In Python 3.x ,

range(0,3) returns a class of immutable iterable objects that lets you iterate over them, it does not produce lists, and they do not store all the elements in the range in memory, instead they produce the elements on the fly (as you are iterating over them) , whereas list(range(0,3)) produces a list (by iterating over all the elements and appending to the list internally) .

Example -

>>> range(0,3)
range(0, 3)
>>> list(range(0,3))
[0, 1, 2]

Ideally, if you only want to iterate over that range of values , range(0,3) would be faster than (list(range(0,3)) because the latter has the overhead of producing a list before you start iterating over it.

In Python 2.x , range(0,3) produces an list, instead we also had an xrange() function that has similar behavior of range() function from Python 3.x (xrange was renamed to range in Python 3.x)

For Python 3.5, From the documentation -

Range objects implement the collections.abc.Sequence ABC, and provide features such as containment tests, element index lookup, slicing and support for negative indices

So you can do things like -

>>> range(0,10)[5]
5
>>> range(0,10)[3:7]
range(3, 7)
>>> 5 in range(6,10)
False
>>> 7 in range(1,8)
True

And all of these are constant time operations , as can be seen from this test -

In [11]: %timeit a = xrange(0,1000000)[1000]
1000000 loops, best of 3: 342 ns per loop

In [12]: %timeit a = xrange(0,1000000)[10000]
1000000 loops, best of 3: 342 ns per loop

In [13]: %timeit a = xrange(0,1000000)[100000]
1000000 loops, best of 3: 342 ns per loop

In [14]: %timeit a = xrange(0,1000000)[999999]
1000000 loops, best of 3: 342 ns per loop

In [15]: %timeit a = xrange(0,10000000)[9999999]
1000000 loops, best of 3: 339 ns per loop

In [16]: %timeit a = xrange(0,1000000000000)[9999999999]
1000000 loops, best of 3: 341 ns per loop

In Python 2.x, range returns a list, but in Python 3.x range returns an immutable sequence, of type range.

Python 2.x:

>>> type(range(10))
<type 'list'>
>>> type(xrange(10))
<type 'xrange'>

Python 3.x:

>>> type(range(10))
<class 'range'>

In Python 2.x, if you want to get an iterable object, like in Python 3.x, you can use xrange function, which returns an immutable sequence of type xrange.

Advantage of xrange over range in Python 2.x:

The advantage of xrange() over range() is minimal (since xrange() still has to create the values when asked for them) except when a very large range is used on a memory-starved machine or when all of the range’s elements are never used (such as when the loop is usually terminated with break).

Note:

Furthermore, the only apparent way to access the integers created by range() is to iterate through them,

Nope. Since range objects in Python 3 are immutable sequences, they support indexing as well. Quoting from the range function documentation,

Ranges implement all of the common sequence operations except concatenation and repetition

...

Range objects implement the collections.abc.Sequence ABC, and provide features such as containment tests, element index lookup, slicing and support for negative indices.

For example,

>>> range(10, 20)[5]
15
>>> range(10, 20)[2:5]
range(12, 15)
>>> list(range(10, 20)[2:5])
[12, 13, 14]
>>> list(range(10, 20, 2))
[10, 12, 14, 16, 18]
>>> 18 in range(10, 20)
True
>>> 100 in range(10, 20)
False

All these are possible with that immutable range sequence.

Recently, I faced a problem and I think it would be appropriate to include here. Consider this Python 3.x code

from itertools import islice
numbers = range(100)
items = list(islice(numbers, 10))
while items:
    items = list(islice(numbers, 10))
    print(items)

One would expect this code to print every ten numbers as a list, till 99. But, it would run infinitely. Can you reason why?

Solution

Because the range returns an immutable sequence, not an iterator object. So, whenever islice is done on a range object, it always starts from the beginning. Think of it as a drop-in replacement for an immutable list. Now the question comes, how will you fix it? Its simple, you just have to get an iterator out of it. Simply change

numbers = range(100)

to

numbers = iter(range(100))

Now, numbers is an iterator object and it remembers how long it has been iterated before. So, when the islice iterates it, it just starts from the place where it previously ended.

list() is a function call, it takes an iterator as input and converts it into a list.

[] simply wraps whatever we put inside the bracket with a list.

So counter to your example, list(32) will throw an error but [32] will make a list with 32 as its element.