Path: csiph.com!usenet.pasdenom.info!gegeweb.org!noc.nerim.net!nerim.net!novso.com!newsfeed.xs4all.nl!newsfeed5.news.xs4all.nl!xs4all!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 0.010 X-Spam-Evidence: '*H*': 0.98; '*S*': 0.00; 'received:mail- vc0-f174.google.com': 0.09; 'sep': 0.09; 'cc:addr:python-list': 0.10; 'better:': 0.16; 'dictionaries': 0.16; 'keys.': 0.16; 'wed,': 0.16; 'wrote:': 0.17; '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; 'am,': 0.27; 'message-id:@mail.gmail.com': 0.27; 'skip:( 20': 0.28; 'received:209.85.220.174': 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; 'received:209': 0.37; 'subject:: ': 0.38; 'where': 0.40; 'header:Received:5': 0.40 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=eskE/c3CoAoR6BtiHeZnf6xJnYh/hJUXFR+R9+4AdJY=; b=DiKmSHUuf0JNp2ALNR3LcZp0GNMHnrNOnpZaLd7yY3xT//OfOvx5sM4oDvqHshhFtX hnVu3kipOZItBIIKPGtNwQaqxW6WPmpQgMbDq1FdLECeLRBfkwq0HmUWsiNyZ3mdN6hu BzK0vQwsKkXPEK+lVD70FcwfetNu8/Jrzpc+/irBvqnw6XKYb9o+hAcMg+X169lbWzdS J2/PQrnB2MnQKlSVlz6p8aIdUo3aA99UM2hQHmiR918m2+Xa3Peh6r/R5//6UNLFz7MW L1AjGXOrvHmvb8lcHXy7oDkOvZpGCZ7aar4rhpnikZmiPWcoMYhhbuvk3UJw+sJAwp+0 BQYw== MIME-Version: 1.0 In-Reply-To: References: Date: Wed, 19 Sep 2012 08:22:27 -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: 1348057739 news.xs4all.nl 6895 [2001:888:2000:d::a6]:35041 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:29489 On Wed, Sep 19, 2012 at 8:01 AM, 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? > This one is better: 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] count = 0 collection_count = 0 search_variable = 1 for dict_key_search in x: if dict_key_search['dict'] == search_variable: print "Match count found: #%i = %i" % (count,search_variable) collection_count += 1 count += 1 print collection_count -- Best Regards, David Hutto CEO: http://www.hitwebdevelopment.com