Backwards Ranges in Python
In Python, if you want to specify a sequence of numbers
from a up to (but excluding) b,
you can write range(a, b).
This generates the sequence a, a+1, a+2, ..., b-1.
You start at a and keep going until the next number would be b.
In Python 3, range is lazy
and the values in the sequence do not materialize
until you consume the range.
>>> range(3,12)
range(3, 12)
>>> list(range(3,12))
[3, 4, 5, 6, 7, 8, 9, 10, 11]
Trey Hunner makes the point that range is a lazy iterable rather than an iterator.
You can also step by an increment other than one:
range(a, b, s).
This generates a, a+s, a+2*s, ..., b-s
(assuming that (b - a) % s == 0;
i.e., a and b are separated by an exact multiple of s.)
>>> list(range(3, 12, 3))
[3, 6, 9]
What if you want to count down?
range(b, a, -s) won't do what you want.
>>> list(range(12,3, -3))
[12, 9, 6]
Why? Because you're starting at b,
a value that doesn't appear in the forward range,
and you're ending before you reach a,
a value that is certainly in the forward range.
You have to subtract s from both b and a:
When you use range(b-s, a-s, -s),
you get b-s, b-2*s, ..., a+s, a.
>>> list(range(12-3,3-3, -3))
[9, 6, 3]
>>> list(range(12-3,3-3, -3)), list(reversed(range(3, 12, 3)))
([9, 6, 3], [9, 6, 3])
