Path: csiph.com!x330-a1.tempe.blueboxinc.net!usenet.pasdenom.info!aioe.org!news.stack.nl!newsfeed.xs4all.nl!newsfeed5.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; 'subject:" ': 0.03; 'subject:test': 0.05; 'dictionary': 0.07; '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; '>>>': 0.12; 'def': 0.13; 'wrote:': 0.14; 'piraeus': 0.16; 'received:dip.t-dialin.net': 0.16; 'received:t-dialin.net': 0.16; 'this?': 0.18; 'code': 0.22; 'testing': 0.28; 'raise': 0.29; 'from:addr:web.de': 0.31; 'to:addr :python-list': 0.32; 'reference': 0.34; 'header:X-Complaints- To:1': 0.34; 'received:org': 0.38; 'anything': 0.38; 'comments': 0.39; 'to:addr:python.org': 0.39; 'header:Mime-Version:1': 0.39; 'i.e.': 0.40; 'header:Received:5': 0.40; 'simple': 0.60 X-Injected-Via-Gmane: http://gmane.org/ To: python-list@python.org From: Peter Otten <__peter__@web.de> Subject: Re: suggestions, comments on an "is_subdict" test Date: Fri, 22 Apr 2011 17:06:03 +0200 Organization: None References: <4DB190DC.8050307@mrabarnett.plus.com> Mime-Version: 1.0 Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 7Bit X-Gmane-NNTP-Posting-Host: p5084c898.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: 16 NNTP-Posting-Host: 82.94.164.166 X-Trace: 1303484755 news.xs4all.nl 41103 [::ffff:82.94.164.166]:48517 X-Complaints-To: abuse@xs4all.nl Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:3880 Zero Piraeus wrote: > : > >>> I'd like to ask for comments or advice on a simple code for testing a >>> "subdict", i.e. check whether all items of a given dictionary are >>> present in a reference dictionary. > > Anything wrong with this? > > def is_subdict(test_dct, base_dct): > return test_dct <= base_dct and all(test_dct[k] == base_dct[k] for > k in test_dct) It may raise a KeyError.