Path: csiph.com!usenet.pasdenom.info!gegeweb.org!usenet-fr.net!nerim.net!novso.com!newsfeed.xs4all.nl!newsfeed1.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; 'from:addr:yahoo.co.uk': 0.04; 'element': 0.07; 'remaining': 0.07; 'subject:two': 0.07; 'lawrence': 0.09; 'received:80.91': 0.09; 'received:80.91.229': 0.09; 'received:gmane.org': 0.09; 'received:list': 0.09; 'jan': 0.12; 'language.': 0.14; 'enough.': 0.16; 'received:80.91.229.3': 0.16; 'received:plane.gmane.org': 0.16; 'skip:[ 40': 0.16; 'skip:{ 40': 0.16; 'subject:based': 0.16; 'subject:key': 0.16; 'index': 0.16; 'sat,': 0.16; 'language': 0.16; 'wrote:': 0.18; 'seems': 0.21; 'code,': 0.22; 'header:User-Agent:1': 0.23; '31,': 0.24; 'header:X-Complaints-To:1': 0.27; 'header:In-Reply-To:1': 0.27; 'chris': 0.29; 'code': 0.31; 'subject:per': 0.31; 'subject:from': 0.34; 'could': 0.34; 'equal': 0.35; 'subject:lists': 0.35; 'but': 0.35; 'adjust': 0.36; 'right?': 0.36; 'skip:[ 10': 0.38; 'to:addr :python-list': 0.38; 'pm,': 0.38; 'to:addr:python.org': 0.39; 'received:org': 0.40; 'easy': 0.60; 'our': 0.64; 'charset:windows-1252': 0.65; 'line,': 0.68; '2015': 0.84; 'convenience,': 0.91 X-Injected-Via-Gmane: http://gmane.org/ To: python-list@python.org From: Mark Lawrence Subject: Re: Create dictionary based of x items per key from two lists Date: Sun, 01 Feb 2015 03:06:21 +0000 References: <0dddee06-233b-436a-be48-3c16e62c1718@googlegroups.com> Mime-Version: 1.0 Content-Type: text/plain; charset=windows-1252; format=flowed Content-Transfer-Encoding: 7bit X-Gmane-NNTP-Posting-Host: host-92-24-222-48.ppp.as43234.net User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; rv:31.0) Gecko/20100101 Thunderbird/31.4.0 In-Reply-To: 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: 55 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1422759990 news.xs4all.nl 2884 [2001:888:2000:d::a6]:36190 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:84973 On 31/01/2015 02:38, Chris Angelico wrote: > On Sat, Jan 31, 2015 at 1:27 PM, wrote: >> l1 = ["a","b","c","d","e","f","g","h","i","j"] >> l2 = ["aR","bR","cR"] >> >> l2 will always be smaller or equal to l1 >> >> numL1PerL2 = len(l1)/len(l2) >> >> I want to create a dictionary that has key from l1 and value from l2 based on numL1PerL2 >> >> So >> >> { >> a:aR, >> b:aR, >> c:aR, >> d:bR, >> e:bR, >> f:bR, >> g:cR, >> h:cR, >> i:cR, >> j:cR >> } >> >> So last item from l2 is key for remaining items from l1 > > So the Nth element of l1 will always be paired with the > (N/numL1PerL2)th element of l2 (with the check at the end)? Seems easy > enough. > > dups = len(l1)/len(l2) > l2.append(l2[-1]) > result = {x:l2[i/dups] for i,x in enumerate(l1)} > > This mutates l2 for convenience, but you could also adjust the index > to take care of the excess. As a one-liner: > > result = {x:l2[min(i/(len(l1)/len(l2)),len(l2)-1)] for i,x in enumerate(l1)} > > But the one-liner is not better code :) > > ChrisA > The one-liner might not be better code, but it must be better speed wise precisely because it's on one line, right? :) -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence