[python] Replacing few values in a pandas dataframe column with another value

I have a pandas dataframe df as illustrated below:

BrandName Specialty
A          H
B          I
ABC        J
D          K
AB         L

I want to replace 'ABC' and 'AB' in column BrandName by A. Can someone help with this?

Replace

DataFrame object has powerful and flexible replace method:

DataFrame.replace(
        to_replace=None,
        value=None,
        inplace=False,
        limit=None,
        regex=False, 
        method='pad',
        axis=None)

Note, if you need to make changes in place, use inplace boolean argument for replace method:

Inplace

inplace: boolean, default False If True, in place. Note: this will modify any other views on this object (e.g. a column form a DataFrame). Returns the caller if this is True.

Snippet

df['BrandName'].replace(
    to_replace=['ABC', 'AB'],
    value='A',
    inplace=True
)

You could also pass a dict to the pandas.replace method:

data.replace({
    'column_name': {
        'value_to_replace': 'replace_value_with_this'
    }
})

This has the advantage that you can replace multiple values in multiple columns at once, like so:

data.replace({
    'column_name': {
        'value_to_replace': 'replace_value_with_this',
        'foo': 'bar',
        'spam': 'eggs'
    },
    'other_column_name': {
        'other_value_to_replace': 'other_replace_value_with_this'
    },
    ...
})

Just wanted to show that there is no performance difference between the 2 main ways of doing it:

df = pd.DataFrame(np.random.randint(0,10,size=(100, 4)), columns=list('ABCD'))

def loc():
    df1.loc[df1["A"] == 2] = 5
%timeit loc
19.9 ns ± 0.0873 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)


def replace():
    df2['A'].replace(
        to_replace=2,
        value=5,
        inplace=True
    )
%timeit replace
19.6 ns ± 0.509 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

loc function can be used to replace multiple values, Documentation for it : loc

df.loc[df['BrandName'].isin(['ABC', 'AB'])]='A'

Created the Data frame:

import pandas as pd
dk=pd.DataFrame({"BrandName":['A','B','ABC','D','AB'],"Specialty":['H','I','J','K','L']})

Now use DataFrame.replace() function:

dk.BrandName.replace(to_replace=['ABC','AB'],value='A')

This solution will change the existing dataframe itself:

mydf = pd.DataFrame({"BrandName":["A", "B", "ABC", "D", "AB"], "Speciality":["H", "I", "J", "K", "L"]})
mydf["BrandName"].replace(["ABC", "AB"], "A", inplace=True)