python sort a list of objects based on attributes in the order of the other list

Question

I am working with Python list sort.

I have two lists: one is a list of integers, the other is a list of objects, and the second object list has the attribute id which is also an integer, I want to sort the object list based on the id attribute, in the order of the same id appears in the first list, well, this is an example:

I got a = [1,2,3,4,5]

and b = [o,p,q,r,s], where o.id = 2, p.id = 1, q.id = 3, r.id = 5, s.id = 4

and I want my list b to be sorted in the order of its id appears in list a, which is like this:

sorted_b = [p, o, q, s, r]

Of course, I can achieve this by using nested loops:

sorted_b = []
for i in a:
    for j in b:
        if j.id == i:
            sorted_b.append(j)
            break

but this is a classic ugly and non-Python way to solve a problem, I wonder if there is a way to solve this in a rather neat way, like using the sort method, but I don't know how.

Answer 1

>>> from collections import namedtuple
>>> Foo = namedtuple('Foo', 'name id') # this represents your class with id attribute
>>> a = [1,2,3,4,5]
>>> b = [Foo(name='o', id=2), Foo(name='p', id=1), Foo(name='q', id=3), Foo(name='r', id=5), Foo(name='s', id=4)]
>>> sorted(b, key=lambda x: a.index(x.id))
[Foo(name='p', id=1), Foo(name='o', id=2), Foo(name='q', id=3), Foo(name='s', id=4), Foo(name='r', id=5)]