Path: csiph.com!x330-a1.tempe.blueboxinc.net!usenet.pasdenom.info!aioe.org!feeder.news-service.com!newsfeed.xs4all.nl!newsfeed6.news.xs4all.nl!xs4all!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 0.000 X-Spam-Evidence: '*H*': 1.00; '*S*': 0.00; 'elif': 0.04; '"""': 0.07; 'python': 0.08; '*is*': 0.09; 'header:In-reply-to:1': 0.09; 'algorithm': 0.13; 'def': 0.15; 'library': 0.15; '"=="': 0.16; '"for"': 0.16; '-john': 0.16; 'alist': 0.16; 'objects:': 0.16; 'received:167.206.4': 0.16; 'received:167.206.4.200': 0.16; 'received:cv.net': 0.16; 'received:hcvlny.cv.net': 0.16; 'received:mta5.srv.hcvlny.cv.net': 0.16; 'received:srv.hcvlny.cv.net': 0.16; 'sublist': 0.16; 'wrote:': 0.16; 'language,': 0.17; 'subject:list': 0.18; 'to:name:python- list': 0.18; 'tue,': 0.23; 'pm,': 0.24; 'aug': 0.24; "python's": 0.24; 'suggestion': 0.26; 'code,': 0.28; 'second': 0.29; 'lines': 0.30; 'match': 0.30; '+0200,': 0.30; 'list': 0.32; 'cases': 0.32; "isn't": 0.33; 'to:addr:python-list': 0.33; 'header:User-Agent:1': 0.34; 'nobody': 0.34; 'totally': 0.35; 'core': 0.36; 'fair': 0.37; 'using': 0.37; 'but': 0.37; 'subject:: ': 0.39; 'received:192': 0.39; 'empty': 0.39; 'to:addr:python.org': 0.39; 'case': 0.39; 'results': 0.61; 'special': 0.67 Date: Tue, 16 Aug 2011 09:57:57 -0400 From: John Posner Subject: Re: testing if a list contains a sublist In-reply-to: To: python-list MIME-version: 1.0 Content-type: text/plain; charset=ISO-8859-1 Content-transfer-encoding: 7BIT X-Enigmail-Version: 1.2 References: User-Agent: Mozilla/5.0 (Windows NT 5.1; rv:5.0) Gecko/20110624 Thunderbird/5.0 X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.12 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: 38 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1313504884 news.xs4all.nl 23874 [2001:888:2000:d::a6]:53105 X-Complaints-To: abuse@xs4all.nl Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:11557 On 2:59 PM, Nobody wrote: > On Tue, 16 Aug 2011 01:26:54 +0200, Johannes wrote: > >> what is the best way to check if a given list (lets call it l1) is >> totally contained in a second list (l2)? > "Best" is subjective. AFAIK, the theoretically-optimal algorithm is > Boyer-Moore. But that would require a fair amount of code, and Python > isn't a particularly fast language, so you're likely to get better results > if you can delegate most of the leg-work to a native library (along the > lines of Roy's suggestion of using regexps). > > How about using Python's core support for "==" on list objects: def contains(alist, slist): """ predicate: is *slist* a sublist of *alist*? """ alist_sz = len(alist) slist_sz = len(slist) # special cases if slist_sz == 0: return True # empty list *is* a sublist elif slist_sz == alist_sz and alist == slist: return True elif slist_sz > alist_sz: return False # standard case for i in range(alist_sz - slist_sz + 1): if slist == alist[i:i+slist_sz]: return True # fell through "for" loop: no match found return False -John