How can I reverse a list in Python?

Question

How can I do the following in Python?

array = [0, 10, 20, 40]
for (i = array.length() - 1; i >= 0; i--)

I need to have the elements of an array, but from the end to the beginning.

Answer 1

Possible ways,

list1 = [3,4,3,545,6,4,34,243]

list1.reverse()

list1[::-1]

Answer 2

>>> L = [1, 2, 3, 4]
>>> L = [L[-i] for i in range(1, len(L) + 1)]
>>> L
[4, 3, 2, 1]

Answer 3

If you want to store the elements of reversed list in some other variable, then you can use revArray = array[::-1] or revArray = list(reversed(array)).

But the first variant is slightly faster:

z = range(1000000)
startTimeTic = time.time()
y = z[::-1]
print("Time: %s s" % (time.time() - startTimeTic))

f = range(1000000)
startTimeTic = time.time()
g = list(reversed(f))
print("Time: %s s" % (time.time() - startTimeTic))

Output:

Time: 0.00489711761475 s
Time: 0.00609302520752 s

Answer 4

Reversing in-place by switching references of opposite indices:

>>> l = [1,2,3,4,5,6,7]    
>>> for i in range(len(l)//2):
...     l[i], l[-1-i] = l[-1-i], l[i]
...
>>> l
[7, 6, 5, 4, 3, 2, 1]

Answer 5

Can be done using __reverse__ , which returns a generator.

>>> l = [1,2,3,4,5]
>>> for i in l.__reversed__():
...   print i
... 
5
4
3
2
1
>>>

Answer 6

I find (contrary to some other suggestions) that l.reverse() is by far the fastest way to reverse a long list in Python 3 and 2. I'd be interested to know if others can replicate these timings.

l[::-1] is probably slower because it copies the list prior to reversing it. Adding the list() call around the iterator made by reversed(l) must add some overhead. Of course if you want a copy of the list or an iterator then use those respective methods, but if you want to just reverse the list then l.reverse() seems to be the fastest way.

Functions

def rev_list1(l):
    return l[::-1]

def rev_list2(l):
    return list(reversed(l))

def rev_list3(l):
    l.reverse()
    return l

List

l = list(range(1000000))

Python 3.5 timings

timeit(lambda: rev_list1(l), number=1000)
# 6.48
timeit(lambda: rev_list2(l), number=1000)
# 7.13
timeit(lambda: rev_list3(l), number=1000)
# 0.44

Python 2.7 timings

timeit(lambda: rev_list1(l), number=1000)
# 6.76
timeit(lambda: rev_list2(l), number=1000)
# 9.18
timeit(lambda: rev_list3(l), number=1000)
# 0.46