Traverse a List in Reverse Order in Python

Traverse a list in reverse order in Python

Use the built-in reversed() function:

>>> a = ["foo", "bar", "baz"]
>>> for i in reversed(a):
...     print(i)
... 
baz
bar
foo

To also access the original index, use enumerate() on your list before passing it to reversed():

>>> for i, e in reversed(list(enumerate(a))):
...     print(i, e)
... 
2 baz
1 bar
0 foo

Since enumerate() returns a generator and generators can't be reversed, you need to convert it to a list first.

How to traverse a list in reverse order in Python (index-style: '... in range(...)' only)

Avoid looping with range, just iterate with a for loop as a general advice.

If you want to reverse a list, use reversed(my_list).

So in your case the solution would be:

my_list = reversed([array0,array1,array2,array3])

That gives you [array3, array2, ...]

For omitting the last array from the reversed list, use slicing:

my_list = reversed([array0,array1,array2,array3])[:-1]

So your code turns to a one liner :-)

How to loop a list in reverse order and skip last element

You need to begin from -2 as -1 starts from the last element:

for i in x[-2::-1]:
    print(i)

Alternatively, you can use reversed with list slicing:

for i in reversed(x[:-1]):
    print(i)

How do I reverse a list or loop over it backwards?

To get a new reversed list, apply the reversed function and collect the items into a list:

>>> xs = [0, 10, 20, 40]
>>> list(reversed(xs))
[40, 20, 10, 0]

To iterate backwards through a list:

>>> xs = [0, 10, 20, 40]
>>> for x in reversed(xs):
...     print(x)
40
20
10
0

Print a list in reverse order with range()?

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.

Need to iterate over a Python list in reverse as fast as possible

>>> from timeit import Timer
>>> t = Timer('[x for x in l[::-1]]', 'l = list(range(100000))')
>>> t.timeit(number=1000)
5.549649953842163
>>> t = Timer('l.reverse(); [x for x in l]', 'l = list(range(100000))')
>>> t.timeit(number=1000)
4.548457145690918
>>> t = Timer('[x for x in reversed(l)]', 'l = list(range(100000))')
>>> t.timeit(number=1000)
4.428632974624634

Conclusion: reversed() is marginally faster than l.reverse() on a list with 100000 items.
This of course is even more true if you don't actually loop over the whole list, and it stops being true if you use the list more than once.

l[::-1] is outdated since 2.4 that introduced reversed().

How to loop backwards in python?

range() and xrange() take a third parameter that specifies a step. So you can do the following.

range(10, 0, -1)

Which gives

[10, 9, 8, 7, 6, 5, 4, 3, 2, 1] 

But for iteration, you should really be using xrange instead. So,

xrange(10, 0, -1)

Note for Python 3 users: There are no separate range and xrange functions in Python 3, there is just range, which follows the design of Python 2's xrange.

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