My Google-fu has failed me.
In Python, are the following two tests for equality equivalent?
n = 5
# Test one.
if n == 5:
print 'Yay!'
# Test two.
if n is 5:
print 'Yay!'
Does this hold true for objects where you would be comparing instances (a list say)?
Okay, so this kind of answers my question:
L = []
L.append(1)
if L == [1]:
print 'Yay!'
# Holds true, but...
if L is [1]:
print 'Yay!'
# Doesn't.
So == tests value where is tests to see if they are the same object?
14 Answers
is will return True if two variables point to the same object (in memory), == if the objects referred to by the variables are equal.
>>> a = [1, 2, 3]
>>> b = a
>>> b is a
True
>>> b == a
True
# Make a new copy of list `a` via the slice operator,
# and assign it to variable `b`
>>> b = a[:]
>>> b is a
False
>>> b == a
True
In your case, the second test only works because Python caches small integer objects, which is an implementation detail. For larger integers, this does not work:
>>> 1000 is 10**3
False
>>> 1000 == 10**3
True
The same holds true for string literals:
>>> "a" is "a"
True
>>> "aa" is "a" * 2
True
>>> x = "a"
>>> "aa" is x * 2
False
>>> "aa" is intern(x*2)
True
Please see this question as well.
There is a simple rule of thumb to tell you when to use == or is.
==is for value equality. Use it when you would like to know if two objects have the same value.isis for reference equality. Use it when you would like to know if two references refer to the same object.
In general, when you are comparing something to a simple type, you are usually checking for value equality, so you should use ==. For example, the intention of your example is probably to check whether x has a value equal to 2 (==), not whether x is literally referring to the same object as 2.
Something else to note: because of the way the CPython reference implementation works, you'll get unexpected and inconsistent results if you mistakenly use is to compare for reference equality on integers:
>>> a = 500
>>> b = 500
>>> a == b
True
>>> a is b
False
That's pretty much what we expected: a and b have the same value, but are distinct entities. But what about this?
>>> c = 200
>>> d = 200
>>> c == d
True
>>> c is d
True
This is inconsistent with the earlier result. What's going on here? It turns out the reference implementation of Python caches integer objects in the range -5..256 as singleton instances for performance reasons. Here's an example demonstrating this:
>>> for i in range(250, 260): a = i; print "%i: %s" % (i, a is int(str(i)));
...
250: True
251: True
252: True
253: True
254: True
255: True
256: True
257: False
258: False
259: False
This is another obvious reason not to use is: the behavior is left up to implementations when you're erroneously using it for value equality.
Is there a difference between
==andisin Python?
Yes, they have a very important difference.
==: check for equality - the semantics are that equivalent objects (that aren't necessarily the same object) will test as equal. As the documentation says:
The operators <, >, ==, >=, <=, and != compare the values of two objects.
is: check for identity - the semantics are that the object (as held in memory) is the object. Again, the documentation says:
The operators
isandis nottest for object identity:x is yis true if and only ifxandyare the same object. Object identity is determined using theid()function.x is not yyields the inverse truth value.
Thus, the check for identity is the same as checking for the equality of the IDs of the objects. That is,
a is b
is the same as:
id(a) == id(b)
where id is the builtin function that returns an integer that "is guaranteed to be unique among simultaneously existing objects" (see help(id)) and where a and b are any arbitrary objects.
