Print a list in reverse order with range()?

Question

use reversed() function:

reversed(range(10))

It's much more meaningful.

Update:

If you want it to be a list (as btk pointed out):

list(reversed(range(10)))

Update:

If you want to use only range to achieve the same result, you can use all its parameters. range(start, stop, step)

For example, to generate a list [5,4,3,2,1,0], you can use the following:

range(5, -1, -1)

It may be less intuitive but as the comments mention, this is more efficient and the right usage of range for reversed list.

Use the 'range' built-in function. The signature is range(start, stop, step). This produces a sequence that yields numbers, starting with start, and ending if stop has been reached, excluding stop.

>>> range(9,-1,-1)   
    [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
>>> range(-2, 6, 2)
    [-2, 0, 2, 4]

In Python 3, this produces a non-list range object, which functions effectively like a read-only list (but uses way less memory, particularly for large ranges).

You could use range(10)[::-1] which is the same thing as range(9, -1, -1) and arguably more readable (if you're familiar with the common sequence[::-1] Python idiom).

For those who are interested in the "efficiency" of the options collected so far...

Jaime RGP's answer led me to restart my computer after timing the somewhat "challenging" solution of Jason literally following my own suggestion (via comment). To spare the curious of you the downtime, I present here my results (worst-first):

Jason's answer (maybe just an excursion into the power of list comprehension):

$ python -m timeit "[9-i for i in range(10)]"
1000000 loops, best of 3: 1.54 usec per loop

martineau's answer (readable if you are familiar with the extended slices syntax):

$ python -m timeit "range(10)[::-1]"
1000000 loops, best of 3: 0.743 usec per loop

Michał Šrajer's answer (the accepted one, very readable):

$ python -m timeit "reversed(range(10))"
1000000 loops, best of 3: 0.538 usec per loop

bene's answer (the very first, but very sketchy at that time):

$ python -m timeit "range(9,-1,-1)"
1000000 loops, best of 3: 0.401 usec per loop

The last option is easy to remember using the range(n-1,-1,-1) notation by Val Neekman.

No sense to use reverse because the range function can return reversed list.

When you have iteration over n items and want to replace order of list returned by range(start, stop, step) you have to use third parameter of range which identifies step and set it to -1, other parameters shall be adjusted accordingly:

1. Provide stop parameter as -1 (it's previous value of stop - 1, stop was equal to 0).
2. As start parameter use n-1.

So equivalent of range(n) in reverse order would be:

n = 10
print range(n-1,-1,-1) 
#[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

for i in range(8, 0, -1)

will solve this problem. It will output 8 to 1, and -1 means a reversed list