Path: csiph.com!usenet.pasdenom.info!gegeweb.org!de-l.enfer-du-nord.net!feeder1.enfer-du-nord.net!feeds.phibee-telecom.net!newsfeed.xs4all.nl!newsfeed2.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 0.003 X-Spam-Evidence: '*H*': 0.99; '*S*': 0.00; 'exist,': 0.07; 'nested': 0.07; 'none)': 0.07; 'default)': 0.09; 'dict': 0.09; 'exception,': 0.09; 'exists,': 0.09; 'get.': 0.09; 'lookup': 0.09; 'throw': 0.09; 'advance': 0.10; "'a',": 0.16; "'b'": 0.16; "'b',": 0.16; "'c',": 0.16; '(rather': 0.16; 'looping': 0.16; 'namedtuple': 0.16; 'none),': 0.16; 'pythonic': 0.16; 'wrote:': 0.17; 'typical': 0.17; 'keys': 0.22; 'second': 0.24; 'header:In-Reply-To:1': 0.25; 'header:User-Agent:1': 0.26; 'am,': 0.27; 'i.e.': 0.27; 'list:': 0.27; "doesn't": 0.28; 'dictionary': 0.29; 'key,': 0.29; 'way?': 0.29; 'wrap': 0.29; "i'm": 0.29; 'print': 0.32; 'getting': 0.33; 'like:': 0.33; 'to:addr:python-list': 0.33; 'list': 0.35; 'data,': 0.35; 'doing': 0.35; 'something': 0.35; 'but': 0.36; 'method': 0.36; 'should': 0.36; 'thank': 0.36; 'item': 0.37; 'subject:: ': 0.38; 'sure': 0.38; 'instead': 0.39; 'to:addr:python.org': 0.39; 'received:192': 0.39; 'received:192.168': 0.40; 'help': 0.40; 'your': 0.60; 'most': 0.61; 'first': 0.61; 'is.': 0.62; 'within': 0.64; 'received:74.208': 0.71; "'2',": 0.84; 'dict.': 0.84; 'received:74.208.4.194': 0.84; 'subject:value': 0.84; 'was:': 0.91 Date: Tue, 05 Mar 2013 02:20:51 -0500 From: Dave Angel User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:17.0) Gecko/20130106 Thunderbird/17.0.2 MIME-Version: 1.0 To: python-list@python.org Subject: Re: Pythonic way for retrieving value for a nested dictionary. References: <4a0888ba-396b-4b6e-af3e-7560962d6750@googlegroups.com> In-Reply-To: <4a0888ba-396b-4b6e-af3e-7560962d6750@googlegroups.com> Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit X-Provags-ID: V02:K0:37P9H1osAIzG6DrXGO19ROyLqmNp4gIFLfAKRibK7x+ 75YRY07wFMcYrN5kgy2u54sSHOSDA7nn/DsWQ3Xt0b814uL8Mm eKPOWhFsMdx7adcECZon9L93b+Zb6uI7AREUGm3wA4LYNDoxP3 txkrhqn/J/8maaCyXx+Yxq4bvqxqIIj2ki999kPtbl9TrPDqw7 xw2uGDpHQiicUUah/Szs6haxGgyhKM1DFQDaYb3JjAuyk7G/jV 5j7EYUNq5JnjC2+rkSvWRm14hQxzikIKwO2BMqaftRnCdONBml mjooEEFz8QXNy3ayNQ1mT7Wb9ly+Hfh+zU3upGo03RwjHDMUA= = 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: 47 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1362468065 news.xs4all.nl 6955 [2001:888:2000:d::a6]:54209 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:40504 On 03/05/2013 01:48 AM, Lowly Minion wrote: > For a typical dict: > > i.e. > d = { '1': 'a', '2', 'b' } > > I might use something like: > > d.get('1', None) > > To get the value of 1. > > What would be the most pythonic way of getting a nested value of a dictionary within a list: > > some_list = [{ 'item': { 'letter': 'b', > 'word': 'bar' }}, > { 'item': { 'letter': 'c', > 'word': 'charcoal' }}] > > Currently I am looping through the list as such: > > for item in some_list: > print item['item']['letter'] > > One other method explored was: > item.get('item').get('letter') > > Is a double get the best way? Doesn't seem right to me. > > Thank you in advance for your help > I'm not sure what the puzzle is. Use [key] if you know the key exists, and use get( key, default) if you don't. But be aware that if the first key doesn't exist, then the default had better be a dict (rather than something like None), or the second lookup will throw an exception, even if you use get. One other possibility, depending on just what you're doing with the data, is to wrap it in a try/catch. However, if you know exactly what the keys are, perhaps you should use a namedtuple instead of a dict. -- DaveA