How to check efficiently two list of dictionary items in Python?

Question

I have two lists:

l1 = [{"a":1, "b":2, "c":"pqr"}, {"a":3, "b":6, "c":"pir"}, {"a":2, "b":4, "c":""}]

l2 = [{"a":1, "b":3, "c":"def"}, {"a":2, "b":7, "c":"xyz"}]

I want to compare l1_item['a'] with l2_item['a']. If they match then I want to print l2_item['c']

I have used nested for loop to check each list of dictionary and then compare the value of l1_item['a'] with l2_item['a'].

I want to know if there is any other efficient way to check it without using nested for loop. I am having time complexity issue for larger data sets.

Answer 1

You could create a dictionary for the lookup where the value of the a-item is the key.

l1 = [{"a": 1, "b": 2, "c": "pqr"}, {"a": 3, "b": 6, "c": "pir"},
      {"a": 2, "b": 4, "c": ""}]
l2 = [{"a": 1, "b": 3, "c": "def"}, {"a": 2, "b": 7, "c": "xyz"}]

lookup = {item['a']: item for item in l2}
for entry in l1:
    item = lookup.get(entry['a'])
    if item:
        print(item['c'])

You need to create the dictionary once with O(n), but accessing it afterwards is O(1).

---

If the value of a could be in l2 more than once you will have to keep a list of items in the lookup dictionary.

from collections import defaultdict

l1 = [{"a": 1, "b": 2, "c": "pqr"}, {"a": 3, "b": 6, "c": "pir"},
      {"a": 2, "b": 4, "c": ""}]
l2 = [{"a": 1, "b": 3, "c": "def"}, {"a": 2, "b": 7, "c": "xyz"}, {"a": 1, "b": 3, "c": "uvw"}, ]

lookup = defaultdict(list)
for item in l2:
    lookup[item['a']].append(item)

for entry in l1:
    items = lookup.get(entry['a'])
    if items:
        for item in items:
            print(item['c'])

This will give you

def
uvw
xyz