0

For example, mylist = [0,1,2,0] I want mylist[0] == mylist[-1] to output False as it is a different instance of 0, but it is true because they are both zeroes. Is there a way to do this?

Jamiu S.
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lin0258
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    0 is not a class. There is no such statement as "different instances of 0". And they are even the same object according to Python's small integer pool optimization. – Mechanic Pig Oct 01 '22 at 07:00
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    It sounds like you want to compare indexes, not values at those indexes. – quamrana Oct 01 '22 at 07:18
  • You can use `is` to check whether it is the same object; but in this case, both occurrences of `0` **actually are** the same object. Please see the linked duplicates. – Karl Knechtel Oct 07 '22 at 21:36

3 Answers3

1

In your case the mylist[0] and mylist[-1] are not only equal but also the same (same object in memory)

>>> mylist = [0,1,2,0]
>>> mylist[0] == mylist[-1]
True
>>> mylist[0] is mylist[-1]
True
>>> id(mylist[0])
140736735811200
>>> id(mylist[-1])
140736735811200
>>>

You should not receive a False.

You can read these articles to better understand this topic: https://realpython.com/python-is-identity-vs-equality/, https://anvil.works/articles/pointers-in-my-python-1

JacekK
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0

I guess you want to compare values and their indexes.

mylist = [0,1,2,3,0]
if mylist[0] == mylist[-1] and mylist.index(mylist[0]) == mylist.index(mylist[-1]):
    print(“True”)
else:
    print(“False”)
Olga
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0

It sounds like you want to compare indexes and not the values at those indexes.

This code does this by ignoring the container:

def compare(container, index1, index2):
    return index1 == index2

mylist = [0,1,2,0]
print(compare(mylist, 0, -1))

Output:

False

quamrana
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