How to sort one list based on another?

I think this answers your question:

>>> [x for x in Ref if x in Input]
>>> [3, 2, 11, 10, 9, 8, 5, 4]

Hope it helps.

UPDATE: Making Input a set for faster access:

>>> Input_Set = set(Input)
>>> [x for x in Ref if x in Input_Set]
[3, 2, 11, 10, 9, 8, 5, 4]

Another approach in addition to dcg's answer would be as follows:

Ref = [3, 2, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4]
Input = [9, 5, 2, 3, 10, 4, 11, 8]

ref = set(Ref)
inp = set(Input)

sorted_list = sorted(ref.intersection(inp), key = Ref.index)

This outputs to:

[3, 2, 11, 10, 9, 8, 5, 4]

Here you convert the lists into sets, find their intersection, and sort them. The set is sorted based on the 'Ref' list's indexing.

You can use the sorted method:

# keep in a dict the index for each value from Ref
ref  = {val: i for i, val in enumerate(Ref)}
# sort by the index value from Ref for each number from Input 
sorted(Input, key=ref.get)

output:

[3, 2, 11, 10, 9, 8, 5, 4]