Path: csiph.com!newsfeed.hal-mli.net!feeder3.hal-mli.net!newsfeed.hal-mli.net!feeder1.hal-mli.net!newsfeed.xs4all.nl!newsfeed6.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 0.005 X-Spam-Evidence: '*H*': 0.99; '*S*': 0.00; 'example:': 0.03; 'algorithm': 0.03; 'nested': 0.07; 'suppose': 0.07; 'tom': 0.07; 'arguments,': 0.09; 'iterate': 0.09; '3):': 0.16; 'argument.': 0.16; 'from:addr:ian': 0.16; 'wrote:': 0.17; 'variable': 0.20; 'subject:problem': 0.22; 'header:In-Reply-To:1': 0.25; 'header :User-Agent:1': 0.26; 'values': 0.26; 'wondering': 0.26; 'realize': 0.27; 'this?': 0.28; 'initial': 0.28; 'received:208.97': 0.29; 'received:208.97.132': 0.29; 'received:dreamhost.com': 0.29; 'received:g.dreamhost.com': 0.29; "i'm": 0.29; 'function': 0.30; '11,': 0.33; 'to:addr:python-list': 0.33; 'sequence': 0.35; 'something': 0.35; 'there': 0.35; 'but': 0.36; 'subject:: ': 0.38; 'some': 0.38; 'things': 0.38; 'to:addr:python.org': 0.39; 'step': 0.39; 'received:192': 0.39; 'called': 0.39; 'received:192.168': 0.40; 'range': 0.60; 'different': 0.63; 'received:208': 0.63; 'more': 0.63; '10,000': 0.65; 'subject:looking': 0.91 DomainKey-Signature: a=rsa-sha1; c=nofws; d=feete.org; h=message-id:date:from :mime-version:to:subject:references:in-reply-to:content-type: content-transfer-encoding; q=dns; s=feete.org; b=Ryg3RvZn0ZUE1yv nNvcewCVnt34j4jAJtzPO4hqzTdy5Ikitdq2pa71EYqYmBT7/A4zyL12ArR6AT1N 9U8nWLkzqqAYlNSJ6cXRU/dDLLMGAm5JjofK+5E6WedR5/7cKn2yZ/BTQdsNuW71 q5oPqqlfh2rqVnhcAlHh8KtGwo1c= DKIM-Signature: v=1; a=rsa-sha1; c=relaxed; d=feete.org; h=message-id :date:from:mime-version:to:subject:references:in-reply-to :content-type:content-transfer-encoding; s=feete.org; bh=A8kYmPf zqXCKniauC3AOJ1lfLCg=; b=ShWy8kmNDYL6yjy45GvWt8nFyqS/RZwuye3ewPl 4zYooSnb9675l/HtuTZhectXj8SjmS2df9EcCJKlc+vXcYkKskQ23ju00ZzEJNi7 RyqznGBFNl9FH9VxSYO6asg9f7b4llEM1TU/ycIslVGvTUAq5GZ6nEgNVZPdJjft pOl0= Date: Mon, 06 Aug 2012 17:07:19 +0100 From: Ian Foote User-Agent: Mozilla/5.0 (X11; Linux i686; rv:14.0) Gecko/20120714 Thunderbird/14.0 MIME-Version: 1.0 To: python-list@python.org Subject: Re: looking for a neat solution to a nested loop problem References: In-Reply-To: Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.12 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: 33 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1344269252 news.xs4all.nl 6952 [2001:888:2000:d::a6]:40450 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:26629 The function range can be called with more than one argument. For example: for i in range(N, N + 10): for j in range(M, M + 100): do_something(i, j) You can also call range with 3 arguments, if want a step size different to 1: for k in range(2, 11, 3): print(k) 2 5 8 Hope this is clear, Ian On 06/08/12 16:52, Tom P wrote: > consider a nested loop algorithm - > > for i in range(100): > for j in range(100): > do_something(i,j) > > Now, suppose I don't want to use i = 0 and j = 0 as initial values, > but some other values i = N and j = M, and I want to iterate through > all 10,000 values in sequence - is there a neat python-like way to > this? I realize I can do things like use a variable for k in > range(10000): and then derive values for i and j from k, but I'm > wondering if there's something less clunky.