How to sort a dictionary based on a list in python

Question

I have a dictionary

a = {\'ground\': obj1, \'floor 1\': obj2, \'basement\': obj3}

I have a list.

a_list = [\'floor 1\', \'grou         


        

Answer 1

The naive way, sorting the list of (key, value) tuples, using the sorted() function and a custom sort key (called for each (key, value) pair produced by dict.items())):

sorted(a.items(), key=lambda pair: a_list.index(pair[0]))

The faster way, creating an index map first:

index_map = {v: i for i, v in enumerate(a_list)}
sorted(a.items(), key=lambda pair: index_map[pair[0]])

This is faster because the dictionary lookup in index_map takes O(1) constant time, while the a_list.index() call has to scan through the list each time, so taking O(N) linear time. Since that scan is called for each key-value pair in the dictionary, the naive sorting option takes O(N^2) quadratic time, while using a map keeps the sort efficient (O(N log N), linearithmic time).

Both assume that a_list contains all keys found in a. However, if that's the case, then you may as well invert the lookup and just retrieve the keys in order:

[(key, a[key]) for key in a_list if key in a]

which takes O(N) linear time, and allows for extra keys in a_list that don't exist in a.

To be explicit: O(N) > O(N log N) > O(N^2), see this cheat sheet for reference.

Demo:

>>> a = {'ground': 'obj1', 'floor 1': 'obj2', 'basement': 'obj3'}
>>> a_list = ('floor 1', 'ground', 'basement')
>>> sorted(a.items(), key=lambda pair: a_list.index(pair[0]))
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
>>> index_map = {v: i for i, v in enumerate(a_list)}
>>> sorted(a.items(), key=lambda pair: index_map[pair[0]])
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
>>> [(key, a[key]) for key in a_list if key in a]
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]