Python Start counting array of length x starting from second item and finishing on the first one

Question

In Python I have an array of length x.

let array = [0, 1, 2, 3, 4, 5]

I want to get from the array above the result like this: [1, 2, 3, 4, 5, 0]

So basically i wan't to skip the first term and loop it after the array ends and stop on the last skipped term. I'm pretty new to python, so couldn't figure it out on on my own + googling didn't help.

Help please, Much appreciated!

Silas Ray · Accepted Answer · 2012-12-14T01:22:37.463

Use slicing and append().

lst = [0, 1, 2, 3, 4, 5]
new_lst = lst[1:]
new_lst.append(lst[0])

You could also use new_lst.extend(lst[:1]), though when the head slice is a single element, append(lst[0]) is probably slightly more efficient, since you don't have to construct another temporary list just to wrap a single value. lst[1:] + list[:1] is probably the worst though, since it has to create yet another throw away list object compared to the extend() version.

score 2 · Answer 2 · answered Dec 14 '12 at 01:09

I would go with slicing, but another option (which is normally overkill for something this simple is using a collections.deque)

Small example:

>>> e =  [0, 1, 2, 3, 4, 5]
>>> from collections import deque
>>> d = deque(e)
>>> d
deque([0, 1, 2, 3, 4, 5])
>>> d.rotate(1)
>>> d
deque([5, 0, 1, 2, 3, 4])
>>> d.rotate(-2)
>>> d
deque([1, 2, 3, 4, 5, 0])

score 2 · Answer 3 · answered Dec 14 '12 at 01:24

2

e =  [0, 1, 2, 3, 4, 5]

e.append(e.pop(0))

Why people don't think immediately to in-place transformations ???

answered Dec 14 '12 at 01:24

eyquem

26,771
7
38
46

Because usually, it's not what you want. But yes, if it works for you, this is a great solution. – Silas Ray Dec 14 '12 at 01:29

score 1 · Answer 4 · answered Dec 14 '12 at 01:18

1

Yet another way, which might be more general. Couldn't think of a good name...

def iterate_from(l, start):
    for i in range(start, len(l)):
        yield l[i]
    for i in range(start):
        yield l[i]

I got the following output:

In [39]: iterate_from(range(7), 3)
Out[39]: <generator object iterate_from at 0x120259b40>

In [40]: list(iterate_from(range(7), 3))
Out[40]: [3, 4, 5, 6, 0, 1, 2]

answered Dec 14 '12 at 01:18

acjay

34,571
6
57
100

Nice. Expensive if all you want is a new list, but if you have to iterate on a bunch of lists using this pattern, this is great. – Silas Ray Dec 14 '12 at 01:33

score 1 · Answer 5 · edited May 23 '17 at 12:10

1

when you loop thru your list, don't even need new variable for that

for i in my_list[1:] + my_list[:1]:
    print i,

print
for i in my_list:
    print i,

will return:

1 2 3 4 5 0
0 1 2 3 4 5

This way you don't make any changes to original my_list.

Also, take a look over here: Explain Python's slice notation

edited May 23 '17 at 12:10

Community

1
1

answered Dec 14 '12 at 01:25

Artem Kisel

21
2

This works despite a major error: `i` is a value of the list, not an index into it. If the contents of the list were anything but 0-5 (where each value was also its own index), it would fail, probably with an index error. You want `print i` in both of your loops, not `print my_list[i]`. – Blckknght Dec 14 '12 at 01:34

score 0 · Answer 6 · answered Dec 14 '12 at 01:09

Use slices to get the portions of the list you are interested in, and then add them together:

>>> lst = [0, 1, 2, 3, 4, 5]
>>> lst[1:] + lst[:1]
[1, 2, 3, 4, 5, 0]

lst[1:] will give you everything from the first index (inclusive) to the end of the list.

lst[:1] will give you everything from the beginning of the list to the first index (exclusive). (This is equivalent to [lst[0]])

Python Start counting array of length x starting from second item and finishing on the first one

6 Answers6