Path: csiph.com!usenet.pasdenom.info!dedibox.gegeweb.org!gegeweb.eu!nntpfeed.proxad.net!proxad.net!feeder1-2.proxad.net!usenet-fr.net!de-l.enfer-du-nord.net!feeder1.enfer-du-nord.net!feeds.phibee-telecom.net!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 0.031 X-Spam-Evidence: '*H*': 0.94; '*S*': 0.00; 'skip:[ 20': 0.03; 'sep': 0.09; 'dct': 0.16; 'dictionaries': 0.16; 'keys.': 0.16; 'wed,': 0.16; 'wrote:': 0.17; 'keys': 0.22; 'header:In-Reply-To:1': 0.25; 'am,': 0.27; 'message-id:@mail.gmail.com': 0.27; 'values.': 0.33; 'to:addr:python-list': 0.33; 'received:google.com': 0.34; 'list': 0.35; 'received:209.85': 0.35; 'item': 0.37; 'received:209': 0.37; 'subject:: ': 0.38; 'to:addr:python.org': 0.39; 'where': 0.40; 'header:Received:5': 0.40; 'pardon': 0.84; 'to:name:python': 0.84 DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:in-reply-to:references:from:date:message-id:subject:to :content-type; bh=M+snlVL2YXhktZuYFWKowMiYq4WULaQRkcBw3IM+9WU=; b=w0yMA+04TLDEgQqpF+CuutoFTvmdTdjyYMzFDWIGE82wyOCVdDl+k3S1Z+ZdNPBUvh CbkIWp/SlLWxDddvqetlaMQ7kqewPsb6XEFD2QlFhQKLKncVetN6mIdsHcijesI1RQBj nYVBSaLB5DNJRnaFghm3G2XNT/Il50Wqo5MyUmIA8euW0tWphG+jVZOSOjjdglXDc7OI unZJnNDvIygdul0DbkiGZ7zCiErupCYxK5wijMP18cxJGfdFIGsJD1bRVTLKe0lsAg2z +Z1cji48va5AoTlDn5kmz0YmYuv4lsacJMwRWX92KyYbZST6850PW94o5cFM2mPenXqK 34fA== MIME-Version: 1.0 In-Reply-To: <5059B6DD.8030100@rece.vub.ac.be> References: <5059B6DD.8030100@rece.vub.ac.be> From: Ian Kelly Date: Wed, 19 Sep 2012 09:16:23 -0600 Subject: Re: A little morning puzzle To: Python Content-Type: text/plain; charset=ISO-8859-1 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: 13 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1348067815 news.xs4all.nl 6927 [2001:888:2000:d::a6]:57437 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:29500 On Wed, Sep 19, 2012 at 6:13 AM, Antoon Pardon wrote: > On 19-09-12 13:17, Neal Becker 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? > common_items = reduce(opereator.__and__, [set(dct.iteritems()) for dct > in lst]) > common_keys = set([item[0] for item in common_items]) You can use dictviews for that: common_items = reduce(operator.__and__, (d.viewitems() for d in ds)) common_keys = [item[0] for item in common_items]