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Home/ Questions/Q 8981047
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T20:14:10+00:00 2026-06-15T20:14:10+00:00

I have created a Pandas DataFrame df = DataFrame(index=[‘A’,’B’,’C’], columns=[‘x’,’y’]) Now, I would like

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I have created a Pandas DataFrame

df = DataFrame(index=['A','B','C'], columns=['x','y'])

Now, I would like to assign a value to particular cell, for example to row C and column x. In other words, I would like to perform the following transformation:

     x    y             x    y
A  NaN  NaN        A  NaN  NaN
B  NaN  NaN   ⟶   B  NaN  NaN
C  NaN  NaN        C   10  NaN

with this code:

df.xs('C')['x'] = 10

However, the contents of df has not changed. The dataframe contains yet again only NaNs. How do I what I want?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T20:14:12+00:00Added an answer on June 15, 2026 at 8:14 pm

    RukTech’s answer, df.set_value('C', 'x', 10), is far and away faster than the options I’ve suggested below. However, it has been slated for deprecation.

    Going forward, the recommended method is .iat/.at.


    Why df.xs('C')['x']=10 does not work:

    df.xs('C') by default, returns a new dataframe with a copy of the data, so

    df.xs('C')['x']=10
    

    modifies this new dataframe only.

    df['x'] returns a view of the df dataframe, so

    df['x']['C'] = 10
    

    modifies df itself.

    Warning: It is sometimes difficult to predict if an operation returns a copy or a view. For this reason the docs recommend avoiding assignments with “chained indexing”.


    So the recommended alternative is

    df.at['C', 'x'] = 10
    

    which does modify df.


    In [18]: %timeit df.set_value('C', 'x', 10)
    100000 loops, best of 3: 2.9 µs per loop
    
    In [20]: %timeit df['x']['C'] = 10
    100000 loops, best of 3: 6.31 µs per loop
    
    In [81]: %timeit df.at['C', 'x'] = 10
    100000 loops, best of 3: 9.2 µs per loop
    
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