Finding list elements in order in another list, but not consecutively? (Python)

Question

I'm trying to figure out a cleaner (Pythonic) way to tell whether elements in one list exist in another list, in order but not consecutively. For example, I might have the following lists:

list1 = ["I", "her"]
list2 = ["I", "hate", "her", "sometimes"]
list3 = ["her", "hat", "I", "found"]

I want to return True when comparing list1 to list2, because "I" and "her" exist in list2 in that order, although not consecutively. False would be returned when comparing list1 to list3, because "her" and "I" do not appear in the correct order.

I could write a series of loops and if statements to achieve this, but I was wondering if there was a simpler, built-in way to do it. I have tried using the all() function:

all(i in list2 for i in list1)

This works to find whether all the elements in list1 exist in list2, but does not account for ordering. I've also tried using join:

" ".join(str(i) for i in list1) in list2

This will only return True if the elements from list1 are consecutive in list2.

Any ideas on a simpler way to do this?

Your example lists are not valid Python (or, at best, the code is incomplete). — Scott Hunter, May 14 '20 at 18:55
@MadPhysicist: Not sure closed questions are the *best* references... — Scott Hunter, May 14 '20 at 18:59
@ScottHunter why not? The answers posted there are quite valid and recognized as such by the community. — Mad Physicist, May 14 '20 at 19:00
That being said, my absolute favorite is https://stackoverflow.com/a/11820887/2988730. It's about as concise and efficient as you can get. — Mad Physicist, May 14 '20 at 19:01
@MadPhysicist Not quite, I do know how to do what I am aiming for using normal loops. I was just wondering if there was a specific Python function to do this, since Python has so much built-in functionality for manipulating lists. — A.S., May 14 '20 at 19:02
@MadPhysicist The second link you included is the one! Sorry I didn't see it. — A.S., May 14 '20 at 19:03
Should I delete this thread? I apologize for posting a repeat, I'm a bit new. — A.S., May 14 '20 at 19:03

LevB · Answer 1 · 2020-05-14T20:31:33.810

-1

Just use the intersection operator for sets:

result = bool(set(list1) & set(list2))

If you want to count only ordered pairs, use zip.

result = any([el1 == el2 for el1, el2 in zip (list1, list2)])

edited May 14 '20 at 20:31

answered May 14 '20 at 19:08

LevB

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This one doesn't quite work, since it returns True regardless of the ordering of the elements in the second list. For example, it would return True when comparing list1 and list3 also, when it should return False. – A.S. May 14 '20 at 20:01
try the second solution I posted for ordered pairs – LevB May 14 '20 at 20:32

Finding list elements in order in another list, but not consecutively? (Python)

1 Answers1