[python] Split / Explode a column of dictionaries into separate columns with pandas

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 dataframe has a dictionary of values inside it. The DataFrame df looks like this:

Station ID     Pollutants
8809           {"a": "46", "b": "3", "c": "12"}
8810           {"a": "36", "b": "5", "c": "8"}
8811           {"b": "2", "c": "7"}
8812           {"c": "11"}
8813           {"a": "82", "c": "15"}

I need to split this column into separate columns, so that the DataFrame `df2 looks like this:

Station ID     a      b       c
8809           46     3       12
8810           36     5       8
8811           NaN    2       7
8812           NaN    NaN     11
8813           82     NaN     15

The major issue I'm having is that the lists are not the same lengths. But all of the lists only contain up to the same 3 values: 'a', 'b', and 'c'. And they always appear in the same order ('a' first, 'b' second, 'c' third).

The following code USED to work and return exactly what I wanted (df2).

objs = [df, pandas.DataFrame(df['Pollutant Levels'].tolist()).iloc[:, :3]]
df2 = pandas.concat(objs, axis=1).drop('Pollutant Levels', axis=1)
print(df2)

I was running this code just last week and it was working fine. But now my code is broken and I get this error from line [4]:

IndexError: out-of-bounds on slice (end) 

I made no changes to the code but am now getting the error. I feel this is due to my method not being robust or proper.

Any suggestions or guidance on how to split this column of lists into separate columns would be super appreciated!

EDIT: I think the .tolist() and .apply methods are not working on my code because it is one Unicode string, i.e.:

#My data format 
u{'a': '1', 'b': '2', 'c': '3'}

#and not
{u'a': '1', u'b': '2', u'c': '3'}

The data is imported from the postgreSQL database in this format. Any help or ideas with this issue? is there a way to convert the Unicode?

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

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...

Merlin's answer is better and super easy, but we don't need a lambda function. The evaluation of dictionary can be safely ignored by either of the following two ways as illustrated below:

Way 1: Two steps

# step 1: convert the `Pollutants` column to Pandas dataframe series
df_pol_ps = data_df['Pollutants'].apply(pd.Series)

df_pol_ps:
    a   b   c
0   46  3   12
1   36  5   8
2   NaN 2   7
3   NaN NaN 11
4   82  NaN 15

# step 2: concat columns `a, b, c` and drop/remove the `Pollutants` 
df_final = pd.concat([df, df_pol_ps], axis = 1).drop('Pollutants', axis = 1)

df_final:
    StationID   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

Way 2: The above two steps can be combined in one go:

df_final = pd.concat([df, df['Pollutants'].apply(pd.Series)], axis = 1).drop('Pollutants', axis = 1)

df_final:
    StationID   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

You can use join with pop + tolist. Performance is comparable to concat with drop + tolist, but some may find this syntax cleaner:

res = df.join(pd.DataFrame(df.pop('b').tolist()))

Benchmarking with other methods:

df = pd.DataFrame({'a':[1,2,3], 'b':[{'c':1}, {'d':3}, {'c':5, 'd':6}]})

def joris1(df):
    return pd.concat([df.drop('b', axis=1), df['b'].apply(pd.Series)], axis=1)

def joris2(df):
    return pd.concat([df.drop('b', axis=1), pd.DataFrame(df['b'].tolist())], axis=1)

def jpp(df):
    return df.join(pd.DataFrame(df.pop('b').tolist()))

df = pd.concat([df]*1000, ignore_index=True)

%timeit joris1(df.copy())  # 1.33 s per loop
%timeit joris2(df.copy())  # 7.42 ms per loop
%timeit jpp(df.copy())     # 7.68 ms per loop

• The fastest method to normalize a column of flat, one-level dicts, as per the timing analysis performed by Shijith in this answer:
  • df.join(pd.DataFrame(df.pop('Pollutants').values.tolist()))
  • It will not resolve other issues with columns of list or dicts that are addressed below, such as rows with NaN, or nested dicts.

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 str type, 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)

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.

How do I split a column of dictionaries into separate columns with pandas?

pd.DataFrame(df['val'].tolist()) is the canonical method for exploding a column of dictionaries

Here's your proof using a colorful graph.

Benchmarking code for reference.

Note that I am only timing the explosion since that's the most interesting part of answering this question - other aspects of result construction (such as whether to use pop or drop) are tangential to the discussion and can be ignored (it should be noted however that using pop avoids the followup drop call, so the final solution is a bit more performant, but we are still listifying the column and passing it to pd.DataFrame either way).

Additionally, pop destructively mutates the input DataFrame, making it harder to run in benchmarking code which assumes the input is not changed across test runs.

