Path: csiph.com!x330-a1.tempe.blueboxinc.net!usenet.pasdenom.info!aioe.org!feeder.news-service.com!news2.euro.net!newsgate.cistron.nl!newsgate.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 0.002 X-Spam-Evidence: '*H*': 1.00; '*S*': 0.00; 'example:': 0.03; 'received:80.91': 0.09; 'received:80.91.229': 0.09; 'received:80.91.229.12': 0.09; 'received:gmane.org': 0.09; 'received:list': 0.09; 'received:lo.gmane.org': 0.09; 'received:dip.t-dialin.net': 0.16; 'received:t-dialin.net': 0.16; 'wrote:': 0.16; 'subject:list': 0.18; 'assume': 0.22; 'assumes': 0.23; 'lists': 0.28; 'problem': 0.28; 'second': 0.29; 'from:addr:web.de': 0.30; 'list': 0.32; 'to:addr:python-list': 0.33; 'header:X-Complaints-To:1': 0.35; 'totally': 0.35; 'another': 0.37; 'example,': 0.37; 'but': 0.37; 'something': 0.37; 'received:org': 0.38; 'subject:: ': 0.39; 'sets': 0.39; 'header :Mime-Version:1': 0.39; 'to:addr:python.org': 0.39 X-Injected-Via-Gmane: http://gmane.org/ To: python-list@python.org From: Peter Otten <__peter__@web.de> Subject: Re: testing if a list contains a sublist Date: Tue, 16 Aug 2011 09:44:13 +0200 Organization: None References: <4E49AB3E.9000801@web.de> Mime-Version: 1.0 Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 7Bit X-Gmane-NNTP-Posting-Host: p5084a68a.dip.t-dialin.net 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: 17 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1313480614 news.xs4all.nl 23864 [2001:888:2000:d::a6]:59562 X-Complaints-To: abuse@xs4all.nl Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:11522 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)? > > for example: > l1 = [1,2], l2 = [1,2,3,4,5] -> l1 is contained in l2 > l1 = [1,2,2,], l2 = [1,2,3,4,5] -> l1 is not contained in l2 > l1 = [1,2,3], l2 = [1,3,5,7] -> l1 is not contained in l2 > > my problem is the second example, which makes it impossible to work with > sets insteads of lists. But something like set.issubset for lists would > be nice. Would [1, 2, 1] be contained in [1, 1, 2]? You have received an answer that assumes yes and another that assume no.