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; 'subject:" ': 0.03; 'subject:test': 0.05; 'dictionary': 0.07; 'python': 0.07; 'builtin': 0.09; 'from:addr:python': 0.09; 'value:': 0.09; '>>>': 0.12; 'solutions,': 0.12; 'def': 0.13; 'wrote:': 0.14; '"""test': 0.16; 'from:addr:mrabarnett.plus.com': 0.16; 'from:name:mrab': 0.16; 'message-id:@mrabarnett.plus.com': 0.16; 'received:84.92': 0.16; 'received:84.92.122': 0.16; 'received:84.92.122.60': 0.16; 'reply-to:addr:python-list': 0.16; 'possibly': 0.16; 'have:': 0.19; 'maybe': 0.21; 'code': 0.22; 'header:In-Reply-To:1': 0.22; 'received:84': 0.25; 'testing': 0.28; 'missed': 0.29; 'this.': 0.30; 'key,': 0.31; 'all,': 0.31; 'to:addr:python-list': 0.32; 'reference': 0.34; 'header:User-Agent:1': 0.35; 'reply- to:addr:python.org': 0.35; 'some': 0.37; 'ways': 0.38; 'comments': 0.39; 'to:addr:python.org': 0.39; 'i.e.': 0.40; 'would': 0.40; 'simple': 0.60; 'reply-to:no real name:2**0': 0.72; 'header:Reply- To:1': 0.72; 'idiomatic': 0.84 X-IronPort-Anti-Spam-Filtered: true X-IronPort-Anti-Spam-Result: AhUIAE6bsU3Unw4S/2dsb2JhbACYEY1Md4hwux6FdgSSNIJV Date: Fri, 22 Apr 2011 16:18:49 +0100 From: MRAB User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-GB; rv:1.9.2.15) Gecko/20110303 Thunderbird/3.1.9 MIME-Version: 1.0 To: python-list@python.org Subject: Re: suggestions, comments on an "is_subdict" test References: <4db19756$0$81473$e4fe514c@news.xs4all.nl> In-Reply-To: <4db19756$0$81473$e4fe514c@news.xs4all.nl> Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.12 Precedence: list Reply-To: python-list@python.org 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: 40 NNTP-Posting-Host: 82.94.164.166 X-Trace: 1303485531 news.xs4all.nl 81478 [::ffff:82.94.164.166]:58314 X-Complaints-To: abuse@xs4all.nl Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:3881 On 22/04/2011 15:57, Irmen de Jong wrote: > On 22-4-2011 15:55, Vlastimil Brom wrote: >> Hi all, >> 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. >> Sofar I have: >> >> def is_subdict(test_dct, base_dct): >> """Test whether all the items of test_dct are present in base_dct.""" >> unique_obj = object() >> for key, value in test_dct.items(): >> if not base_dct.get(key, unique_obj) == value: >> return False >> return True >> >> I'd like to ask for possibly more idiomatic solutions, or more obvious >> ways to do this. Did I maybe missed some builtin possibility? > > > I would use: > > test_dct.items()<= base_dct.items() > In Python 2: >>> test_dct = {"foo": 0, "bar": 1} >>> base_dct = {"foo": 0, "bar": 1, "baz": 2} >>> >>> test_dct.items() <= base_dct.items() False In Python 3: >>> test_dct = {"foo": 0, "bar": 1} >>> base_dct = {"foo": 0, "bar": 1, "baz": 2} >>> test_dct.items() <= base_dct.items() True YMMV