Python enumerate reverse index only

Question

I am trying to reverse the index given by enumerate whilst retaining the original order of the list being enumerated.

Assume I have the following:

Answer 1

Python 2

import itertools

def reversed_enumerate(seq):
    return itertools.izip(reversed(range(len(seq))), reversed(seq))

Python 3

Substitute zip for itertools.izip :)

Answer 2

Actually I'm using the same logic as @RemcoGerlich did, but I use list comprehension directly, which make the code now become 1-liner:

def generatelist(x):
    return [(x-1-i,n) for i,n in enumerate(range(x))]

Regarding the dilemma of choosing generator or list comprehension, here is the suggested way:

Basically, use a generator expression if all you're doing is iterating once. If you want to store and use the generated results, then you're probably better off with a list comprehension.

Answer 3

We can use enumerate with len:

$ cat enumerate.py 


arr = ['stone', 'cold', 'steve', 'austin']
for i, val in enumerate(arr):
    print ("enu {} val {}".format(i, val))
for i, val in enumerate(arr):
    print ("enu {} val {}".format(len(arr) - i - 1, val))
$  python enumerate.py 
enu 0 val stone
enu 1 val cold
enu 2 val steve
enu 3 val austin
enu 3 val stone
enu 2 val cold
enu 1 val steve
enu 0 val austin
$  

Answer 4

values = 'abcde'

for i, value in zip(reversed(range(len(values))), values):
    print(i, value)

Explanation:

values = 'abcde'

values_len = len(values) # 5
indexes = range(values_len) # [0, 1, 2, 3, 4]
reversed_indexes = reversed(indexes) # [4, 3, 2, 1, 0]

# combine reversed indexes and values
reversed_enumerator = zip(reversed_indexes, values)

for i, value in reversed_enumerator:
    print(i, value)

Answer 5

If you're going to re-use it several times, you can make your own generator:

def reverse_enum(lst):
    for j, item in enumerate(lst):
        yield len(lst)-1-j, item

print list(reverse_enum(range(5)))
# [(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)]

or

def reverse_enum(lst):
    return ((len(lst)-1-j, item) for j, item in enumerate(lst))

Answer 6

Just take the length of your list and subtract the index from that...

L = range(5)

for i, n in L:
    my_i = len(L) -1 - i
    ...

Or if you really need a generator:

def reverse_enumerate(L):
   # Only works on things that have a len()
   l = len(L)
   for i, n in enumerate(L):
       yield l-i-1, n

enumerate() can't possibly do this, as it works with generic iterators. For instance, you can pass it infinite iterators, that don't even have a "reverse index".