Understanding Python's "is" operator

Question

The is operator does not match the values of the variables, but the instances themselves.

What does it really mean?

I declared two variables named x and y assigning the same values in both variables, but it returns false when I use the is operator.

I need a clarification. Here is my code.

x = [1, 2, 3]
y = [1, 2, 3]

print(x is y)  # It prints false!

Solution 1

You misunderstood what the is operator tests. It tests if two variables point the same object, not if two variables have the same value.

From the documentation for the is operator:

The operators is and is not test for object identity: x is y is true if and only if x and y are the same object.

Use the == operator instead:

print(x == y)

This prints True. x and y are two separate lists:

x[0] = 4
print(y)  # prints [1, 2, 3]
print(x == y)   # prints False

If you use the id() function you'll see that x and y have different identifiers:

>>> id(x)
4401064560
>>> id(y)
4401098192

but if you were to assign y to x then both point to the same object:

>>> x = y
>>> id(x)
4401064560
>>> id(y)
4401064560
>>> x is y
True

and is shows both are the same object, it returns True.

Remember that in Python, names are just labels referencing values; you can have multiple names point to the same object. is tells you if two names point to one and the same object. == tells you if two names refer to objects that have the same value.

Solution 2

Another duplicate was asking why two equal strings are generally not identical, which isn't really answered here:

>>> x = 'a' 
>>> x += 'bc'
>>> y = 'abc'
>>> x == y
True
>>> x is y
False

So, why aren't they the same string? Especially given this:

>>> z = 'abc'
>>> w = 'abc'
>>> z is w
True

---

Let's put off the second part for a bit. How could the first one be true?

The interpreter would have to have an "interning table", a table mapping string values to string objects, so every time you try to create a new string with the contents 'abc', you get back the same object. Wikipedia has a more detailed discussion on how interning works.

And Python has a string interning table; you can manually intern strings with the sys.intern method.

In fact, Python is allowed to automatically intern any immutable types, but not required to do so. Different implementations will intern different values.

CPython (the implementation you're using if you don't know which implementation you're using) auto-interns small integers and some special singletons like False, but not strings (or large integers, or small tuples, or anything else). You can see this pretty easily:

>>> a = 0
>>> a += 1
>>> b = 1
>>> a is b
True
>>> a = False
>>> a = not a
>>> b = True
a is b
True
>>> a = 1000
>>> a += 1
>>> b = 1001
>>> a is b
False

---

OK, but why were z and w identical?

That's not the interpreter automatically interning, that's the compiler folding values.

If the same compile-time string appears twice in the same module (what exactly this means is hard to define—it's not the same thing as a string literal, because r'abc', 'abc', and 'a' 'b' 'c' are all different literals but the same string—but easy to understand intuitively), the compiler will only create one instance of the string, with two references.

In fact, the compiler can go even further: 'ab' + 'c' can be converted to 'abc' by the optimizer, in which case it can be folded together with an 'abc' constant in the same module.

Again, this is something Python is allowed but not required to do. But in this case, CPython always folds small strings (and also, e.g., small tuples). (Although the interactive interpreter's statement-by-statement compiler doesn't run the same optimization as the module-at-a-time compiler, so you won't see exactly the same results interactively.)

---

So, what should you do about this as a programmer?

Well… nothing. You almost never have any reason to care if two immutable values are identical. If you want to know when you can use a is b instead of a == b, you're asking the wrong question. Just always use a == b except in two cases:

• For more readable comparisons to the singleton values like x is None.
• For mutable values, when you need to know whether mutating x will affect the y.

Solution 3

Prompted by a duplicate question, this analogy might work:

# - Darling, I want some pudding!
# - There is some in the fridge.

pudding_to_eat = fridge_pudding
pudding_to_eat is fridge_pudding
# => True

# - Honey, what's with all the dirty dishes?
# - I wanted to eat pudding so I made some. Sorry about the mess, Darling.
# - But there was already some in the fridge.

pudding_to_eat = make_pudding(ingredients)
pudding_to_eat is fridge_pudding
# => False

Solution 4

is only returns true if they're actually the same object. If they were the same, a change to one would also show up in the other. Here's an example of the difference.

