When processing a list a specific value is skipped

Question

def checkio(array: list) -> int:
    """
    Sums even-indexes elements and multiply at the last
    """
    a = []
    if array:
        for i in array:
            if array.index(i) == 0 or array.index(i) % 2 == 0:
                a.append(i)
        print(a)
        return sum(a) * array[-1]
    else:
        return 0


checkio([-37, -36, -19, -99, 29, 20, 3, -7, -64, 84, 36, 62, 26, -76, 55, -24, 84, 49, -65, 41])

The output for print(a) is [-37, -19, 29, 3, -64, 36, 26, 55, -65], but 84 is not included in the list. Why is it doing so?

`array.index(i)` will return index of 1st match element, since you have two `84` but for both it'll return index of 1st `84` which is 9 odd — sittsering, Oct 08 '21 at 17:08
Why it's not working as expected: [Index of duplicates items in a python list](https://stackoverflow.com/questions/5419204/index-of-duplicates-items-in-a-python-list) — mkrieger1, Oct 08 '21 at 17:09
What you want to do instead: https://stackoverflow.com/questions/23130300/split-list-into-2-lists-corresponding-to-every-other-element — mkrieger1, Oct 08 '21 at 17:09
I don't think you're using `index()` quite correctly. `s.index(x)` returns the "index of the *first* occurrence of `x` in `s`: https://docs.python.org/3/library/stdtypes.html#typesseq-common Note that "84" occurs in your list more than once. — jjramsey, Oct 08 '21 at 17:11

Alain T. · Answer 1 · 2021-10-08T17:43:40.350

The index(i) method will always return the index of the first occurence of a value. Since the first occurrence of 84 is at an odd index (9) it is excluded every time it is encountered even if farther down the list there is an 84 at an even index (16).

If you want to select values at even indexes you could use the enumerate() function to get each item's index along with its value:

for index,value in enumerate(array):
    if index%2 == 0: 
        a.append(value)

If you're only going to add up the values, you don't need to place them in a separate list, you can simply keep a running total:

total = 0
for index,value in enumerate(array):
    if index%2 == 0: 
        total += value

An even simpler way would be to use a striding subscript on the list to sum only the values at even indexes (i.e stepping by 2). Your function would then have only a single line:

return sum(array[::2]) * sum(array[-1:]) # zero if array is empty

score 0 · Answer 2 · answered Oct 08 '21 at 17:12

Because the index(x) method returns the position of the first value that is equal to x as written in Python documentation, you can find here:

https://docs.python.org/3/tutorial/datastructures.html

so the array.index(84) will return 9 and so it will not be appended to "a".

score 0 · Accepted Answer · answered Oct 08 '21 at 17:17

0

array.index(i) will return the index of 1st occurrence of i, since you have two 84 but for both, it'll return index of 1st 84 which is 9 odd.
use enumerate()

for i,n in enumerate(array):
    if i % 2 == 0:
        a.append(n)

answered Oct 08 '21 at 17:17

sittsering

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When processing a list a specific value is skipped

3 Answers3