How can I check if an element is equal to none in another list of elements in Python?

Question

I'm just getting started to Python, and I'm having an hard time figuring out something with if statements.

Say I have the following two lists:

one = [{'id': '2'}, {'id': 3}]
two = [{'id': '4'}, {'id': 5}]

I want to loop through one and compare the id s of one to two . Basically, what I want to find is if the two lists share an id with the same value. In this example, my code should return a Not found , because 2 is different from 4 and 5, and 3 is different from 4 and 5.

Basically, what i want to do is: check if 2 is equal to any id of two , if not, print something, then the same with 3 . In pseudo code it would be like this: IF there is no id in two EQUAL TO 2 , PRINT something. Then the same with 3 and so on.

I tried the following:

for x in one:
  for y in two:
    if x['id'] != y['id']:
      print('Not found')

The problem with this code is that it will return four Not found , because it's comparing every element. Instead, I simply want to check if there is a 2 or a 3 in two .

Answer 1

Note that you've got what looks like an error in your code, you're using a string for the first ID and an integer for the second one.

There's several ways to check for intersection in python.Here's an article that goes over several different approaches. I like using the set() type, because it has a helpful built-in method called intersection . Sets are a bit different than lists, in that they cannot contain duplicates and are automatically sorted.

For example, you if you had a set that contained 1 and 2, ie set1 = set([1,2]) , you could use set1.intersection([2]) to check if 2 was within that set.

Here's how I would implement such functionality in your case (with all keys as integers):

one = [{'id': 2}, {'id': 3}]
two = [{'id': 4}, {'id': 5}]

def flatten(list_of_dicts):
    return [item['id'] for item in list_of_dicts]

intersecting_ids = set(flatten(one)).intersection(flatten(two))