Path: csiph.com!usenet.pasdenom.info!news.redatomik.org!newsfeed.xs4all.nl!newsfeed7.news.xs4all.nl!post.news.xs4all.nl!not-for-mail Return-Path: X-Original-To: python-list@python.org Delivered-To: python-list@mail.python.org X-Spam-Status: OK 0.004 X-Spam-Evidence: '*H*': 0.99; '*S*': 0.00; 'smallest': 0.07; 'variant': 0.07; 'received:80.91': 0.09; 'received:80.91.229': 0.09; 'received:gmane.org': 0.09; 'received:list': 0.09; 'def': 0.13; 'subject: \n ': 0.15; 'beginning.': 0.16; 'element,': 0.16; 'lukas': 0.16; 'pythonic': 0.16; 'received:80.91.229.3': 0.16; 'received:plane.gmane.org': 0.16; 'rotation': 0.16; 'status?': 0.16; 'wrote:': 0.16; 'circular': 0.18; 'element': 0.18; 'defined': 0.23; 'seems': 0.23; 'elements': 0.23; 'header:In- Reply-To:1': 0.24; 'header:User-Agent:1': 0.26; 'subject:list': 0.26; 'appear': 0.26; 'header:X-Complaints-To:1': 0.26; 'skip:" 20': 0.26; 'compare': 0.27; 'followed': 0.27; 'index,': 0.29; 'seemingly': 0.29; 'unique,': 0.29; 'code': 0.30; 'skip:[ 10': 0.31; "i'd": 0.31; 'another': 0.32; 'problem': 0.33; 'this?': 0.34; 'list': 0.34; 'so,': 0.35; 'could': 0.35; 'i.e.': 0.35; 'something': 0.35; 'problem.': 0.35; 'but': 0.36; 'list,': 0.36; 'should': 0.36; 'there': 0.36; 'to:addr:python-list': 0.36; 'pm,': 0.36; 'subject:: ': 0.37; 'say': 0.37; 'received:org': 0.37; 'minimum': 0.38; 'to:addr:python.org': 0.40; 'determine': 0.61; 'charset:windows-1252': 0.62; 'above,': 0.63; 'positions': 0.64; 'treat': 0.72; 'received:12': 0.81; 'front.': 0.84; 'nice,': 0.84 X-Injected-Via-Gmane: http://gmane.org/ To: python-list@python.org From: Emile van Sebille Subject: Re: Most pythonic way of rotating a circular list to a canonical point Date: Sat, 1 Aug 2015 13:49:58 -0700 References: Mime-Version: 1.0 Content-Type: text/plain; charset=windows-1252; format=flowed Content-Transfer-Encoding: 7bit X-Gmane-NNTP-Posting-Host: www.westernstatesglass.com User-Agent: Mozilla/5.0 (Windows NT 6.2; WOW64; rv:38.0) Gecko/20100101 Thunderbird/38.1.0 In-Reply-To: X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.20+ Precedence: list List-Id: General discussion list for the Python programming language List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Newsgroups: comp.lang.python Message-ID: Lines: 25 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1438462212 news.xs4all.nl 2918 [2001:888:2000:d::a6]:48486 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:94861 On 8/1/2015 1:34 PM, Lukas Barth wrote: > Hi! > > 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? Well, I have not understood the problem for such a seemingly simple one. :) Is the problem to determine if one list of circular numbers 'matches' another one despite rotation status? If so, I'd do something like: def matchcircularlists(L1,L2): #return True if L1 is a rotation variant of L2 return "".join(map(str,L1)) in "".join(map(str,L2+L2)) Emile