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The easiest way is to use the replace method on the column. The arguments are a list of the things you want to replace (here ['ABC', 'AB'] ) and what you want to replace them with (the string 'A' in this case):

>>> df['BrandName'].replace(['ABC', 'AB'], 'A')
0    A
1    B
2    A
3    D
4    A
This creates a new Series of values so you need to assign this new column to the correct column name:
df['BrandName'] = df['BrandName'].replace(['ABC', 'AB'], 'A')
                One tricky thing if your datatypes are messed up in the dataframe (ie they look like strings but are not), use: df['BrandName'] = df['BrandName'].str.replace(['ABC', 'AB'], 'A')
– ski_squaw
                Sep 15, 2017 at 17:28
                If you would like to extend this to an entire dataframe, it will be df = df.replace(['ABC', 'AB'], 'A')
– user42
                Apr 14, 2022 at 12:31
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
                thanks for the snippet example, but it does not work. For one, if there is no = in the to_replace portion it errors out. For another, it is not making any replacements. Is there anyway to get a working example of the replace functionality in v 0.20.1?
– Alison S
                May 21, 2017 at 17:57
                Does replace not scale well? It seems to crash my machine when replacing ~5 million rows of integers. Any way around this?
– guy
                Sep 27, 2017 at 14:29
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'
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)
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)
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')
You can use loc for replacing based on condition and specifying the column name
df = pd.DataFrame([['A','H'],['B','I'],['ABC','ABC'],['D','K'],['AB','L']],columns=['BrandName','Col2'])
df.loc[df['BrandName'].isin(['ABC', 'AB']),'BrandName'] = 'A'
Output 

        
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