Remove duplicate of a list via list matching in Python

I have chosen a slightly different procedure to remove duplicates of a list. I want to keep a new list in parallel, in which each duplicate is added. Afterwards I check if the element is present in the "newly created list" in order to delete it.

The code looks like this:

# nums = [1,1,2] or [0,0,1,1,1,2,2,3,3,4]

t = []
nums_new = nums
for i in nums:
    if nums[i] not in t:
        t.append(nums[i])
    else:
        nums_new.remove(nums[i])
nums = nums_new
print(nums)

For the case when nums = [1,1,2] this works fine and returns [1,2].

However, for nums = [0,0,1,1,1,2,2,3,3,4] this case does not seem to work as I get the following output: [0, 1, 2, 2, 3, 3, 4].

Why is this? Can someone explain me the steps?

>Solution :

There are two issues with your code:

1. Since you are iterating over a list, for i in nums, i is your actual number, not the indexer, so you should just use i instead of nums[i].

2. nums_new = nums will not actually make a copy of nums, but instead it will make a second binding to the very same list. So you should write nums_new = nums.copy() to avoid changing your original list while you iterate over it.

With those two changes your code works as you wish:

nums = [0,0,1,1,1,2,2,3,3,4]

t = []
nums_new = nums.copy()
for i in nums:
    if i not in t:
        t.append(i)
    else:
        nums_new.remove(i)

nums = nums_new

print(nums)

Returns [0,1,2,3,4].

Of course this is an academic exercise of some kind, but note that the Pythonic way to remove duplicates from a list is:

• list(dict.fromkeys(nums)) if you want to preserve the order, and
• list(set(nums)) for a slightly faster method if you do not care about order.