Reversing a list using slice notation

Question

in the following example:

foo = [\'red\', \'white\', \'blue\', 1, 2, 3]

where: foo[0:6:1] will print all elements in foo. Howe

Answer 1

Slice notation in short:

[ <first element to include> : <first element to exclude> : <step> ]

If you want to include the first element when reversing a list, leave the middle element empty, like this:

foo[::-1]

You can also find some good information about Python slices in general here:
Explain Python's slice notation

Answer 2

...why foo[6:0:-1] doesn't print the entire list?

Because the middle value is the exclusive, rather than inclusive, stop value. The interval notation is [start, stop).

This is exactly how [x]range works:

>>> range(6, 0, -1)
[6, 5, 4, 3, 2, 1]

Those are the indices that get included in your resulting list, and they don't include 0 for the first item.

>>> range(6, -1, -1)
[6, 5, 4, 3, 2, 1, 0]

Another way to look at it is:

>>> L = ['red', 'white', 'blue', 1, 2, 3]
>>> L[0:6:1]
['red', 'white', 'blue', 1, 2, 3]
>>> len(L)
6
>>> L[5]
3
>>> L[6]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

The index 6 is beyond (one-past, precisely) the valid indices for L, so excluding it from the range as the excluded stop value:

>>> range(0, 6, 1)
[0, 1, 2, 3, 4, 5]

Still gives you indices for each item in the list.

Answer 3

Use

>>>foo[::-1]

This displays the reverse of the list from the end element to the start,

Answer 4

If you are having trouble remembering slice notation, you could try doing the Hokey Cokey:

[In: Out: Shake it all about]

[First element to include: First element to leave out: The step to use]

YMMV

Answer 5

formalizing the answer from andrew-clark a little bit more:

Suppose the list v and v[n1:n2:n3] slice. n1 is initial position, n2 is final position and n3 is step

Let's write some pseucode in a Python way:

n3 = 1  if (n3 is missing) else n3
if n3==0:
   raise exception # error, undefined step

First part: n3 positive

if n3>0:
   slice direction is from left to right, the most common direction         

   n1 is left slice position in `v` 
   if n1 is missing: 
      n1 = 0   # initial position
   if n1>=0:
      n1 is a normal position
   else: 
     (-n1-1) is the position in the list from right to left 

   n2 is right slice position in `v` 
   if n2 is missing: 
      n2 = len(x)  # after final position
   if n2>=0:
      n2 is a normal final position (exclusive)
   else: 
      -n2-1 é the final position in the list from right to left 
       (exclusive)

Second part: n3 negative

else: 
  slice direction is from right to left (inverse direction)

  n1 is right slice position in `v` 
  if n1 is missing: 
     n1 = -1   # final position is last position in the list.
  if n1>=0:
     n1 is a normal position
  else: 
     (-n1-1) is the position in the list from right to left 

  n2 is  left slice position in `v` 
  if n2 is missing: 
     n2 = -len(x)-1   # before 1st character  (exclusive)
  if n2>=0:
     n2 is a normal final position (exclusive)
  else: 
     -n2-1 is the ending position in the list from right to left 
     (exclusive)

Now the original problema: How to reverse a list with slice notation?

L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print(L(::-1)) # [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]

Why?
n1 is missing and n3<0 => n1=0
n2 is missing and n3<0 => n2 = -len(x)-1

So L(::-1) == L(-1:-11:-1)

Answer 6

Complement. to reverse step by 2:

A = [1,2,2,3,3]
n = len(A)
res = [None] * n
mid = n//2 + 1 if n%2 == 1 else n//2

res[0::2] = A[0:mid][::-1]
res[1::2] = A[0:mid][::-1]
print(res)

[2, 3, 2, 3, 1]