Path: csiph.com!usenet.pasdenom.info!gegeweb.org!de-l.enfer-du-nord.net!feeder1.enfer-du-nord.net!newsfeed.eweka.nl!eweka.nl!feeder3.eweka.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 0.041 X-Spam-Evidence: '*H*': 0.92; '*S*': 0.00; 'received:mail- vc0-f174.google.com': 0.09; 'cc:addr:python-list': 0.10; 'dictionaries': 0.16; 'keys.': 0.16; 'keys': 0.22; 'cc:2**0': 0.23; 'cc:no real name:2**0': 0.24; 'cc:addr:python.org': 0.25; 'header:In-Reply-To:1': 0.25; 'message-id:@mail.gmail.com': 0.27; 'received:209.85.220.174': 0.29; 'though.': 0.29; 'print': 0.32; 'values.': 0.33; 'skip:d 20': 0.34; 'received:google.com': 0.34; 'list': 0.35; 'received:209.85.220': 0.35; 'received:209.85': 0.35; 'ones': 0.37; 'received:209': 0.37; 'subject:: ': 0.38; 'where': 0.40; 'header:Received:5': 0.40; 'here': 0.65 DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :cc:content-type; bh=PIpON0B/pDcVKMzCY7e2RoRWMinpeZ+3D99R9M1iLoQ=; b=XHfwdO6DchZ2GiBHqvsGk34QGSYj8DtRw3IyQaj22OMWDRYiRbjMOQbv/c8jGHTINW oHj5spOdRVlCvkYv/bWofnmtIEHCPkbZleZOHM7OnbUgfC3JpMv7jdLw2Ks/Yvv6D7Um AJU2tCoY71QEHjejgk5RpWA/afkFkziCQcAwW1gQkefRHU3Tk2wAZ8IqUjsjLpHTGoye g5mHikY9j+kCntW/nvyvnNrqKSgAjoFzft3ET/HXNHdF5kWOD+9/Q0AZvi+Db3NZa1yy 966Fk9FSrhGWAqsZShHNQuR1T/4Rs4SkNTc5WSGJ5T67P974iC8xzemoQVibD99TzSpA OUhA== MIME-Version: 1.0 In-Reply-To: References: Date: Wed, 19 Sep 2012 08:01:11 -0400 Subject: Re: A little morning puzzle From: Dwight Hutto To: Neal Becker Content-Type: text/plain; charset=ISO-8859-1 Cc: python-list@python.org 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: 38 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1348056073 news.xs4all.nl 6863 [2001:888:2000:d::a6]:52687 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:29483 > 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. -- Best Regards, David Hutto CEO: http://www.hitwebdevelopment.com