How to find first value in a list having no duplicates?

Question

l1 = ['A','B','C','D','A','B']
l2 = []

'C' is the first value in list l1, i want to create a function so that it returns C in l2.

Answer 1

In 3.6 and higher, this is very easy. Now that dicts preserve insertion order, collections.Counter can be used to efficiently count all elements in a single pass, then you can just scan the resulting Counter in order to find the first element with a count of 1:

from collections import Counter

l1 = ['A','B','C','D','A','B']
l2 = [next(k for k, v in Counter(l1).items() if v == 1)]

Work is strictly O(n), with only one pass of the input required (plus a partial pass of the unique values in the Counter itself), and the code is incredibly simple. In modern Python, Counter even has a C accelerator for counting inputs that pushes all the Counter construction work to the C layer, making it impossible to beat. If you want to account for the possibility that no such element exists, just wrap the l2 initialization to make it:

try:
    l2 = [next(k for k, v in Counter(l1).items() if v == 1)]
except StopIteration:
    l2 = []
    # ... whatever else makes sense for your scenario ...

or avoid exception handling with itertools.islice (so l2 is 0-1 items, and it still short-circuits once a hit is found):

from itertools import islice

l2 = list(islice((k for k, v in Counter(l1).items() if v == 1), 1))