Conditional Replace Pandas

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灰色年华
灰色年华 2020-11-21 10:17

I have a DataFrame, and I want to replace the values in a particular column that exceed a value with zero. I had thought this was a way of achieving this:

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  • 2020-11-21 10:29

    The reason your original dataframe does not update is because chained indexing may cause you to modify a copy rather than a view of your dataframe. The docs give this advice:

    When setting values in a pandas object, care must be taken to avoid what is called chained indexing.

    You have a few alternatives:-

    loc + Boolean indexing

    loc may be used for setting values and supports Boolean masks:

    df.loc[df['my_channel'] > 20000, 'my_channel'] = 0
    

    mask + Boolean indexing

    You can assign to your series:

    df['my_channel'] = df['my_channel'].mask(df['my_channel'] > 20000, 0)
    

    Or you can update your series in place:

    df['my_channel'].mask(df['my_channel'] > 20000, 0, inplace=True)
    

    np.where + Boolean indexing

    You can use NumPy by assigning your original series when your condition is not satisfied; however, the first two solutions are cleaner since they explicitly change only specified values.

    df['my_channel'] = np.where(df['my_channel'] > 20000, 0, df['my_channel'])
    
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  • 2020-11-21 10:30

    np.where function works as follows:

    df['X'] = np.where(df['Y']>=50, 'yes', 'no')
    

    In your case you would want:

    import numpy as np
    df['my_channel'] = np.where(df.my_channel > 20000, 0, df.my_channel)
    
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  • 2020-11-21 10:33

    Try this:

    df.my_channel = df.my_channel.where(df.my_channel <= 20000, other= 0)

    or

    df.my_channel = df.my_channel.mask(df.my_channel > 20000, other= 0)

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  • 2020-11-21 10:36

    I would use lambda function on a Series of a DataFrame like this:

    f = lambda x: 0 if x>100 else 1
    df['my_column'] = df['my_column'].map(f)
    

    I do not assert that this is an efficient way, but it works fine.

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  • 2020-11-21 10:39

    Try

    df.loc[df.my_channel > 20000, 'my_channel'] = 0
    

    Note: Since v0.20.0, ix has been deprecated in favour of loc / iloc.

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  • 2020-11-21 10:45

    .ix indexer works okay for pandas version prior to 0.20.0, but since pandas 0.20.0, the .ix indexer is deprecated, so you should avoid using it. Instead, you can use .loc or iloc indexers. You can solve this problem by:

    mask = df.my_channel > 20000
    column_name = 'my_channel'
    df.loc[mask, column_name] = 0
    

    Or, in one line,

    df.loc[df.my_channel > 20000, 'my_channel'] = 0
    

    mask helps you to select the rows in which df.my_channel > 20000 is True, while df.loc[mask, column_name] = 0 sets the value 0 to the selected rows where maskholds in the column which name is column_name.

    Update: In this case, you should use loc because if you use iloc, you will get a NotImplementedError telling you that iLocation based boolean indexing on an integer type is not available.

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