Path: csiph.com!usenet.pasdenom.info!weretis.net!feeder4.news.weretis.net!feeder2.ecngs.de!ecngs!feeder.ecngs.de!xlned.com!feeder1.xlned.com!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 0.007 X-Spam-Evidence: '*H*': 0.99; '*S*': 0.00; 'received:80.91': 0.09; 'received:80.91.229': 0.09; 'received:gmane.org': 0.09; 'received:list': 0.09; 'dictionaries': 0.16; 'keys.': 0.16; 'received:80.91.229.3': 0.16; 'received:dip.t-dialin.net': 0.16; 'received:plane.gmane.org': 0.16; 'received:t-dialin.net': 0.16; 'wrote:': 0.17; 'keys': 0.22; 'header:User-Agent:1': 0.26; 'header:X-Complaints-To:1': 0.28; 'though.': 0.29; 'print': 0.32; 'values.': 0.33; 'to:addr:python-list': 0.33; 'skip:d 20': 0.34; 'list': 0.35; 'received:org': 0.36; 'ones': 0.37; 'subject:: ': 0.38; 'to:addr:python.org': 0.39; 'where': 0.40; 'header:Received:5': 0.40; 'here': 0.65 X-Injected-Via-Gmane: http://gmane.org/ To: python-list@python.org From: Peter Otten <__peter__@web.de> Subject: Re: A little morning puzzle Date: Wed, 19 Sep 2012 14:09:01 +0200 Organization: None References: Mime-Version: 1.0 Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 7Bit X-Gmane-NNTP-Posting-Host: p50848fa6.dip.t-dialin.net User-Agent: KNode/4.7.3 X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.15 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: 40 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1348056502 news.xs4all.nl 6961 [2001:888:2000:d::a6]:55619 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:29484 Dwight Hutto wrote: >> I have a list of dictionaries. They all have the same keys. I want to >> find the >> set of keys where all the dictionaries have the same values. >> Suggestions? > > Here is my solution: > > > a = {} > a['dict'] = 1 > > b = {} > b['dict'] = 2 > > c = {} > c['dict'] = 1 > > d = {} > d['dict'] = 3 > > e = {} > e['dict'] = 1 > > > x = [a,b,c,d,e] > collection_count = 0 > > for dict_key_search in x: > if dict_key_search['dict'] == 1: > collection_count += 1 > print dict_key_search['dict'] > > > Might be better ones though. Unlikely.