Critique of other solutions

• df['val'].apply(pd.Series) is extremely slow for large N as pandas constructs Series objects for each row, then proceeds to construct a DataFrame from them. For larger N the performance dips to the order of minutes or hours.

• pd.json_normalize(df['val'])) is slower simply because json_normalize is meant to work with a much more complex input data - particularly deeply nested JSON with multiple record paths and metadata. We have a simple flat dict for which pd.DataFrame suffices, so use that if your dicts are flat.

• Some answers suggest df.pop('val').values.tolist() or df.pop('val').to_numpy().tolist(). I don't think it makes much of a difference whether you listify the series or the numpy array. It's one operation less to listify the series directly and really isn't slower so I'd recommend avoiding generating the numpy array in the intermediate step.

Try this: The data returned from SQL has to converted into a Dict. or could it be "Pollutant Levels" is now Pollutants'

   StationID                   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"}


df2["Pollutants"] = df2["Pollutants"].apply(lambda x : dict(eval(x)) )
df3 = df2["Pollutants"].apply(pd.Series )

    a    b   c
0   46    3  12
1   36    5   8
2  NaN    2   7
3  NaN  NaN  11
4   82  NaN  15


result = pd.concat([df, df3], axis=1).drop('Pollutants', axis=1)
result

   StationID    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

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

>>> 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"}

speed comparison for a large dataset of 10 million rows

>>> df = pd.concat([df]*100000).reset_index(drop=True)
>>> df = pd.concat([df]*20).reset_index(drop=True)
>>> print(df.shape)
(10000000, 2)

def apply_drop(df):
    return df.join(df['Pollutants'].apply(pd.Series)).drop('Pollutants', axis=1)  

def json_normalise_drop(df):
    return df.join(pd.json_normalize(df.Pollutants)).drop('Pollutants', axis=1)  

def tolist_drop(df):
    return df.join(pd.DataFrame(df['Pollutants'].tolist())).drop('Pollutants', axis=1)  

def vlues_tolist_drop(df):
    return df.join(pd.DataFrame(df['Pollutants'].values.tolist())).drop('Pollutants', axis=1)  

def pop_tolist(df):
    return df.join(pd.DataFrame(df.pop('Pollutants').tolist()))  

def pop_values_tolist(df):
    return df.join(pd.DataFrame(df.pop('Pollutants').values.tolist()))

>>> %timeit apply_drop(df.copy())
1 loop, best of 3: 53min 20s per loop
>>> %timeit json_normalise_drop(df.copy())
1 loop, best of 3: 54.9 s per loop
>>> %timeit tolist_drop(df.copy())
1 loop, best of 3: 6.62 s per loop
>>> %timeit vlues_tolist_drop(df.copy())
1 loop, best of 3: 6.63 s per loop
>>> %timeit pop_tolist(df.copy())
1 loop, best of 3: 5.99 s per loop
>>> %timeit pop_values_tolist(df.copy())
1 loop, best of 3: 5.94 s per loop

+---------------------+-----------+
| apply_drop          | 53min 20s |
| json_normalise_drop |    54.9 s |
| tolist_drop         |    6.62 s |
| vlues_tolist_drop   |    6.63 s |
| pop_tolist          |    5.99 s |
| pop_values_tolist   |    5.94 s |
+---------------------+-----------+

df.join(pd.DataFrame(df.pop('Pollutants').values.tolist())) is the fastest

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.

To convert the string to an actual dict, you can do df['Pollutant Levels'].map(eval). Afterwards, the solution below can be used to convert the dict to different columns.

Using a small example, you can use .apply(pd.Series):

In [2]: df = pd.DataFrame({'a':[1,2,3], 'b':[{'c':1}, {'d':3}, {'c':5, 'd':6}]})

In [3]: df
Out[3]:
   a                   b
0  1           {u'c': 1}
1  2           {u'd': 3}
2  3  {u'c': 5, u'd': 6}

In [4]: df['b'].apply(pd.Series)
Out[4]:
     c    d
0  1.0  NaN
1  NaN  3.0
2  5.0  6.0

To combine it with the rest of the dataframe, you can concat the other columns with the above result:

In [7]: pd.concat([df.drop(['b'], axis=1), df['b'].apply(pd.Series)], axis=1)
Out[7]:
   a    c    d
0  1  1.0  NaN
1  2  NaN  3.0
2  3  5.0  6.0

Using your code, this also works if I leave out the iloc part:

In [15]: pd.concat([df.drop('b', axis=1), pd.DataFrame(df['b'].tolist())], axis=1)
Out[15]:
   a    c    d
0  1  1.0  NaN
1  2  NaN  3.0
2  3  5.0  6.0

df = pd.concat([df['a'], df.b.apply(pd.Series)], axis=1)