Path: csiph.com!news.swapon.de!fu-berlin.de!uni-berlin.de!not-for-mail From: Ian Kelly Newsgroups: comp.lang.python Subject: Re: Extract the middle N chars of a string Date: Wed, 18 May 2016 10:34:12 -0600 Lines: 84 Message-ID: References: <573c8e97$0$1596$c3e8da3$5496439d@news.astraweb.com> Mime-Version: 1.0 Content-Type: text/plain; charset=UTF-8 X-Trace: news.uni-berlin.de sO/dP7D4vchReMt9tJibvwO0UVqO+HofbEmDAtnTQrGg== Return-Path: X-Original-To: python-list@python.org Delivered-To: python-list@mail.python.org X-Spam-Status: OK 0.001 X-Spam-Evidence: '*H*': 1.00; '*S*': 0.00; 'elif': 0.04; 'seemed': 0.07; 'cc:addr:python-list': 0.09; '"""return': 0.09; 'ambiguity': 0.09; 'subject:string': 0.09; 'received:209.85.218': 0.10; 'def': 0.13; 'wed,': 0.15; '2016': 0.16; 'cc:name:python': 0.16; 'received:io': 0.16; 'received:psf.io': 0.16; 'slicing:': 0.16; 'wrote:': 0.16; 'string': 0.17; 'odd': 0.18; 'cc:2**0': 0.20; 'cc:addr:python.org': 0.20; 'function,': 0.22; 'am,': 0.23; 'seems': 0.23; 'this:': 0.23; 'header:In-Reply-To:1': 0.24; 'message-id:@mail.gmail.com': 0.27; 'print': 0.30; 'getting': 0.33; 'problem': 0.33; "d'aprano": 0.33; 'errors,': 0.33; 'steven': 0.33; 'received:google.com': 0.35; 'but': 0.36; 'received:209.85': 0.36; 'subject:: ': 0.37; 'received:209': 0.38; 'skip:p 20': 0.38; 'subject:the': 0.39; 'easy': 0.60; 'different': 0.63; 'to:none': 0.91 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:cc; bh=QrXLNZnWkwNkqVxlIHBYKHijZONcrrkfLfPlEHfHzrE=; b=bMm0G2XOFCLLzU7xH4FE+U2obFD4Hvr3Ti/5vK9jXsVsRLAQ9gPsM2PZMxU70pmuUq HFvwcodA3FtaIdA6UZxqXnd3yNiDJChDLZiylGed2HvLTxUmU6GoGk2al2DIoRjOEMZH 7UGV+A9kSkrehQ4C2dJdsigmQZvJYiVDt2cks2cOUsueT8o9sjgzajfwlugrQnkbf4cr kvHZzcwtGs4eKJKY0C0kEl87N1Hoqvc5BwtDQ3XQbuD1C8REL7BA/RweTzLNLH0u6fAa QgiUrKHx2i1JOdZ5+S9Srhy8xYHv0MGBdE0Eb/q17OEGlPa6/rv5/SNIBwifSBHS4whe kuUg== X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=1e100.net; s=20130820; h=x-gm-message-state:mime-version:in-reply-to:references:from:date :message-id:subject:cc; bh=QrXLNZnWkwNkqVxlIHBYKHijZONcrrkfLfPlEHfHzrE=; b=IW4YaLfwIUu1BKIUbLpN/RdLd2lrHaQJrNYqAHYOSdy7abU47xIow5djcHaW+f8zbI FUw7O3ZM3rjfsABWpaGnAKs1et+nBzIwkb2yhA8FUtpzat1NtF97wwZoU8VjSsN0zvBp Mh8kYRMrnnocbATcBPJj5Pk8io4lPkXd0WibfSwu4ZzBAWGtLPd6Go50hDBZsDX/D3d2 c8Ixavlk07+QEK3QOZmfZ9YSXt805M6ctUdZFGmtbf18lYhdAJCGITV/v4C4Tvqfzk2e OW6ZShXVmIfb2CWYPUIVCkYJiM17QtBSnybF1GzyTFoFYl5IW048wo0jnRYpjIhkDewN 322A== X-Gm-Message-State: AOPr4FVOP6TbpUpAm93NU7n6p+J7j5j9WS90e0o0KvTSI/Dgn1bt3IojIhoWUIs1okCPX9D/LsZScV23+U5JNg== X-Received: by 10.202.87.12 with SMTP id l12mr1674935oib.54.1463589292257; Wed, 18 May 2016 09:34:52 -0700 (PDT) In-Reply-To: <573c8e97$0$1596$c3e8da3$5496439d@news.astraweb.com> X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.22 Precedence: list List-Id: General discussion list for the Python programming language List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-Mailman-Original-Message-ID: X-Mailman-Original-References: <573c8e97$0$1596$c3e8da3$5496439d@news.astraweb.com> Xref: csiph.com comp.lang.python:108773 On Wed, May 18, 2016 at 9:47 AM, Steven D'Aprano wrote: > Extracting the first N or last N characters of a string is easy with > slicing: > > s[:N] # first N > s[-N:] # last N > > Getting the middle N seems like it ought to be easy: > > s[N//2:-N//2] > > but that is wrong. It's not even the right length! > > py> s = 'aardvark' > py> s[5//2:-5//2] > 'rdv' > > > So after spending a ridiculous amount of time on what seemed like it ought > to be a trivial function, and an embarrassingly large number of off-by-one > and off-by-I-don't-even errors, I eventually came up with this: > > def mid(string, n): > """Return middle n chars of string.""" > L = len(string) > if n <= 0: > return '' > elif n < L: > Lr = L % 2 > a, ar = divmod(L-n, 2) > b, br = divmod(L+n, 2) > a += Lr*ar > b += Lr*br > string = string[a:b] > return string > > > which works for me: > > > # string with odd number of characters > py> for i in range(1, 8): > ... print mid('abcdefg', i) > ... > d > de > cde > cdef > bcdef > bcdefg > abcdefg > # string with even number of characters > py> for i in range(1, 7): > ... print mid('abcdef', i) > ... > c > cd > bcd > bcde > abcde > abcdef > > > > Is this the simplest way to get the middle N characters? There's an ambiguity in the problem description when N and the length of the string have different parity, but how about: def mid(string, n): L = len(string) a = (L - n) // 2 return string[a:a+n] py> for i in range(1, 8): ... print(mid('abcdefg', i)) ... d cd cde bcde bcdef abcdef abcdefg