Split / Explode a column of dictionaries into separate columns with pandas
PythonJsonPandasDictionaryJson NormalizePython Problem Overview
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?
Python Solutions
Solution 1 - Python
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
Solution 2 - Python
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...
Solution 3 - Python
- 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
ordicts
, that are addressed below, such as rows withNaN
, or nesteddicts
.
pd.json_normalize(df.Pollutants)
is significantly faster thandf.Pollutants.apply(pd.Series)
- See the
%%timeit
below. For 1M rows,.json_normalize
is 47 times faster than.apply
.
- See the
- 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 hasdict
orstr
type.- If the dictionaries in the column are
str
type, they must be converted back to adict
type, usingast.literal_eval
, orjson.loads(…)
.
- If the dictionaries in the column are
- Use
pd.json_normalize
to convert thedicts
, withkeys
as headers andvalues
for rows.- There are additional parameters (e.g.
record_path
&meta
) for dealing with nesteddicts
.
- There are additional parameters (e.g.
- Use
pandas.DataFrame.join
to combine the original DataFrame,df
, with the columns created usingpd.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.
- If the index isn't integers (as in the example), first use
- Finally, use
pandas.DataFrame.drop
, to remove the unneeded column ofdicts
- As a note, if the column has any
NaN
, they must be filled with an emptydict
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?.
- If the
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)
Solution 4 - Python
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
Solution 5 - Python
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.
Solution 6 - Python
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
Solution 7 - Python
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
Solution 8 - Python
> ### 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 becausejson_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 whichpd.DataFrame
suffices, so use that if your dicts are flat. -
Some answers suggest
df.pop('val').values.tolist()
ordf.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.
Solution 9 - Python
> Note : for dictionary with depth=1 (one-level)
>>> 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]*2000000).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
Solution 10 - Python
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
Solution 11 - Python
df = pd.concat([df['a'], df.b.apply(pd.Series)], axis=1)
Solution 12 - Python
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
Solution 13 - Python
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.