Delete Starting Even Numbers

Question

This function will repeatedly remove the first element of a list until it finds an odd number or runs out of elements. It will accept a list of numbers as an input parameter and return the modified list where any even numbers at the beginning of the list are removed. To do this, we will need the following steps:

Define our function to accept a single input parameter my_list which is a list of numbers
Loop through every number in the list if there are still numbers in the list and if we haven’t hit an odd number yet
Within the loop, if the first number in the list is even, then remove the first number of the list
Once we hit an odd number or we run out of numbers, return the modified list

I wrote a function called delete_starting_evens() that has a parameter named my_list.

The function should remove elements from the front of my_list until the front of the list is not even. The function should then return my_list.

For example, if my_list started as [4, 8, 10, 11, 12, 15], then delete_starting_evens(my_list) should return [11, 12, 15].

My code:

def delete_starting_evens(my_list):
    for i in my_list:
        if i % 2 == 0:
            my_list.remove(i)
        else:
            break
    return my_list
a=[4,8,10,12,14,15,16,17,19,20]
print(delete_starting_evens(a))

The problem is that the function was removing the even indexes. I was expecting that the function should delete the even numbers until it finds the odd number or runs out of elements. How to solve it?

@PranavHosangadi I agree, he should probably make a second list maybe — LiterallyGutsFromBerserk, Jul 27 '23 at 13:32
Just to show the power of Python, a solution with `itertools` from the standard library. You need only one line of code: `a[:] = list(itertools.dropwhile((lambda x: x % 2 == 0), a))`. If it is not necessary to change the original list - as the question suggests - `a = ...` is sufficient. — Matthias, Jul 27 '23 at 14:03

LiterallyGutsFromBerserk · Accepted Answer · 2023-07-27T13:47:21.527

-1

def delete_starting_evens(my_list):
    idx=-1
    for i in my_list:
        if i % 2 == 1:
            idx = my_list.index(i)
            break
    if(idx==-1):
       return []
    else:
        second_list = my_list[idx:]
        return second_list
a=[4,5,8,10,12]
print(delete_starting_evens(a))

edited Jul 27 '23 at 13:47

answered Jul 27 '23 at 13:30

LiterallyGutsFromBerserk

58
9

2

With `for i in my_list`, `i` _is_ a value not an index – Pranav Hosangadi Jul 27 '23 at 13:33
oh yeah, sorry i mistook it for a in range loop. gimme a second I'll change it – LiterallyGutsFromBerserk Jul 27 '23 at 13:33
Now the new list only includes odd numbers, which is not what OP wanted. Please test your suggestions before posting them. – Pranav Hosangadi Jul 27 '23 at 13:36
@PranavHosangadi I think it's what he wanted now, I'm not sure though – LiterallyGutsFromBerserk Jul 27 '23 at 13:40
A couple improvements: 1. Use `enumerate` to get the index and the value in your for loop. That way you don't need to call `my_list.index` because you already have the index of the current item. 2. You don't need to initialize `second_list = []`, since that list is discarded when you do `second_list = my_list[idx:]` anyway. – Pranav Hosangadi Jul 27 '23 at 13:42
Also this fails on a list containing only even numbers, because it returns `my_list[-1:]` in that case. – Pranav Hosangadi Jul 27 '23 at 13:44
Thanks for pointing out the failure. However, I first went with enumerate but it showed me an error. As for the second list, you're right : I'd only need to initialize it if my index is different from -1. – LiterallyGutsFromBerserk Jul 27 '23 at 13:48

Delete Starting Even Numbers

1 Answers1