map() function on pandas DataFrame in python does not work as expected

Question

I tried the following code and worked on it for a long time to correct it. But I do not know what I am doing wrong.

    def hiLow(x):
    if x > 10:
        print(x, 'is greater than 10')
        df['High/Low'] = 'High'
    else:
        print(x, 'is less than 10' )
        df['High/Low'] = 'Low'

df['total'].map(hiLow)

I expected it will add 'High' in the second and third rows of the 'High/Low' column.

When I replaced df['High/Low'] = 'High' and df['High/Low'] = 'Low' with df['High/Low'] = x it prints 15 for all rows.

In the image,

This output is correct
Why is this all returned 'None'
Why is this showing 'object' when at point 4 it is showing as 'int64'

The problem is that df['High/Low'] = 'Low' replaces the entire column High/Low with 'Low' (the same for df['High/Low'] = 'High') and then the functions returns None (therefore it produces a Series of Nones as you see). You want to return a single value (return 'Low'), not mutating directly the column of df inside the function. — Rodalm
– Rodalm, Commented May 29, 2022 at 20:13

Christian Eslabon · Accepted Answer · 2022-05-29 20:01:34Z

1

def hiLow(x):
    if x > 10:
        print(x, 'is greater than 10')
        str_answer = 'High'
    else:
        print(x, 'is less than 10' )
        str_answer = 'Low'

    return str_answer

df['High/Low'] =  df['total'].apply(lambda x: hiLow(x) )

answered May 29, 2022 at 20:01

Christian Eslabon

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1 Comment

Rodalm Over a year ago

using lambda in .apply(lambda x: hiLow(x) ) is redundant, you can and should simply do .apply(hiLow)

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