Python generator function to loop over iterable sequence while eliminating duplicates

I am trying to create a generator function that loops over an iterable sequence while eliminating duplicates and then returns each result in order one at a time (not as a set or list), but I am having difficulty getting it to work. I have found similar questions here, but the responses pretty uniformly result in a list being produced.

I would like the output to be something like:

>>> next(i)
2
>>> next(i)
8
>>> next(i)
4....

I was able to write it as a regular function that produces a list:

def unique(series):
    new_series = []
    for i in series:
        if i not in new_series:
            new_series.append(i)
    return new_series

series = ([2,8,4,5,5,6,6,6,2,1])
print(unique(series))

I then tried rewriting it as a generator function by eliminating the lines that create a blank list and that append to that list, and then using "yield" instead of "return"; but I’m not getting it to work:

def unique(series):
    for i in series:
        if i not in new_series:
            yield new_series

I don't know if I'm leaving something out or putting too much in. Thank you for any assistance.

Solution

Well, to put it simply, you need something to "remember" the values you find. In your first function you were using the new list itself, but in the second one you don't have it, so it fails. You can use a set() for this purpose.

def unique(series):
    seen = set()

    for i in series:
        if i not in seen:
            seen.add(i)
            yield i

Also, yield should "yield" a single value at once, not the entire new list.

To print out the elements, you'll have to iterate on the generator. Simply doing print(unique([1, 2, 3])) will print the resulting generator object.

>>> print(unique([1, 1, 2, 3]))
<generator object unique at 0x1023bda98>

>>> print(*unique([1, 1, 2, 3]))
1 2 3

>>> for x in unique([1, 1, 2, 3]):
    print(x)
1
2
3

Note: * in the second example is the iterable unpack operator.