Apply pandas function to column to create multiple new columns?

Question

How to do this in pandas:

I have a function extract_text_features on a single text column, returning multiple output columns. Specifically, the function returns 6 values.

The function works, however there doesn't seem to be any proper return type (pandas DataFrame/ numpy array/ Python list) such that the output can get correctly assigned df.ix[: ,10:16] = df.textcol.map(extract_text_features)

So I think I need to drop back to iterating with df.iterrows(), as per this?

UPDATE: Iterating with df.iterrows() is at least 20x slower, so I surrendered and split out the function into six distinct .map(lambda ...) calls.

UPDATE 2: this question was asked back around v0.11.0, before the useability of df.apply was improved or df.assign() was added in v0.16. Hence much of the question and answers are not too relevant since then.

Answer 1

Just to add to this, for me it was also necessary in some cases to use the unstack() method, because otherwise I'd just get a new column that contained a dictionary.

It works like this:

df.groupby('variable')['value'].apply(lambda grp: {
    'Min': grp.min(),
    'Median': grp.median(),
    'Max': grp.max()
}).unstack()

Answer 2

In 2020, I use apply() with argument result_type='expand'

applied_df = df.apply(lambda row: fn(row.text), axis='columns', result_type='expand')
df = pd.concat([df, applied_df], axis='columns')

fn() should return a dict; its keys will be the new column names.

Alternatively you can do a one-liner by also specifying the column names:

df[["col1", "col2", ...]] = df.apply(lambda row: fn(row.text), axis='columns', result_type='expand')

Answer 3

Although the question specifies that the function should be applied to a Series, most of the answers seem to be applying the function to a DataFrame, with the function getting the relevant column from each row. This seems somewhat inelegant and potentially slow.

Say the function f takes a value in column df["argument"] and returns two values. The nicest way I've found to do it by applying to the column Series is this:

df[["value_1", "value_2"]] = df["argument"].apply(f).to_list()

Unlike DataFrame.apply, unfortunately Series.apply has no result_type parameter to expand the result into a DataFrame to assign to. But pandas understands just as well if you assign to a list of tuples.

Answer 4

This works for me:

import pandas as pd
import numpy as np
future = pd.DataFrame(
    pd.date_range('2022-09-01',periods=360),
    columns=['date']
)

def featurize(datetime):
    return pd.Series({
        'month':datetime.month,
        'year':datetime.year,
        'dayofweek':datetime.dayofweek,
        'dayofyear':datetime.dayofyear
    })
    
future.loc[
    :,['month','year','dayofweek','dayofyear']
    ] = future.date.apply(featurize)

future.head()

Output:

    date    month   year    dayofweek   dayofyear
0   2022-09-01  9   2022    3           244
1   2022-09-02  9   2022    4           245
2   2022-09-03  9   2022    5           246
3   2022-09-04  9   2022    6           247
4   2022-09-05  9   2022    0           248

Answer 5

For me this worked:

Input df

df = pd.DataFrame({'col x': [1,2,3]})
   col x
0      1
1      2
2      3

Function

def f(x):
    return pd.Series([x*x, x*x*x])

Create 2 new columns:

df[['square x', 'cube x']] = df['col x'].apply(f)

Output:

   col x  square x  cube x
0      1         1       1
1      2         4       8
2      3         9      27

Answer 6

def extract_text_features(feature):
    ...
    ...
    return pd.Series((feature1, feature2)) 

df[['NewFeature1', 'NewFeature1']] = df[['feature']].apply(extract_text_features, axis=1)

Here the a dataframe with a single feature is being converted to two new features. Give this a try too.

Answer 7

I have a more complicated situation, the dataset has a nested structure:

import json
data = '{"TextID":{"0":"0038f0569e","1":"003eb6998d","2":"006da49ea0"},"Summary":{"0":{"Crisis_Level":["c"],"Type":["d"],"Special_Date":["a"]},"1":{"Crisis_Level":["d"],"Type":["a","d"],"Special_Date":["a"]},"2":{"Crisis_Level":["d"],"Type":["a"],"Special_Date":["a"]}}}'
df = pd.DataFrame.from_dict(json.loads(data))
print(df)

output:

        TextID                                            Summary
0  0038f0569e  {'Crisis_Level': ['c'], 'Type': ['d'], 'Specia...
1  003eb6998d  {'Crisis_Level': ['d'], 'Type': ['a', 'd'], 'S...
2  006da49ea0  {'Crisis_Level': ['d'], 'Type': ['a'], 'Specia...

