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?
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]
>>>
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.
None (see my answer).Omit the end index:
print a[15::-1]
print a[low:high+1][::-1]
None.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
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]
>>> 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
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]
print(list(reversed(range(5))))