Python & Pandas: Structuring conditionals

Question

Probably a rookie question, but I'm new to Python and Pandas and still learning the syntax. I'm developing a script where I'd like to structure a conditional that multiplies a cell value over an entire column.

Here's the structure of the test.csv:

I'd like to create a column error_count_score that multiplies the value in the error_count cell by 3 if it is greater than 3 (as an Excel formula, it would be: =IF(C2<=3,0,(C2*3)) ). The values for the new column in this case would be 0 and 12.

I suspect there are multiple ways to solve this, but I'm uncertain how to structure the code. Here's my working code attempt:

import pandas as pd

df = pd.read_csv("test.csv")
    
df.loc[df['error_count'] >= 3, 'error_count_score'] = #do I put an object here? * 3

Any assistance or advice greatly appreciated.

jfaccioni · Accepted Answer · 2021-07-27 14:44:23Z

You can create a function that does exactly what you want for a single element in the column, and then apply that function over every element of the column, generating your new column.

This is done via the DataFrame.apply method.

In your case:

def my_function(element):
    if element <= 3:
        return 0
    else: 
        return element * 3

df['error_count_score'] = df['error_count'].apply(my_function)

Note that pandas expects my_function to take an argument (each element of a column), and then apply it through some logic, returning a new element in the end. This is exactly what's happening here.

As you get used to pandas, you'll see that a lot of times people don't bother defining a function just to apply it a single time. In these cases, people will use an anonymous function - basically a function that only exists for one line - using the lambda keyword.

In your case:

df['error_count_score'] = df['error_count'].apply(lambda element: 0 if element <= 3 else element * 3)

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