Lukas Barth <m...@tinloaf.de>:
> I have a list of numbers that I treat as "circular", i.e. [1,2,3] and
> [2,3,1] should be the same. Now I want to rotate these to a well defined
> status, so that I can can compare them.
>
> If all elements are unique, the solution is easy: find the minimum
> element, find its index, then use mylist[:index] + mylist[index:], i.e.
> the minimum element will always be at the beginning.
>
> But say I have [0,1,0,2,0,3]. I can in fact guarantee that no *pair*
> will appear twice in that list, i.e. I could search for the minimum, if
> that is unique go on as above, otherwise find *all* positions of the
> minimum, see which is followed by the smallest element, and then rotate
> that position to the front.
>
> Now that seems an awful lot of code for a (seemingly?) simple problem.
> Is there a nice, pythonic way to do this?
How about:
========================================================================
def circularly_equal(l1, l2):
length = len(l1)
if length != len(l2):
return False
twice = l1 + l1
for i in range(length):
if twice[i:i + length] == l2:
return True
return False
========================================================================
Marko
--
https://mail.python.org/mailman/listinfo/python-list
