将“熊猫”列中的字典/列表拆分为单独的列

Question 1

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 (or list?) of values within it. The DataFrame looks like this:

[1] df
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 looks like this:

[2] df2
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).

[3] df 
[4] objs = [df, pandas.DataFrame(df['Pollutant Levels'].tolist()).iloc[:, :3]]
[5] df2 = pandas.concat(objs, axis=1).drop('Pollutant Levels', axis=1)
[6] 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 importing from the postgreSQL database in this format. Any help or ideas with this issue? is there a way to convert the Unicode?

Question 2

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

Question 3

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…

Question 4

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

Question 5

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

Question 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.

Question 7

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

Question 8

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

Question 9

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.

Question 10

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

Question 11

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

将“熊猫”列中的字典/列表拆分为单独的列

问题：将“熊猫”列中的字典/列表拆分为单独的列

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将“熊猫”列中的字典/列表拆分为单独的列

问题：将“熊猫”列中的字典/列表拆分为单独的列

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