand: str) -> pd.DataFrame:
    """
    dw: DataFrame with some column which contain a dict to expand
        in columns
    column_to_expand: String with column name of dw
    """
    import pandas as pd

    def convert_to_dict(sequence: str) -> Dict:
        import json
        s = sequence
        json_acceptable_string = s.replace("'", "\"")
        d = json.loads(json_acceptable_string)
        return d    

    expanded_dataframe = pd.concat([dw.drop([column_to_expand], axis=1),
                                    dw[column_to_expand]
                                    .apply(convert_to_dict)
                                    .apply(pd.Series)],
                                    axis=1)
    return expanded_dataframe

Answer 2

I know the question is quite old, but I got here searching for answers. There is actually a better (and faster) way now of doing this using json_normalize:

import pandas as pd

df2 = pd.json_normalize(df['Pollutant Levels'])

This avoids costly apply functions...

Answer 3

One line solution is following:

>>> df = pd.concat([df['Station ID'], df['Pollutants'].apply(pd.Series)], axis=1)
>>> print(df)
   Station ID    a    b   c
0        8809   46    3  12
1        8810   36    5   8
2        8811  NaN    2   7
3        8812  NaN  NaN  11
4        8813   82  NaN  15

Answer 4

my_df = pd.DataFrame.from_dict(my_dict, orient='index', columns=['my_col'])

.. would have parsed the dict properly (putting each dict key into a separate df column, and key values into df rows), so the dicts would not get squashed into a single column in the first place.

Answer 5

1. pd.json_normalize(df.Pollutants) is significantly faster than df.Pollutants.apply(pd.Series)
   • See the %%timeit below. For 1M rows, .json_normalize is 47 times faster than .apply.
2. Whether reading data from a file, or from an object returned by a database, or API, it may not be clear if the dict column has dict or str type.
   • If the dictionaries in the column are strings, they must be converted back to a dict type, using ast.literal_eval.
3. Use pd.json_normalize to convert the dicts, with keys as headers and values for rows.
   • Has additional parameters (e.g. record_path & meta) for dealing with nested dicts.
4. Use pandas.DataFrame.join to combine the original DataFrame, df, with the columns created using pd.json_normalize
   • If the index isn't integers (as in the example), first use df.reset_index() to get an index of integers, before doing the normalize and join.
5. Finally, use pandas.DataFrame.drop, to remove the unneeded column of dicts

• As a note, if the column has any NaN, they must be filled with an empty dict
  • df.Pollutants = df.Pollutants.fillna({i: {} for i in df.index})
    • If the 'Pollutants' column is strings, use '{}'.
    • Also see How to json_normalize a column with NaNs?.

import pandas as pd
from ast import literal_eval
import numpy as np

data = {'Station ID': [8809, 8810, 8811, 8812, 8813, 8814],
        'Pollutants': ['{"a": "46", "b": "3", "c": "12"}', '{"a": "36", "b": "5", "c": "8"}', '{"b": "2", "c": "7"}', '{"c": "11"}', '{"a": "82", "c": "15"}', np.nan]}

df = pd.DataFrame(data)

# display(df)
   Station ID                        Pollutants
0        8809  {"a": "46", "b": "3", "c": "12"}
1        8810   {"a": "36", "b": "5", "c": "8"}
2        8811              {"b": "2", "c": "7"}
3        8812                       {"c": "11"}
4        8813            {"a": "82", "c": "15"}
5        8814                               NaN

# replace NaN with '{}' if the column is strings, otherwise replace with {}
# df.Pollutants = df.Pollutants.fillna('{}')  # if the NaN is in a column of strings
df.Pollutants = df.Pollutants.fillna({i: {} for i in df.index})  # if the column is not strings

# Convert the column of stringified dicts to dicts
# skip this line, if the column contains dicts
df.Pollutants = df.Pollutants.apply(literal_eval)

# reset the index if the index is not unique integers from 0 to n-1
# df.reset_index(inplace=True)  # uncomment if needed

# normalize the column of dictionaries and join it to df
df = df.join(pd.json_normalize(df.Pollutants))

# drop Pollutants
df.drop(columns=['Pollutants'], inplace=True)

# display(df)
   Station ID    a    b    c
0        8809   46    3   12
1        8810   36    5    8
2        8811  NaN    2    7
3        8812  NaN  NaN   11
4        8813   82  NaN   15
5        8814  NaN  NaN  NaN

%%timeit

# dataframe with 1M rows
dfb = pd.concat([df]*200000).reset_index(drop=True)

%%timeit
dfb.join(pd.json_normalize(dfb.Pollutants))
[out]:
5.44 s ± 32.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
pd.concat([dfb.drop(columns=['Pollutants']), dfb.Pollutants.apply(pd.Series)], axis=1)
[out]:
4min 17s ± 2.44 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answer 6

I strongly recommend the method extract the column 'Pollutants':

df_pollutants = pd.DataFrame(df['Pollutants'].values.tolist(), index=df.index)

it's much faster than

df_pollutants = df['Pollutants'].apply(pd.Series)

when the size of df is giant.