How to find the location of the nth occurrence of an element in a python list

Question

I need to create a function that takes two arguments: a list lst and a number num. If an element occurs in lst more than num times, remove the extra occurrence(s) and return the result.

So far I have:

def delete_occurrences(lst, num):
    for x in lst:
        if lst.count(x) > num:
            lst.pop(lst.index(x, lst.index(x)+1))
    return lst

However for an case such as ([1, 1, 3, 3, 7, 2, 2, 2, 2], 3) it doesn't delete the correct repetitions.

Cheers for the edit. I was just doing it myself but you beat me to it! — UntilComputersTakeOver, Jan 29 '23 at 12:17
As a general rule its bad to modify lists while iterating over it i.e. [Modifying list while iterating [duplicate] ](https://stackoverflow.com/questions/1637807/modifying-list-while-iterating#:~:text=The%20general%20rule%20of%20thumb,loop%20after%20your%20initial%20loop.) — DarrylG, Jan 29 '23 at 12:19
Agreed, I reckon there's a way to get it in one return statement but I can't figure out what that is — UntilComputersTakeOver, Jan 29 '23 at 12:22
You could always keep your logic, but iterating over the list indices, in reverse order. — Swifty, Jan 29 '23 at 12:23
@DarrylG: won't this remove ALL occurrences of the surnumerary elements ? — Swifty, Jan 29 '23 at 12:30

Timeless · Answer 1 · 2023-01-29T12:59:50.760

3

IIUC, use list.count with a listcomp to slice the extra-occurrences :

L = [1, 1, 3, 3, 7, 2, 2, 2, 2]

def delete_occurrences(lst, num):
    return [x for i, x in enumerate(lst) if lst[:i+1].count(x) <= num]

Output :

delete_occurrences(L, 3)
#[1, 1, 3, 3, 7, 2, 2, 2]

edited Jan 29 '23 at 12:59

answered Jan 29 '23 at 12:38

Timeless

22,580
4
12
30

1

Perhaps `<= num` since only remove if "lst *more than* num times" – DarrylG Jan 29 '23 at 12:42
Also, we want to include the current index in the count, so should be `if lst[:i+1].count(x) <= num` (i.e. posted solution incorrect for `delete_occurrences(L, 1)`) – DarrylG Jan 29 '23 at 12:57
Fair enough, answer updated. Thanks DarryIG ;) – Timeless Jan 29 '23 at 13:00
1

Actually, if you ignore both of my suggestions then your original answer is also correct. – DarrylG Jan 29 '23 at 13:08

score 1 · Answer 2 · answered Jan 29 '23 at 12:26

1

def delete_occurrences(lst, num):
    i = 0
    while i < len(lst) :
        if lst.count(lst[i]) > num:
            lst.pop(i)
            i-=1
        i+=1
    return lst

answered Jan 29 '23 at 12:26

Nehal Birla

142
1
14

score 1 · Answer 3 · answered Jan 29 '23 at 12:28

Here's a solution that keeps your general logic, parsing the list in reverse order so as not to mess up the iteration:

def delete_occurrences(lst, num):
    for i in range(len(lst)-1,-1,-1):
        if lst.count(lst[i]) > num:
            lst.pop(i)
    return lst

RifloSnake · Answer 4 · 2023-01-29T15:25:45.613

This works in the way you like:

def delete_occurrences(lst, num):
    element_work_done = []
    for element in lst:
        if lst.count(element) > num and element not in element_work_done:
            elm_indices = [i for i, x in enumerate(lst) if x == element]
            abundant_occurrences = elm_indices[-(lst.count(element)-num):]
            index = 0
            for occurrence in abundant_occurrences:
                lst.pop(occurrence-index)
                index += 1
            element_work_done.append(element)
    return lst

For bigger lists and data samples, this function out-performs other methods for speed of execution by 10x.

How to find the location of the nth occurrence of an element in a python list

4 Answers4