Path: csiph.com!v102.xanadu-bbs.net!xanadu-bbs.net!feeder.erje.net!eu.feeder.erje.net!xlned.com!feeder1.xlned.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 0.012 X-Spam-Evidence: '*H*': 0.98; '*S*': 0.00; 'else:': 0.03; 'subject:two': 0.07; 'subject:into': 0.09; 'cc:addr:python-list': 0.11; 'thread': 0.14; 'posted': 0.15; 'cc:name:python list': 0.16; 'reedy': 0.16; 'rewriting': 0.16; 'subject:based': 0.16; 'wrote:': 0.18; 'cc:addr:python.org': 0.22; 'cc:2**0': 0.24; 'header:In- Reply-To:1': 0.27; 'statement': 0.30; 'subject:list': 0.30; 'message-id:@mail.gmail.com': 0.30; 'code': 0.31; 'this.': 0.32; "we're": 0.32; 'but': 0.35; 'received:google.com': 0.35; 'subject:?': 0.36; 'skip:o 20': 0.38; 'back': 0.62; 'skip:n 10': 0.64; 'started.': 0.68; 'oscar': 0.84; '2013': 0.98 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 :cc:content-type; bh=ykzSaKEzVsasHe9NnUF0jZ2UKAPrrZBp2uoAU6Ku53g=; b=lwisW0vWxLZvh3JTOG+oykKMYQnJPD6/XQ9eb2qUkEEmDf+FBIlS9/Uwks/N03eTsO LzzWaDU4sjmiF56VA0Phgjla0+LzFt+OffVwkC6Xx1EvlZZriuglKBVY5bVHC1DmYxkC eAErj5Mr/pVF3pefrZ5CU5wHPCZ0rBYPuQYjpqAiqKtdCPhkd3id2lWGH9xoJiXauZt+ gCI/XRvN+WtlILiLgljknRD5J8EAdLYg79TU2cCtSXB/xCRro5V1FDYeQCvOmKW/iOxt 5XZ6UCB+CBBgZNLikfmaY9I7cpNvuj7rrqORpFQPeeKnvuNxGQBOLAX2y1bwo4w5dNF/ pxEQ== X-Received: by 10.52.27.20 with SMTP id p20mr9524342vdg.123.1371116613803; Thu, 13 Jun 2013 02:43:33 -0700 (PDT) MIME-Version: 1.0 In-Reply-To: References: From: Oscar Benjamin Date: Thu, 13 Jun 2013 10:43:13 +0100 Subject: Re: Split a list into two parts based on a filter? To: Terry Reedy Content-Type: text/plain; charset=ISO-8859-1 Cc: Python List 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: 21 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1371116623 news.xs4all.nl 15882 [2001:888:2000:d::a6]:43813 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:47929 On 12 June 2013 19:47, Terry Reedy wrote: > The proper loop statement > > for s in songs: > (new_songs if s.is_new() else old_songs).append(s) I think I would just end up rewriting this as for s in songs: if s.is_new(): new_songs.append(s) else: old_songs.append(s) but then we're back where we started. I don't think any of the solutions posted in this thread have been better than this. If you want to make this a nice one-liner then just put this code in a function. Oscar