Identify distinct elements in a list and map it to a corresponding index in another list in python

Question

I have 2 different lists:

l1 = ['a','b','a','e','b','c','a','d']
l2 = ['t1','t2','t3','t4','t5','t6','t7','t8']

The lengths of l1 and l2 will always be the same. They're in fact logical mappings - each item in l1 corresponds to a value in l2.

I wanted to identify distinct elements in l1. I did that using set and list comprehension as follows:

used = set()
distl1 = [x for x in l1 if x not in used and (used.add(x) or True)]

Here, the output will be:

distl1 = ['a','b','e','c','d']

which is nothing but the first occurrence of every distinct element.

Now, how do I build a list distl2 so that I get the output as the value in l2 that corresponds to the first occurrence's value i.e., distl1?

distl2 = ['t1','t2','t4','t6','t8']

Answer 1

My idea is to use an OrderedDict to build a mapping of (key, value) pairs corresponding to the elements of l1 and l2 and then extract the values from that dict as a list.

>>> from collections import OrderedDict
>>> 
>>> l1 = ['a','b','a','e','d','c','a','b']
>>> l2 = ['t1','t2','t3','t4','t5','t6','t7','t8']
>>> 
>>> d = OrderedDict()
>>> for k, v in zip(l1, l2):
...:    if k not in d: # <--- check if this key has already been seen!
...:        d[k] = v
...:        
>>> distl2 = list(d.values())
>>> distl2
>>> ['t1', 't2', 't4', 't5', 't6']

Note for Python 3.7+ users: regular dicts are guaranteed to remember their key insertion order, so you can omit importing the OrderedDict.

Answer 2

You can also do this:

distl2 = [l2[l1.index(key)] for key in distl1]