lists comparing with nested values

Question

I have a list of objects, that among other attributes have an attribute named "id"


listA = [{"id":"5","age":"44"},{"id":"8","age":"34"},{"id":"15","age":"84"}]

I have a list of strings such as :


listB = ["9","1","7","5","10","15","20"]

I wish to extract into a third list, all object ids from listA that have an id value in listB

the expected result would be :


listResult = ["5","15"]

How can I do this without using loops?

`[obj["id"] for obj in listA if obj["id"] in listB]`? That's a loop, but then so would be any other reasonable solution to the problem. — jasonharper, Mar 07 '23 at 16:39
@PranavHosangadi maybe they wan't to avoid the use of `for` and `while`, although I agree that with `map` and `filter` you are still looping. — Victor Oliveira, Mar 07 '23 at 16:46

score 3 · Accepted Answer · answered Mar 07 '23 at 16:39

listA = [{"id":"5","age":"44"},{"id":"8","age":"34"},{"id":"15","age":"84"}]
listB = ["9","1","7","5","10","15","20"]
listC = [ x['id'] for x in listA if x['id'] in listB ]

print(listC) 
# ['5', '15']

Get each x['id'] for each x in listA where x['id'] in listB
The in means if x['id'] exists in listB:
Fastest way to check if a value exists in a list

Adrien Riaux · Answer 2 · 2023-03-07T17:26:18.593

3

Another way is using set and the intersection method:

listA = [{"id":"5","age":"44"},{"id":"8","age":"34"},{"id":"15","age":"84"}]
listB = ["9","1","7","5","10","15","20"]

listC = list(set(listB).intersection(set(d['id'] for d in listA)))
print(listC)

The result is ['15', '5'].

edited Mar 07 '23 at 17:26

answered Mar 07 '23 at 16:46

Adrien Riaux

266
9

1

The `set` constructor takes generator expressions so you can avoid the list comprehension around `d['id'] for ...` – Pranav Hosangadi Mar 07 '23 at 16:52

score 0 · Answer 3 · answered Mar 07 '23 at 16:45

You can use filter and list to achieve it, although it would make more sense to simply use list comprehension like in @0stone0 answer.

list_a = [{"id":"5","age":"44"},{"id":"8","age":"34"},{"id":"15","age":"84"}]
list_b = ["9","1","7","5","10","15","20"]
list_a_ids = map(lambda object: object['id'], list_a)
list_result = list(filter(lambda id: id in list_b, list_a_ids))

score 0 · Answer 4 · answered Mar 07 '23 at 17:37

0

Combining the filter() and map() function with an unpacking assignment will do it without an explicit loop (although these function do perform loops internally):

*listResult, = filter(listB.__contains__,map(lambda d:d["id"],listA))

answered Mar 07 '23 at 17:37

Alain T.

40,517
4
31
51

lists comparing with nested values

4 Answers4