为什么列表理解与没有理解的代码的工作方式不同？

Question

I tried to create a simple function to remove duplicates from the list.

x = [7, 7, 5, 6, 8, 9, 9, 0]
for n, i in enumerate(x):
  if i in x[n + 1:]:
    x.remove(i)
print(x)

Output:

[7, 5, 6, 8, 9, 0]

This code works fine as far as I know.

But when I'm trying to convert it in comprehension list form, I'm getting wrong output:

def unique_list(lst):
  return [lst.remove(i) for n, i in enumerate(lst) if i in lst[n + 1:]]

x = [7, 7, 5, 6, 8, 9, 9, 0]
print(unique_list(x))

Output:

[None, None]

So, the question is - why the output is different?

Answer 1

a = [1, 2, 3]
deleted = a.remove(1)
print(deleted)
# returns None
print(a)
# returns [2, 3]

.remove() changes the list in place and returns None

Answer 2

Your two ways of writing the functionality of set(x) are not identical. While the first is using a side-effect of x.remove(..) to "return" your modified list in place, the second form returns a newly created list.

The elements of that list are formed by what is returned by the expression lst.remove(i) , which is None as Manuel already has pointed out in their answer.

You receive [None, None] because your code calls lst.remove 2 times. So your function unique_list could be called count_duplicates , just with a peculiar output format (with the length of a list of None encoding the result instead of a straightforward int ).