The Summary column contains dict objects, so I use apply with from_dict and stack to extract each row of dict:

df2 = df.apply(
    lambda x: pd.DataFrame.from_dict(x[1], orient='index').stack(), axis=1)
print(df2)

output:

    Crisis_Level Special_Date Type     
                0            0    0    1
0            c            a    d  NaN
1            d            a    a    d
2            d            a    a  NaN

Looks good, but missing the TextID column. To get TextID column back, I've tried three approach:

1. Modify apply to return multiple columns:

   df_tmp = df.copy()
   
   df_tmp[['TextID', 'Summary']] = df.apply(
       lambda x: pd.Series([x[0], pd.DataFrame.from_dict(x[1], orient='index').stack()]), axis=1)
   print(df_tmp)
   

   output:

       TextID                                            Summary
   0  0038f0569e  Crisis_Level  0    c
   Type          0    d
   Spec...
   1  003eb6998d  Crisis_Level  0    d
   Type          0    a
       ...
   2  006da49ea0  Crisis_Level  0    d
   Type          0    a
   Spec...
   

   But this is not what I want, the Summary structure are flatten.

2. Use pd.concat:

   df_tmp2 = pd.concat([df['TextID'], df2], axis=1)
   print(df_tmp2)
   

   output:

       TextID (Crisis_Level, 0) (Special_Date, 0) (Type, 0) (Type, 1)
   0  0038f0569e                 c                 a         d       NaN
   1  003eb6998d                 d                 a         a         d
   2  006da49ea0                 d                 a         a       NaN
   

   Looks fine, the MultiIndex column structure are preserved as tuple. But check columns type:

   df_tmp2.columns
   

   output:

   Index(['TextID', ('Crisis_Level', 0), ('Special_Date', 0), ('Type', 0),
       ('Type', 1)],
       dtype='object')
   

   Just as a regular Index class, not MultiIndex class.

3. use set_index:

   Turn all columns you want to preserve into row index, after some complicated apply function and then reset_index to get columns back:

   df_tmp3 = df.set_index('TextID')
   
   df_tmp3 = df_tmp3.apply(
       lambda x: pd.DataFrame.from_dict(x[0], orient='index').stack(), axis=1)
   
   df_tmp3 = df_tmp3.reset_index(level=0)
   print(df_tmp3)
   

   output:

       TextID Crisis_Level Special_Date Type     
                           0            0    0    1
   0  0038f0569e            c            a    d  NaN
   1  003eb6998d            d            a    a    d
   2  006da49ea0            d            a    a  NaN
   

   Check the type of columns

   df_tmp3.columns
   

   output:

   MultiIndex(levels=[['Crisis_Level', 'Special_Date', 'Type', 'TextID'], [0, 1, '']],
           codes=[[3, 0, 1, 2, 2], [2, 0, 0, 0, 1]])
   

So, If your apply function will return MultiIndex columns, and you want to preserve it, you may want to try the third method.

Answer 8

Building off of user1827356 's answer, you can do the assignment in one pass using df.merge:

df.merge(df.textcol.apply(lambda s: pd.Series({'feature1':s+1, 'feature2':s-1})), 
    left_index=True, right_index=True)

    textcol  feature1  feature2
0  0.772692  1.772692 -0.227308
1  0.857210  1.857210 -0.142790
2  0.065639  1.065639 -0.934361
3  0.819160  1.819160 -0.180840
4  0.088212  1.088212 -0.911788

EDIT: Please be aware of the huge memory consumption and low speed: https://ys-l.github.io/posts/2015/08/28/how-not-to-use-pandas-apply/ !

