Use dictionary.get() function to output item in imbedded list

Question

In the Python Data structures section on Dictionaries, right before the assignment "fuzzy search" it has a description of how the dictionary.get() function works. The assignment itself involves a dictionary where each key is tied to a two item list. Maybe I'm wrong, but based on how the assignment was on the same page as the get() function lesson, I assumed I was supposed to use the get() function in that assignment. I managed to do it in a very roundabout way, but is there any way to do that in a single line of code?

Say I have dict = {"a":[1, 2, 3], "b":[4, 5, 6]}. Is there a way to use dict.get() to return one of the items inside of one of the lists, rather than the full list?

I tried dict.get("a"[0], "Not here") but just returned the full [1, 2, 3] list.

I also tried dict.get("a", "Not here")[0] and that works when the item is in dict, but if it's not in the dict I'll just get the "N" from "Not here" because it's the first letter.

I want it to return 1 on its own, or the full phrase "Not here". Is that possible?

Answer 1

Hi Joel!
Of course, you can do it in a single line of code. For that, you can change "Not here" as a list element. So, get()[0] returns the first element which is the full phrase "Not here" if the key is not found in dict.
Here is it what I mentioned above

dict = {"a":[1, 2, 3], "b":[4, 5, 6]}
print(dict.get("a", ["Not here"])[0])
print(dict.get("c", ["Not here"])[0])

Answer 2

https://code.sololearn.com/cFqK5ZRmnpqD/?ref=app

Answer 3

@Python Learner thanks, that makes sense.