在您的第一个函数中，mylist[0::n]是[1, 3]，因为0::n表示第一个元素为0，其他元素每第n个 元素在第一个之后.正如Daniel所建议的，您可以使用mylist[::n]，这意味着每第n个 个元素.

在您的第二个函数索引中，索引以0开头，而0 % 0为0，因此它不会复制第一个元素.第三个元素(2 % 2为0)相同.所以您要做的就是new_list = [item for index, item in enumerate(mylist) if (index + 1) % n != 0]

提示:在此类功能中，您可能希望使用return而不是print().

I had already looked through this post: Python: building new list from existing by dropping every n-th element, but for some reason it does not work for me:

I tried this way:

def drop(mylist, n):
    del mylist[0::n]
    print(mylist)

This function takes a list and n. Then it removes every n-th element by using n-step from list and prints result.

Here is my function call:

drop([1,2,3,4],2)

Wrong output:
[2, 4] instead of [1, 3]

---

Then I tried a variant from the link above:

def drop(mylist, n):
    new_list = [item for index, item in enumerate(mylist) if index % n != 0]
    print(new_list)

Again, function call:

drop([1,2,3,4],2)

Gives me the same wrong result: [2, 4] instead of [1, 3]

---

How to correctly remove/delete/drop every n-th item from a list?

In your first function mylist[0::n] is [1, 3] because 0::n means first element is 0 and other elements are every nth element after first. As Daniel suggested you could use mylist[::n] which means every nth element.

In your second function index is starting with 0 and 0 % 0 is 0, so it doesn't copy first element. It's same for third element (2 % 2 is 0). So all you need to do is new_list = [item for index, item in enumerate(mylist) if (index + 1) % n != 0]

Tip: you may want to use return instead of print() in functions like these.