pyhon list reverse slicing excludes 1st item-CodePudding

I would like to go through on a list in reverse order using slices of the list, but I am having trouble to have the 1st (index 0) item of the list on the output.

l=[0,1,2,3,4,5,6]
s=len(l)

for i in range(s):
    print(l[s:i:-1])

The output is missing the 1st item of the list:

 [6, 5, 4, 3, 2, 1]
 [6, 5, 4, 3, 2]
 [6, 5, 4, 3]
 [6, 5, 4]
 [6, 5]
 [6]
 []

It looks like just a shifting problem, but i-1 doesn't work as an index when i=0.

The only solution so far I could figure out is ugly:

l=[0,1,2,3,4,5,6]
s=len(l)

for i in range(s):
    if i == 0:
        print(l[s::-1])
    else:
        print(l[s:i-1:-1])

Output (what I really wanted):

 [6, 5, 4, 3, 2, 1, 0]
 [6, 5, 4, 3, 2, 1]
 [6, 5, 4, 3, 2]
 [6, 5, 4, 3]
 [6, 5, 4]
 [6, 5]
 [6]

For me it looks like a design failure in python as in a reverse slicing there is no other way to have the 1st item of the list then leaving the field empty, what is a problem when you have a variable there... Tell me how to do it better.

CodePudding user response：

You could slice it in reverse then iterate through the reversed slice

l=[0,1,2,3,4,5,6]
s=len(l)

for i in range(s):
    print(l[::-1][0:s-i])
    
print(l)

and the original list wouldn't be updated

CodePudding user response：

You can avoid any checking or slicing by just inverting the list as already suggested.

After defining your list you can do: l.reverse() OR l = l[::-1] (which is the same thing)

For the future, just know that, when slicing lists, the first index is included, but the last isn't. So if you have mylist = [0, 1, 2, 3, 4] doing mylist[1:3] will return [1, 2]

CodePudding user response：

You could use None.

for i in range(s):
    print(l[: i-1 if i else None : -1])