>>> x = [1, 2, 3]
>>> y = [1, 2, 3]
>>> print x is y
False
>>> z = y
>>> print y is z
True
>>> print x is z
False
>>> y[0] = 5
>>> print z
[5, 2, 3]

Solution 5

is and is not are the two identity operators in Python. is operator does not compare the values of the variables, but compares the identities of the variables. Consider this:

>>> a = [1,2,3]
>>> b = [1,2,3]
>>> hex(id(a))
'0x1079b1440'
>>> hex(id(b))
'0x107960878'
>>> a is b
False
>>> a == b
True
>>>

The above example shows you that the identity (can also be the memory address in Cpython) is different for both a and b (even though their values are the same). That is why when you say a is b it returns false due to the mismatch in the identities of both the operands. However when you say a == b, it returns true because the == operation only verifies if both the operands have the same value assigned to them.

Interesting example (for the extra grade):

>>> del a
>>> del b
>>> a = 132
>>> b = 132
>>> hex(id(a))
'0x7faa2b609738'
>>> hex(id(b))
'0x7faa2b609738'
>>> a is b
True
>>> a == b
True
>>>

In the above example, even though a and b are two different variables, a is b returned True. This is because the type of a is int which is an immutable object. So python (I guess to save memory) allocated the same object to b when it was created with the same value. So in this case, the identities of the variables matched and a is b turned out to be True.

This will apply for all immutable objects:

>>> del a
>>> del b
>>> a = "asd"
>>> b = "asd"
>>> hex(id(a))
'0x1079b05a8'
>>> hex(id(b))
'0x1079b05a8'
>>> a is b
True
>>> a == b
True
>>>

Hope that helps.

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Comments

[David Heffernan]

Python doesn't have pointers. You need to tighten up your terminology.

[Neko]

It does internally, just like Java and so many other languages. In fact, the is operator's functionality shows this.

[David Heffernan]

The implementation details are not what matters. The documentation uses the terminology "object identity". So should you. "The operators is and is not test for object identity: x is y is true if and only if x and y are the same object. x is not y yields the inverse truth value."

[abarnert]

@Neko: CPython internally uses pointers. But obviously Jython (implemented in Java) and PyPy (implemented in a subset of Python) don't use pointers. In PyPy, some objects won't even have an id unless you ask for it.

[Tom Close]

Could be just personal taste (no pun intended) but I found this analogy more confusing than helpful and has got me wanting to eat pudding when I don't have any in my fridge :( I think Mark Ransom's answer, although more boring, is probably more instructive

[Amadan]

@TomClose: There are many fine answers on this question, enough so that there is space for levity. Also, I want pudding too.

[imallett]

So, A is B is the same as id(A) == id(B).

[Martijn Pieters]

@imallett: that's a proxy for the same test, provided you don't store id(A) in a variable and later expect variable == id(B) to still work; if A was deleted in the meantime then B could have been given the same memory location.

[imallett]

Makes sense, and it's also the Right Thing; variable is storing a property of something that formerly existed. There's no way for the runtime to detect that the later usage is erroneous. The key part of the standard is "[id() ]is guaranteed to be unique and constant for this object during its lifetime. Two objects with non-overlapping lifetimes may have the same id() value."

[user2183078]

But try a=123456789 b=123456789

[Tomasz Kurgan]

Not, it doesn't. It might behave similar in most cases, but it's not always true. See this - the very bottom of the page, bullet 6.: > (...), you may notice seemingly unusual behaviour in certain uses of the is operator, like those involving comparisons between instance methods, or constants And the minimal working example: ` class A(object): def foo(self): pass a = A() print a.foo is a.foo print id(a.foo) == id(a.foo) `

[smart]

Everything less than -5 or higher than 256 in Python will be False. Python caches numbers in range [-5, 256].

[Chi-chi]

Thank you for that detailed explanation. Does somebody know: if w and z are identical because of the compiler folding values, why does this also work in the REPL, even using id() to check the references? Using the REPL on Python 3.7

[David Buck]

is is not 'an English version of =='

[Mark Ransom]

Not all immutable objects will be shared as you show, that's an optimization applied by the Python runtime for some objects but not others. The process of sharing small integers is well documented, but I don't think it is for string interning.