Python - getting a list of numbers from N to 0

Question

What is the best way to get a list [N, N-1, ..., 0]in python? I have 2 ways to think

>>> range(10)[::-1]
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

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

Answer 1

range(N, -1, -1) is better

You can see that it takes much less time:

N = 10000

%timeit range(N+1)[::-1]
1000000 loops, best of 3: 767 ns per loop

%timeit range(N, -1, -1)
1000000 loops, best of 3: 334 ns per loop

In range(N+1)[::-1]you first do the same thing range(N, -1, -1)as and then invert the list, so it takes more time.

Answer 2

If your list of numbers is potentially large or you really don’t need this list, but just sorting through something by range, you can avoid actually creating the list with

xrange(N, -1, -1)

Rob , Python 3 ( , ), Python 2 range(), xrange() .

Answer 3

import timeit

print timeit.timeit("range(10)[::-1]", number=10000000)
print timeit.timeit("range(9, -1, -1)", number=10000000)

Exit on my car

2.99922299385
2.24587512016

It seems that the second one is a bit faster, so use range(9, -1, -1)

Answer 4

If you want to use a range, do the following:

range(n, -1, -1)

As others have shown, this is faster.

Personally, for the loopback, I prefer to use the built-in functions reversedin combination with enumerate:

for i, v in enumerate(reversed(mylist))

This is a bit more readable and you get both index and value :)

Answer 5

Your second method is to touch faster:

%timeit range(9,-1,-1)
> 1000000 loops, best of 3: 496 ns per loop

%timeit range(10)[::-1]
> 1000000 loops, best of 3: 659 ns per loop