Answer 9

Just use result_type="expand"

df = pd.DataFrame(np.random.randint(0,10,(10,2)), columns=["random", "a"])
df[["sq_a","cube_a"]] = df.apply(lambda x: [x.a**2, x.a**3], axis=1, result_type="expand")

Answer 10

Have posted the same answer in two other similar questions. The way I prefer to do this is to wrap up the return values of the function in a series:

def f(x):
    return pd.Series([x**2, x**3])

And then use apply as follows to create separate columns:

df[['x**2','x**3']] = df.apply(lambda row: f(row['x']), axis=1)

Answer 11

you can return the entire row instead of values:

df = df.apply(extract_text_features,axis = 1)

where the function returns the row

def extract_text_features(row):
      row['new_col1'] = value1
      row['new_col2'] = value2
      return row

Answer 12

Summary: If you only want to create a few columns, use df[['new_col1','new_col2']] = df[['data1','data2']].apply( function_of_your_choosing(x), axis=1)

For this solution, the number of new columns you are creating must be equal to the number columns you use as input to the .apply() function. If you want to do something else, have a look at the other answers.

Details Let's say you have two-column dataframe. The first column is a person's height when they are 10; the second is said person's height when they are 20.

Suppose you need to calculate both the mean of each person's heights and sum of each person's heights. That's two values per each row.

You could do this via the following, soon-to-be-applied function:

def mean_and_sum(x):
    """
    Calculates the mean and sum of two heights.
    Parameters:
    :x -- the values in the row this function is applied to. Could also work on a list or a tuple.
    """

    sum=x[0]+x[1]
    mean=sum/2
    return [mean,sum]

You might use this function like so:

 df[['height_at_age_10','height_at_age_20']].apply(mean_and_sum(x),axis=1)

(To be clear: this apply function takes in the values from each row in the subsetted dataframe and returns a list.)

However, if you do this:

df['Mean_&_Sum'] = df[['height_at_age_10','height_at_age_20']].apply(mean_and_sum(x),axis=1)

you'll create 1 new column that contains the [mean,sum] lists, which you'd presumably want to avoid, because that would require another Lambda/Apply.

Instead, you want to break out each value into its own column. To do this, you can create two columns at once:

df[['Mean','Sum']] = df[['height_at_age_10','height_at_age_20']]
.apply(mean_and_sum(x),axis=1)

Answer 13

This is the correct and easiest way to accomplish this for 95% of use cases:

>>> df = pd.DataFrame(zip(*[range(10)]), columns=['num'])
>>> df
    num
0    0
1    1
2    2
3    3
4    4
5    5

>>> def example(x):
...     x['p1'] = x['num']**2
...     x['p2'] = x['num']**3
...     x['p3'] = x['num']**4
...     return x

>>> df = df.apply(example, axis=1)
>>> df
    num  p1  p2  p3
0    0   0   0    0
1    1   1   1    1
2    2   4   8   16
3    3   9  27   81
4    4  16  64  256

Answer 14

The accepted solution is going to be extremely slow for lots of data. The solution with the greatest number of upvotes is a little difficult to read and also slow with numeric data. If each new column can be calculated independently of the others, I would just assign each of them directly without using apply.

Example with fake character data

Create 100,000 strings in a DataFrame

df = pd.DataFrame(np.random.choice(['he jumped', 'she ran', 'they hiked'],
                                   size=100000, replace=True),
                  columns=['words'])
df.head()
        words
0     she ran
1     she ran
2  they hiked
3  they hiked
4  they hiked

Let's say we wanted to extract some text features as done in the original question. For instance, let's extract the first character, count the occurrence of the letter 'e' and capitalize the phrase.

df['first'] = df['words'].str[0]
df['count_e'] = df['words'].str.count('e')
df['cap'] = df['words'].str.capitalize()
df.head()
        words first  count_e         cap
0     she ran     s        1     She ran
1     she ran     s        1     She ran
2  they hiked     t        2  They hiked
3  they hiked     t        2  They hiked
4  they hiked     t        2  They hiked

Timings

%%timeit
df['first'] = df['words'].str[0]
df['count_e'] = df['words'].str.count('e')
df['cap'] = df['words'].str.capitalize()
127 ms ± 585 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

def extract_text_features(x):
    return x[0], x.count('e'), x.capitalize()

%timeit df['first'], df['count_e'], df['cap'] = zip(*df['words'].apply(extract_text_features))
101 ms ± 2.96 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Surprisingly, you can get better performance by looping through each value

