一个简单的方法:

n = int(input())
arr = list(map(int, input().split()))

for i in reversed(range(0, n)):
    print("%d %d" %(i, arr[i]))

无论如何，你也可以这样做。非常简单。

a = [1, 2, 3, 4, 5, 6, 7]
for x in xrange(len(a)):
    x += 1
    print a[-x]

def reverse(spam):
    k = []
    for i in spam:
        k.insert(0,i)
    return "".join(k)

其他答案都不错，但如果你想这么做 列表理解风格

collection = ['a','b','c']
[item for item in reversed( collection ) ]

假设任务是在列表中找到最后一个满足某些条件的元素(即向后看时的第一个元素)，我得到以下数字。

Python 2:

>>> min(timeit.repeat('for i in xrange(len(xs)-1,-1,-1):\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))
4.6937971115112305
>>> min(timeit.repeat('for i in reversed(xrange(0, len(xs))):\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))
4.809093952178955
>>> min(timeit.repeat('for i, x in enumerate(reversed(xs), 1):\n    if 128 == x: break', setup='xs, n = range(256), 0', repeat=8))
4.931743860244751
>>> min(timeit.repeat('for i, x in enumerate(xs[::-1]):\n    if 128 == x: break', setup='xs, n = range(256), 0', repeat=8))
5.548468112945557
>>> min(timeit.repeat('for i in xrange(len(xs), 0, -1):\n    if 128 == xs[i - 1]: break', setup='xs, n = range(256), 0', repeat=8))
6.286104917526245
>>> min(timeit.repeat('i = len(xs)\nwhile 0 < i:\n    i -= 1\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))
8.384078979492188

所以，最丑的选项xrange(len(xs)-1，-1，-1)是最快的。

Python 3(不同机器):

>>> timeit.timeit('for i in range(len(xs)-1,-1,-1):\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', number=400000)
4.48873088900001
>>> timeit.timeit('for i in reversed(range(0, len(xs))):\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', number=400000)
4.540959084000008
>>> timeit.timeit('for i, x in enumerate(reversed(xs), 1):\n    if 128 == x: break', setup='xs, n = range(256), 0', number=400000)
1.9069805409999958
>>> timeit.timeit('for i, x in enumerate(xs[::-1]):\n    if 128 == x: break', setup='xs, n = range(256), 0', number=400000)
2.960720073999994
>>> timeit.timeit('for i in range(len(xs), 0, -1):\n    if 128 == xs[i - 1]: break', setup='xs, n = range(256), 0', number=400000)
5.316207007999992
>>> timeit.timeit('i = len(xs)\nwhile 0 < i:\n    i -= 1\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', number=400000)
5.802550058999998

这里，enumerate(reversed(xs)， 1)是最快的。

>>> l = ["a","b","c","d"]
>>> l.reverse()
>>> l
['d', 'c', 'b', 'a']

OR

>>> print l[::-1]
['d', 'c', 'b', 'a']