Reverse Indexing in Python?

Question

I know that a[end:start:-1] slices a list in a reverse order.

For example

a = range(20)
print a[15:10:-1] # prints [15, ..., 11]
print a[15:0:-1] # prints [15, ..., 1]

but you cannot get to the first element (0 in the example). It seems that -1 is a special value.

print a[15:-1:-1] # prints []

Any ideas?

[The Python Slice Notation](http://stackoverflow.com/questions/509211/the-python-slice-notation) — Ashwini Chaudhary, Jul 12 '13 at 07:55
This could also be interesting for you: `print(list(reversed(range(5))))` — Martin Thoma, Jul 12 '13 at 07:56

score 22 · Accepted Answer · answered Jul 12 '13 at 08:13

22

You can assign your variable to None:

>>> a = range(20)
>>> a[15:None:-1]
[15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
>>>

answered Jul 12 '13 at 08:13

zhangyangyu

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1

This (using `None`) is probably the best solution, but note that you can also use negative indices all the way around: `a[-5:-21:-1]` gets you the same slice, including the final `a[0]` element. – torek Jul 12 '13 at 08:35
@torek I also thought of using negative indices, but I think that is the best solution, because you can avoid the extra assignment to `None` (see my answer). – Paulo Almeida Jul 12 '13 at 10:07

score 15 · Answer 2 · answered Jul 12 '13 at 07:53

15

Omit the end index:

print a[15::-1]

answered Jul 12 '13 at 07:53

Pablo

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start and end are actually variables. – Mohammad Moghimi Jul 12 '13 at 07:54
do I need a special if statement for this? – Mohammad Moghimi Jul 12 '13 at 07:54
you can do it without an if statement if you don't mind doing this in two steps: `print a[low:high+1][::-1]` – Pablo Jul 12 '13 at 07:59
@MohammadMoghimi You can use `None`. – glglgl Jul 12 '13 at 08:00
My list is very long. I think this slices the list first and then reverses it. – Mohammad Moghimi Jul 12 '13 at 08:00
Well, I guess you can go with an if statement: `print a[high:low-1:-1] if low>0 else a[high::-1]` – Pablo Jul 12 '13 at 08:07

score 8 · Answer 3 · answered Jul 19 '16 at 03:41

In Python2.x, the simplest solution in terms of number of characters should probably be :

>>> a=range(20)

>>> a[::-1]
[19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Though i want to point out that if using xrange(), indexing won't work because xrange() gives you an xrange object instead of a list.

>>> a=xrange(20)
>>> a[::-1]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: sequence index must be integer, not 'slice'

After in Python3.x, range() does what xrange() does in Python2.x but also has an improvement accepting indexing change upon the object.

>>> a = range(20)
>>> a[::-1]
range(19, -1, -1)
>>> b=a[::-1]
>>> for i in b:
...     print (i)
... 
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
>>>

the difference between range() and xrange() learned from source: http://pythoncentral.io/how-to-use-pythons-xrange-and-range/ by author: Joey Payne

My direct answer is just a[::-1]. The following part is some possible complications people may find if they are using different version of python. Sorry about the confusion. I am new here, do you think i should delete the non-direct answer part and where should i put it? Thanks! — cookie_chu, Jul 21 '16 at 14:51

0x90 · Answer 4 · 2013-07-16T19:30:43.533

>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> print a[:6:-1]
[19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7]
>>> a[7] == a[:6:-1][-1]
True
>>> a[1] == a[:0:-1][-1]
True

So as you can see when subsitute a value in start label :end: it will give you from start to end exclusively a[end].

As you can see in here as well:

>>> a[0:2:]
[0, 1]

-1 is the last value in a:

>>> a[len(a)-1] == a[-1]
True

Ludo · Answer 5 · 2013-07-12T08:01:36.157

0

EDIT: begin and end are variables

I never realized this, but a (slightly hacky) solution would be:

>>> a = range(5)
>>> s = 0
>>> e = 3
>>> b = a[s:e]
>>> b.reverse()
>>> print b
[2, 1, 0]

edited Jul 12 '13 at 08:01

answered Jul 12 '13 at 07:55

Ludo

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start and end are variables – Mohammad Moghimi Jul 12 '13 at 07:56

score 0 · Answer 6 · answered Jul 12 '13 at 08:51

0

If you use negative indexes you can avoid extra assignments, using only your start and end variables:

a = range(20)
start = 20
for end in range(21):
    a[start:-(len(a)+1-end):-1]

answered Jul 12 '13 at 08:51

Paulo Almeida

7,803
28
36

Reverse Indexing in Python?

6 Answers6

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