It is my understanding that the
range() function, which is actually an object type in Python 3, generates its contents on the fly, similar to a generator.
This being the case, I would have expected the following line to take an inordinate amount of time because, in order to determine whether 1 quadrillion is in the range, a quadrillion values would have to be generated:
1_000_000_000_000_000 in range(1_000_000_000_000_001)
Furthermore: it seems that no matter how many zeroes I add on, the calculation more or less takes the same amount of time (basically instantaneous).
I have also tried things like this, but the calculation is still almost instant:
# count by tens 1_000_000_000_000_000_000_000 in range(0,1_000_000_000_000_000_000_001,10)
If I try to implement my own range function, the result is not so nice!
def my_crappy_range(N): i = 0 while i < N: yield i i += 1 return
What is the
range() object doing under the hood that makes it so fast?
Martijn Pieters’s answer was chosen for its completeness, but also see abarnert’s first answer for a good discussion of what it means for
range to be a full-fledged sequence in Python 3, and some information/warning regarding potential inconsistency for
__contains__ function optimization across Python implementations. abarnert’s other answer goes into some more detail and provides links for those interested in the history behind the optimization in Python 3 (and lack of optimization of
xrange in Python 2). Answers by poke and by wim provide the relevant C source code and explanations for those who are interested.
The Python 3
range() object doesn’t produce numbers immediately; it is a smart sequence object that produces numbers on demand. All it contains is your start, stop and step values, then as you iterate over the object the next integer is calculated each iteration.
The object also implements the
object.__contains__ hook, and calculates if your number is part of its range. Calculating is a (near) constant time operation *. There is never a need to scan through all possible integers in the range.
range() object documentation:
The advantage of the
rangetype over a regular
tupleis that a range object will always take the same (small) amount of memory, no matter the size of the range it represents (as it only stores the
stepvalues, calculating individual items and subranges as needed).
So at a minimum, your
range() object would do:
class my_range: def __init__(self, start, stop=None, step=1, /): if stop is None: start, stop = 0, start self.start, self.stop, self.step = start, stop, step if step < 0: lo, hi, step = stop, start, -step else: lo, hi = start, stop self.length = 0 if lo > hi else ((hi - lo - 1) // step) + 1 def __iter__(self): current = self.start if self.step < 0: while current > self.stop: yield current current += self.step else: while current < self.stop: yield current current += self.step def __len__(self): return self.length def __getitem__(self, i): if i < 0: i += self.length if 0 <= i < self.length: return self.start + i * self.step raise IndexError('my_range object index out of range') def __contains__(self, num): if self.step < 0: if not (self.stop < num <= self.start): return False else: if not (self.start <= num < self.stop): return False return (num - self.start) % self.step == 0
This is still missing several things that a real
range() supports (such as the
.count() methods, hashing, equality testing, or slicing), but should give you an idea.
I also simplified the
__contains__ implementation to only focus on integer tests; if you give a real
range() object a non-integer value (including subclasses of
int), a slow scan is initiated to see if there is a match, just as if you use a containment test against a list of all the contained values. This was done to continue to support other numeric types that just happen to support equality testing with integers but are not expected to support integer arithmetic as well. See the original Python issue that implemented the containment test.
* Near constant time because Python integers are unbounded and so math operations also grow in time as N grows, making this a O(log N) operation. Since it’s all executed in optimised C code and Python stores integer values in 30-bit chunks, you’d run out of memory before you saw any performance impact due to the size of the integers involved here.
Answered By – Martijn Pieters
Answer Checked By – Mildred Charles (BugsFixing Admin)