%%timeit
a,b,c = [], [], []
for s in df['words']:
    a.append(s[0]), b.append(s.count('e')), c.append(s.capitalize())

df['first'] = a
df['count_e'] = b
df['cap'] = c
79.1 ms ± 294 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Another example with fake numeric data

Create 1 million random numbers and test the powers function from above.

df = pd.DataFrame(np.random.rand(1000000), columns=['num'])


def powers(x):
    return x, x**2, x**3, x**4, x**5, x**6

%%timeit
df['p1'], df['p2'], df['p3'], df['p4'], df['p5'], df['p6'] = \
       zip(*df['num'].map(powers))
1.35 s ± 83.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Assigning each column is 25x faster and very readable:

%%timeit 
df['p1'] = df['num'] ** 1
df['p2'] = df['num'] ** 2
df['p3'] = df['num'] ** 3
df['p4'] = df['num'] ** 4
df['p5'] = df['num'] ** 5
df['p6'] = df['num'] ** 6
51.6 ms ± 1.9 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

I made a similar response with more details here on why apply is typically not the way to go.

Answer 15

I usually do this using zip:

>>> df = pd.DataFrame([[i] for i in range(10)], columns=['num'])
>>> df
    num
0    0
1    1
2    2
3    3
4    4
5    5
6    6
7    7
8    8
9    9

>>> def powers(x):
>>>     return x, x**2, x**3, x**4, x**5, x**6

>>> df['p1'], df['p2'], df['p3'], df['p4'], df['p5'], df['p6'] = \
>>>     zip(*df['num'].map(powers))

>>> df
        num     p1      p2      p3      p4      p5      p6
0       0       0       0       0       0       0       0
1       1       1       1       1       1       1       1
2       2       2       4       8       16      32      64
3       3       3       9       27      81      243     729
4       4       4       16      64      256     1024    4096
5       5       5       25      125     625     3125    15625
6       6       6       36      216     1296    7776    46656
7       7       7       49      343     2401    16807   117649
8       8       8       64      512     4096    32768   262144
9       9       9       81      729     6561    59049   531441

Answer 16

I've looked several ways of doing this and the method shown here (returning a pandas series) doesn't seem to be most efficient.

If we start with a largeish dataframe of random data:

# Setup a dataframe of random numbers and create a 
df = pd.DataFrame(np.random.randn(10000,3),columns=list('ABC'))
df['D'] = df.apply(lambda r: ':'.join(map(str, (r.A, r.B, r.C))), axis=1)
columns = 'new_a', 'new_b', 'new_c'

The example shown here:

# Create the dataframe by returning a series
def method_b(v):
    return pd.Series({k: v for k, v in zip(columns, v.split(':'))})
%timeit -n10 -r3 df.D.apply(method_b)

10 loops, best of 3: 2.77 s per loop

An alternative method:

# Create a dataframe from a series of tuples
def method_a(v):
    return v.split(':')
%timeit -n10 -r3 pd.DataFrame(df.D.apply(method_a).tolist(), columns=columns)

10 loops, best of 3: 8.85 ms per loop

By my reckoning it's far more efficient to take a series of tuples and then convert that to a DataFrame. I'd be interested to hear people's thinking though if there's an error in my working.

Answer 17

This is what I've done in the past

df = pd.DataFrame({'textcol' : np.random.rand(5)})

df
    textcol
0  0.626524
1  0.119967
2  0.803650
3  0.100880
4  0.017859

df.textcol.apply(lambda s: pd.Series({'feature1':s+1, 'feature2':s-1}))
   feature1  feature2
0  1.626524 -0.373476
1  1.119967 -0.880033
2  1.803650 -0.196350
3  1.100880 -0.899120
4  1.017859 -0.982141

Editing for completeness

pd.concat([df, df.textcol.apply(lambda s: pd.Series({'feature1':s+1, 'feature2':s-1}))], axis=1)
    textcol feature1  feature2
0  0.626524 1.626524 -0.373476
1  0.119967 1.119967 -0.880033
2  0.803650 1.803650 -0.196350
3  0.100880 1.100880 -0.899120
4  0.017859 1.017859 -0.982141