Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]


Groups > comp.lang.python > #26876

Re: dictionary into desired variable....

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!newsfeed5.news.xs4all.nl!xs4all!post.news.xs4all.nl!not-for-mail
Return-Path <d@davea.name>
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; 'value,': 0.03; 'attribute': 0.05; 'class,': 0.07; 'nested': 0.07; 'suppose': 0.07; 'python': 0.09; '__init__': 0.09; 'fetch': 0.09; 'none.': 0.09; 'subject:into': 0.09; 'cc:addr:python-list': 0.10; 'student': 0.15; 'value.': 0.15; 'assumptions': 0.16; 'dictionary,': 0.16; 'pointer,': 0.16; 'pointers.': 0.16; 'variable.': 0.16; 'write,': 0.16; 'wrote:': 0.17; 'pointed': 0.17; 'bit': 0.21; 'supposed': 0.21; 'keys': 0.22; "i'd": 0.22; 'cc:2**0': 0.23; 'programming': 0.23; 'this:': 0.23; 'cc:no real name:2**0': 0.24; 'least': 0.25; 'cc:addr:python.org': 0.25; 'header:In-Reply-To:1': 0.25; 'header:User-Agent:1': 0.26; 'looks': 0.26; 'am,': 0.27; 'language.': 0.27; 'question': 0.27; "doesn't": 0.28; 'fixed': 0.28; 'actual': 0.28; 'represent': 0.28; 'dictionary': 0.29; 'points': 0.29; 'that.': 0.30; 'saves': 0.30; 'function': 0.30; 'code': 0.31; 'gets': 0.32; 'could': 0.32; 'certain': 0.33; 'values.': 0.33; "can't": 0.34; 'list': 0.35; 'data,': 0.35; 'generic': 0.35; 'returning': 0.35; 'so,': 0.35; 'doing': 0.35; 'something': 0.35; 'there': 0.35; 'really': 0.36; 'but': 0.36; 'method': 0.36; 'useful': 0.36; 'should': 0.36; 'two': 0.37; 'subject:: ': 0.38; 'store': 0.38; 'mean': 0.38; 'some': 0.38; 'takes': 0.39; 'received:192': 0.39; 'list,': 0.39; 'where': 0.40; 'received:192.168': 0.40; 'first': 0.61; 'kind': 0.61; 'more': 0.63; 'choose': 0.65; 'subject:....': 0.65; 'header :Reply-To:1': 0.68; 'receive': 0.71; 'received:74.208': 0.71; 'sounds': 0.71; 'reply-to:no real name:2**0': 0.72; 'presumably': 0.84; 'routines': 0.84
Date Fri, 10 Aug 2012 11:14:05 -0400
From Dave Angel <d@davea.name>
User-Agent Mozilla/5.0 (X11; Linux x86_64; rv:14.0) Gecko/20120714 Thunderbird/14.0
MIME-Version 1.0
To Tamer Higazi <th982a@googlemail.com>
Subject Re: dictionary into desired variable....
References <50251477.7010507@googlemail.com>
In-Reply-To <50251477.7010507@googlemail.com>
Content-Type text/plain; charset=ISO-8859-1
Content-Transfer-Encoding 7bit
X-Provags-ID V02:K0:tQvco1aAQl/eb4nXCRxOs8zj7H2eyIekwePwsfA6tVn sVBraPeYdS76vzCZRVFL+t1DUWGtTj/7kwqSCT7xOYRRxqXoic tdjep1QeGlve4LYbSuGs7P0zOf+UgpNK5o1kacz2dZE3yx8aDT K1C1rIskaMa3MID0AquTG95bdWVRsMVz/3gzprTau2SwIYzyzF qOU7YShlPPFo+Gdc5YMfZgbJjLttjMBRNl/fMRzGRa1kZb2fZ1 4ozUdougG1XTpqrwgq9Oj3fF2sbw/w5/9W1hEKI9ibiCOLElYp V5yEaQQogYXW+ka5LSnrysSGCigyP7SovdEVOQuUVdhdNGMvQ= =
Cc python-list@python.org
X-BeenThere python-list@python.org
X-Mailman-Version 2.1.12
Precedence list
Reply-To d@davea.name
List-Id General discussion list for the Python programming language <python-list.python.org>
List-Unsubscribe <http://mail.python.org/mailman/options/python-list>, <mailto:python-list-request@python.org?subject=unsubscribe>
List-Archive <http://mail.python.org/pipermail/python-list>
List-Post <mailto:python-list@python.org>
List-Help <mailto:python-list-request@python.org?subject=help>
List-Subscribe <http://mail.python.org/mailman/listinfo/python-list>, <mailto:python-list-request@python.org?subject=subscribe>
Newsgroups comp.lang.python
Message-ID <mailman.3165.1344611670.4697.python-list@python.org> (permalink)
Lines 47
NNTP-Posting-Host 2001:888:2000:d::a6
X-Trace 1344611670 news.xs4all.nl 6970 [2001:888:2000:d::a6]:48053
X-Complaints-To abuse@xs4all.nl
Xref csiph.com comp.lang.python:26876

Show key headers only | View raw


On 08/10/2012 10:02 AM, Tamer Higazi wrote:
> Hi!
> suppose you have a dictionary that looks like this:
>
> x = [1,3,6,1,1] which should represent a certain other variable.
>
> in reality it would represent:
>
> y[1][3][6][1][1]
>
> Now, how do I write a python routine, that points in this dictionary,
> where I should receive or set other values.
>

Others have pointed out how you could "receive" a value from that. 
Doing a setter would be tricker.

But I'd want to start with the question of just what you're really
looking for.  Is this an assignment, so it doesn't have to actually make
sense?  If so, could we please have the actual wording.   x is not a
dictionary, and Python doesn't have routines nor pointers.  Perhaps it's
a generic programming class, and the student gets to choose the language.

If it's indeed intended to model some kind of useful data, could we know
a bit more about the constraints?  is y supposed to be a nested list, or
a nested dictioary, or something else? How is this nested list/dict
first created?  Do you need to have other parts of the code access the
raw, nested Something.  Are there always going to be 5 subscripts, and
are each constrained to be in a fixed range?  If it's a nested
dictionary, are the keys always integers?

Since a python function or method can't return a pointer, presumably you
mean for a function or method to fetch or store a value.

Sounds like what you want is a class, where the __init__ method takes a
nested something, y and saves it as an attribute _y.  And there are at
least two more methods in that class, fetch() that takes a list of ints,
and returns a value.  And store() that takes a list of ints and a value,
and mutates the _y, returning None.  The store() method is a good bit
trickier to write, depending on the assumptions above.



-- 

DaveA

Back to comp.lang.python | Previous | Next | Find similar | Unroll thread


Thread

Re: dictionary into desired variable.... Dave Angel <d@davea.name> - 2012-08-10 11:14 -0400

csiph-web