Question

Given n lists with m dictionaries as their elements, I would like to produce a new list, with a joined set of dictionaries. Each dictionary is guaranteed to have a key called “index”, but could have an arbitrary set of keys beyond that. The non-index keys will never overlap across lists. For example, imagine the following two lists:

l1 = [{"index":1, "b":2}, {"index":2, "b":3}, {"index":3, "green":"eggs"}]
l2 = [{"index":1, "c":4}, {"index":2, "c":5}]

("b" would never appear in l2, since it appeared in l1, and similarly, "c" would never appear in l1, since it appeared in l2)

I would like to produce a joined list:

l3 = [{"index":1, "b":2, "c":4}, 
      {"index":2, "b":3, "c":5}, 
      {"index":3, "green":"eggs"}]

What is the most efficient way to do this in Python?

Answer 1

from collections import defaultdict

l1 = [{"index":1, "b":2}, {"index":2, "b":3}, {"index":3, "green":"eggs"}]
l2 = [{"index":1, "c":4}, {"index":2, "c":5}]

d = defaultdict(dict)
for l in (l1, l2):
    for elem in l:
        d[elem['index']].update(elem)
l3 = d.values()

# l3 is now:

[{'b': 2, 'c': 4, 'index': 1},
 {'b': 3, 'c': 5, 'index': 2},
 {'green': 'eggs', 'index': 3}]

EDIT: Since l3 is not guaranteed to be sorted (.values() returns items in no specific order), you can do as @user560833 suggests:

from operator import itemgetter

...

l3 = sorted(d.values(), key=itemgetter("index"))