Replacing a None value in a list with an index in Python

Question

In this question I have to substitute all None values in a list with the mean of the number before and after the None value. So if i have a list with

[1, None, 3, None, 7]

The desired output is

[1, 2, 3, 5, 7]

My current code:

def substitution(L):
    
    result = []
    
    for i in L:
        if i is None:
            result.append(L[3])
            
        else:
            result.append(i)
            
    return result

list = [4, None, 6, 2, 3, None, 8, 11]
test = substitution(list)
print(test)

My output makes no sense, I only came as far as to substitute the Nones with an index but can’t figure out how to get the mean of the index before and after it and then print it.

[4, 2, 6, 2, 3, 2, 8, 11]

The desired output is as follows:

[4, 5, 6, 2, 3, 5.5, 8, 11]

Thanks in advance 🙂

Asked By: albrit

||

Source

Answer 1

If you want to use a for loop, you can loop over the indices (the position of a given element in the list) instead of the actual elements of the list:

def substitution(L):
    
    result = []
    
    for i in range(len(L)):
        if L[i] is None:
            result.append((L[i-1] + L[i+1])/2)
            
        else:
            result.append(L[i])
            
    return result

Answered By: nathan liang

Replacing a None value in a list with an index in Python

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