[Tutor] Data structure question / PyGoogle

[Python]

(forwarding to list)
On Sat, 2006-09-16 at 10:31 -0500, Brian Edward wrote:
> Thanks for the quick reply!  I really appreciate your assistance.  Of
> course, it will take some time to get this worked out, but your
> explanation is very clear.  
>  
> Best,
> Brian
> 
>  
> On 9/16/06, Python <python at venix.com> wrote: 
>         On Sat, 2006-09-16 at 09:44 -0500, Brian Edward wrote:
>         > Hello all,
>         >
>         > I am new to Python (and programming in general) and am
>         trying to get 
>         > PyGoogle figured out for some specific research
>         interests.  Basically,
>         > I have done a simple search using PyGoogle and have some
>         sitting in
>         > memory.  I have an object data.results, which is apparently
>         a list: 
>         >
>         > >>> type(data.results)
>         > <type 'list'>
>         >
>         > In this list, I have ten URL saved, which I can access by
>         using the
>         > brackets and noting the specific elements.  For example: 
>         >
>         > >>> data.results[0].URL
>         > 'http://www.psychguides.com/gl-
>         treatment_of_schizophrenia_1999.html'
>         >
>         > >>> data.results [1].URL
>         > 'http://www.psychguides.com/sche.pdf'
>         >
>         > My question is, how can I access all ten URLs in a single
>         command.
>         > Specifically, why does the following statement not work: 
>         >
>         > >>> data.results[0:10].URL
>         You need to extract the URL from each item in the result list.
>         Something like:
>         
>                urls = [r.URL for r in data.results]
>         
>         will extract a list of urls from your list of results. 
>         >
>         > Traceback (most recent call last):
>         >   File "<pyshell#78>", line 1, in -toplevel-
>         >     data.results[0:10].URL
>         > AttributeError: 'list' object has no attribute 'URL'
>         > 
>         >
>         > Again, I am new to Python, so a watered-down, conceptual
>         response to
>         > this would be greatly appreciated.  Thanks in advance.
>         
>         data.results[0:10] simply copies the first 10 results from
>         your original 
>         list into a new list.  The new list does not have a URL
>         attribute, as
>         the error message tells us.  The URL attribute belongs to the
>         individual
>         items in the list.
>         
>         You need to process each result in data.results to extract the
>         URL.
>         Your choices boil down to:
>                for statement
>         or the more functionally oriented
>                map
>                list comprehension
>                generator expression (python 2.4 or later)
>         
>         For creating a new list from an existing list, a list
>         comprehension is 
>         usually the best bet.  The for statement approach would look
>         something
>         like:
>                urls = []
>                for r in data.results:
>                        urls.append(r.URL)
>         
>         list comprehensions provide a simpler, more direct syntax. 
>         
>         
>         
>         >
>         > Brian
>         >
>         >
>         >
>         >
>         > _______________________________________________
>         > Tutor maillist  -  Tutor at python.org
>         > http://mail.python.org/mailman/listinfo/tutor
>         --
>         Lloyd Kvam
>         Venix Corp
>         
> 
-- 
Lloyd Kvam
Venix Corp