Number of occurrences of a number in a list

Question

This program below is supposed to count the number of occurrences of x in a list. Can not identify the error in the code

def count_x( items, x ):
    if items==[]:
        return 0
    first = items.pop(0)
    if first == x:
        return 1 + count_x(items, x)

Asked By: Murshida Mouree

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Source

Answer 1

There are better ways to do this, but it’s worth addressing why your code is throwing an error.

You don’t have a case for when the item popped is not equal to your search item. This causes the function to return a None. Since it is working recursively, it tries to compute int + None, which leads to an error.

The other issue is that you are modifying the list with the function, which you may not want. For example:

def count_x( items, x ):
    print(items)
    if items==[]:
        return 0
    first = items.pop(0)
    if first == x:
        return 1 + count_x(items, x)
    
    else:
        return count_x(items, x)

items = [1, 2, 1, 1]

print(count_x(items, 1))

print(items)

Your items will become an empty list after you run the function.

Answered By: Michael Cao

Answer 2

I think it should work as following:

def count_x(items:list, x ):
    count = 0
    for item in items:
        if item == x:
            count = count + 1
    return count

count = count_x(items, x)

You could also simply use count = items.count(x) tho.

Answered By: kaliiiiiiiii

Answer 3

There are three cases in this recursion:

The list is empty
The popped item is the value you are counting
The popped item is not the value you are counting

You have treated the first two cases, but not the third.

def count_x( items, x ):
    if items==[]:
        return 0
    first = items.pop(0)
    if first == x:
        return 1 + count_x(items, x)

    return count_x(items, x) # Fix

Answered By: thomaoc

Number of occurrences of a number in a list

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