Python Pandas: Apply a Function with Arguments to a Series

python pandas: apply a function with arguments to a series

Newer versions of pandas do allow you to pass extra arguments (see the new documentation). So now you can do:

my_series.apply(your_function, args=(2,3,4), extra_kw=1)

The positional arguments are added after the element of the series.

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For older version of pandas:

The documentation explains this clearly. The apply method accepts a python function which should have a single parameter. If you want to pass more parameters you should use functools.partial as suggested by Joel Cornett in his comment.

An example:

>>> import functools
>>> import operator
>>> add_3 = functools.partial(operator.add,3)
>>> add_3(2)
5
>>> add_3(7)
10

You can also pass keyword arguments using partial.

Another way would be to create a lambda:

my_series.apply((lambda x: your_func(a,b,c,d,...,x)))

But I think using partial is better.

pandas apply function with arguments

Pass the arguments as kwargs to apply:

df['events'] = df['data'].apply(count_ones, total_bits=60, group_size=12)

Pandas apply function treating input as a series rather than a single cell in the passed series

You have to pass the row, not the entire column:

bank_exp['Description'] = bank_exp.apply(lambda row : preprocess_cell(row['Description'], stopwords_all), axis = 1)

Or run apply on the Series:

bank_exp['Description'] = bank_exp['Description'].apply(preprocess_cell, args=(stopwords_all,))

python pandas- apply function with two arguments to columns

Why not just do this?

df['NewCol'] = df.apply(lambda x: segmentMatch(x['TimeCol'], x['ResponseCol']), 
                        axis=1)

Rather than trying to pass the column as an argument as in your example, we now simply pass the appropriate entries in each row as argument, and store the result in 'NewCol'.

Passing a function with multiple arguments to DataFrame.apply

It's just the way you think it would be, apply accepts args and kwargs and passes them directly to some_func.

df.apply(some_func, var1='DOG', axis=1)

Or,

df.apply(some_func, args=('DOG', ), axis=1)

0    foo-x-DOG
1    bar-y-DOG
dtype: object

Pass arguments to function while using apply to pandas series

IIUC you need pass like argument dropna=False:

print (df.apply(pd.Series.value_counts, dropna=False))
       a    b    c
1.0  1.0  NaN  NaN
2.0  2.0  1.0  NaN
3.0  NaN  2.0  NaN
4.0  NaN  1.0  2.0
6.0  NaN  NaN  1.0
NaN  1.0  NaN  1.0

Or lambda function:

print (df.apply(lambda x: x.value_counts(dropna=False)))
       a    b    c
1.0  1.0  NaN  NaN
2.0  2.0  1.0  NaN
3.0  NaN  2.0  NaN
4.0  NaN  1.0  2.0
6.0  NaN  NaN  1.0
NaN  1.0  NaN  1.0

python pandas: apply a function with arguments to a series. Update

The TypeError is saying that you passed the wrong type to the lambda function x + y. It's expecting the args to be a sequence, but it got an int. You may have thought that (100) was a tuple (a sequence), but in python it's the comma that makes a tuple:

In [10]: type((100))
Out[10]: int

In [11]: type((100,))
Out[11]: tuple

So change your last line to

In [12]: a['x'].apply(lambda x, y: x + y, args=(100,))
Out[12]: 
0    101
1    102
Name: x, dtype: int64

Applying function with multiple arguments to create a new pandas column

Alternatively, you can use numpy underlying function:

>>> import numpy as np
>>> df = pd.DataFrame({"A": [10,20,30], "B": [20, 30, 10]})
>>> df['new_column'] = np.multiply(df['A'], df['B'])
>>> df
    A   B  new_column
0  10  20         200
1  20  30         600
2  30  10         300

or vectorize arbitrary function in general case:

>>> def fx(x, y):
...     return x*y
...
>>> df['new_column'] = np.vectorize(fx)(df['A'], df['B'])
>>> df
    A   B  new_column
0  10  20         200
1  20  30         600
2  30  10         300

Pandas .apply() function with multiple args

You can use np.where and reduce this to 1 line.

s = rocksfile['Song_Clean'] 
    + ' was released by ' 
    + rocksfile['Artist_Clean'] 
    + pd.Series(np.where(rocksfile['Release_Year'] < 1970, 'before', 'after'))
    + ' 1970'

rocksfile['new'] = s

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