How does `is` really work with primitive types in python?

Question

My understanding is that all primitive types are referentially equal if they are equal in value.

With integers, this seems to be the case shown from the following example:

v1 = 2
v2 = 2
v1 is v2 # prints True

With booleans, this also seems to be the case shown from the following example:

v1 = False
v2 = False
v1 is v2 # prints True

With strings, it starts to get a little confusing. The following example follows the procedure:

v1 = 'testtest'
v2 = 'testtest'
v1 is v2 # prints True

However, as soon as I add a space to the string, it fails:

v1 = 'test test'
v2 = 'test test'
v1 is v2 # prints False

With floats, the pattern just doesn't exist:

v1 = 2.0
v2 = 2.0
v1 is v2 # prints False

Why do strings follow the rule except when spaces are added? Why do floats not follow the rule at all? Any help is appreciated.

"My understanding is that all primitive types are referentially equal if they are equal in value." - not sure where you got that idea, but your tests clearly show it's false. — user2357112, Jul 08 '21 at 09:04
Does this answer your question? [Is there a difference between "==" and "is"?](https://stackoverflow.com/questions/132988/is-there-a-difference-between-and-is) — Alex Waygood, Jul 08 '21 at 09:04
Python doesn't even have a concept of "primitive" types, anyway. This isn't Java. — user2357112, Jul 08 '21 at 09:04
TL;DR: certain small values are internally cached, *interned*. That's an optimisation detail you shouldn't worry about much. Simply never use `is` in this way, it's not supposed to be used for these operations. — deceze, Jul 08 '21 at 09:22

score 0 · Answer 1 · answered Jul 08 '21 at 09:10

With integers, this seems to be the case shown from the following example:

v1 = 2
v2 = 2
v1 is v2 # prints True

You are spoiled by implementation detail which made certain integers to behave that way, consider counter-example:

t1 = 1000000
t2 = 1000000
print(t1 is t2)  # False

1 Answers1