Custom sort order in Python [duplicate]

Question

Am I able to define my own sort order and still use the Python sort function? For example, say my desired sorting order is:

list1 = ['Stella', 'Bob', 'Joe', 'Betty']

I'd hoped to do something like:

list2 = ['Joe', 'Stella']
sorted(list2, key=list1)

and get:

['Stella', 'Joe']

where the sorted order stays within the custom order in list1. I can do it in multiple steps by swapping out for numeric values but thought there may be a way to do it similar to above.

Answer 1

Note that for the example you have given you are wasting time sorting list2 as you can simply use the already sorted list1 if the items in list2 are unique:

print([e for e in list1 if e in list2])

should print

['Stella', 'Joe']

To improve it even more, you can use a set for faster lookups:

set2 = set(list2)
print([e for e in list1 if e in set2])

This second variant works in O(n) which is faster than any sorting algorithm.

Edit: The answers given by Nick and Jab are close to O(n*m + n*log n). This is a O(n * log n) solution that works for non-unique cases:

lookup_dict = {k: i for i, k in enumerate(list1)}
print(sorted(list2, key=lookup_dict.get))