Path: csiph.com!usenet.pasdenom.info!news.albasani.net!newsfeed.freenet.ag!news2.euro.net!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 0.000 X-Spam-Evidence: '*H*': 1.00; '*S*': 0.00; 'example:': 0.03; 'essentially': 0.04; 'explicitly': 0.04; 'attribute': 0.05; 'function,': 0.07; 'python': 0.09; 'lookup': 0.09; 'object)': 0.09; 'python:': 0.09; 'subject:while': 0.09; 'unnamed': 0.09; 'def': 0.10; 'programmer': 0.11; '(the': 0.15; 'called,': 0.16; 'dictionary).': 0.16; 'function?': 0.16; 'hard-code': 0.16; 'lambda': 0.16; 'merely': 0.16; 'naming': 0.16; 'received:74.55.86': 0.16; 'received:74.55.86.74': 0.16; 'received:smtp.webfaction.com': 0.16; 'received:webfaction.com': 0.16; 'string': 0.17; 'wrote:': 0.17; 'exists': 0.17; 'creates': 0.18; 'tim': 0.18; '>>>': 0.18; 'code.': 0.20; 'skip:p 30': 0.20; 'define': 0.20; 'names.': 0.22; 'questions:': 0.22; 'defined': 0.22; 'this:': 0.23; 'header:In-Reply-To:1': 0.25; 'header:User- Agent:1': 0.26; 'skip:" 20': 0.26; 'am,': 0.27; 'compiled': 0.27; 'functions.': 0.27; 'object,': 0.27; "doesn't": 0.28; 'dictionary': 0.29; 'name?': 0.29; 'prints': 0.29; 'class': 0.29; "i'm": 0.29; 'that.': 0.30; 'normally': 0.30; 'thursday,': 0.30; 'function': 0.30; 'expect': 0.31; 'code': 0.31; 'asking': 0.32; 'december': 0.32; 'could': 0.32; 'print': 0.32; 'defining': 0.33; 'like:': 0.33; 'to:addr:python-list': 0.33; 'knowledge': 0.33; 'doing': 0.35; 'something': 0.35; 'there': 0.35; 'but': 0.36; 'skip:p 20': 0.36; 'bad': 0.37; 'does': 0.37; 'two': 0.37; 'being': 0.37; 'quite': 0.37; 'subject:: ': 0.38; 'object': 0.38; 'some': 0.38; 'to:addr:python.org': 0.39; 'received:192': 0.39; 'received:192.168': 0.40; 'your': 0.60; 'more': 0.63; 'dont': 0.64; 'grab': 0.64; 'here': 0.65; 'answer.': 0.71; 'why?': 0.84; 'abc': 0.91 Date: Thu, 27 Dec 2012 03:03:34 -0500 From: Mitya Sirenef User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:16.0) Gecko/20121011 Thunderbird/16.0.1 MIME-Version: 1.0 To: python-list@python.org Subject: Re: Finding the name of a function while defining it References: <0kknd8tbg7knqa1ng6igbj8u82mqb720oi@4ax.com> 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: 2.1.15 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: 87 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1356595418 news.xs4all.nl 6986 [2001:888:2000:d::a6]:41064 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:35598 On 12/27/2012 02:45 AM, Abhas Bhattacharya wrote: > On Thursday, 27 December 2012 10:22:15 UTC+5:30, Tim Roberts wrote: >> Abhas Bhattacharya wrote: >> >>> While I am defining a function, how can I access the name (separately as >>> string as well as object) of the function without explicitly naming >>> it(hard-coding the name)? >>> For eg. I am writing like: >>> def abc(): >>> #how do i access the function abc here without hard-coding the name? >> >> >> Why? Of what value would that be? >> >> >> >> Note that I'm not merely being obstructionist here. What you're asking >> >> here is not something that a Python programmer would normally ask. The >> >> compiled code in a function, for example, exists as an object without a >> >> name. That unnamed object can be bound to one or more function names, but >> >> the code doesn't know that. Example: >> >> >> >> def one(): >> >> print( "Here's one" ) >> >> >> >> two = one >> >> >> >> That creates one function object, bound to two names. What name would you >> >> expect to grab inside the function? >> >> >> >> Even more obscure: >> >> >> >> two = lamba : "one" >> >> one = two >> >> >> >> Which one of these is the "name" of the function? >> >> -- >> >> Tim Roberts, timr@probo.com >> >> Providenza & Boekelheide, Inc. > It is of quite value to me. > Because I have this situation: > I have used a dictionary with "function_name":value pair in the top of the code. Now when some function is called, I need to print the value assigned to its name in the dictionary (the functions are defined after the dictionary). Now there is only one bad way-around for me: I need to hard-code the name in the function like this: > def function_name(): > print(dict_name.get("function_name")) > but ofcourse it is a bad thing to do because I have a lot of this type of functions. It would be better if I can can use the same code for all of them, because they are all essentially doing the same thing. > > Now, for your questions: > If i call one() and two() respectively, i would like to see "one" and "two". > I dont have much knowledge of lambda functions, neither am i going to use them, so that's something I cant answer. How about defining a function that prints value and then calls a function? def call(func_name): print(mydict[func_name]) globals()[func_name]() You could also define a custom class that does the same thing on attribute lookup and do something like Call.func_name() . -m -- Lark's Tongue Guide to Python: http://lightbird.net/larks/