[Tutor] Flatten a list in tuples and remove doubles

[Francesco Loffredo]

Il 28/07/2012 20:41, eryksun wrote:
> On Sat, Jul 28, 2012 at 11:12 AM, Francesco Loffredo<fal at libero.it>  wrote:
>> I had to study carefully your present and desired lists, and I understood
>> what follows (please, next time explain !):
>> - each 7-tuple in your present list is a record for some measure relative to
>> a person. Its fields are as follows:
>>      - field 0: code (I think you want that in growing order)
>>      - field 1: group code (could be a class or a group to which both of your
>> example persons belong)
>>      - fields 2, 3: surname and name of the person
>>      - field 4: progressive number of the measure (these are in order
>> already, but I think you want to enforce this) that you want to exclude from
>> the output list while keeping the order
>>      - field 5, 6: numerator and denominator of a ratio that is the measure.
>> you want the ratio to be written as a single string: "%s/%s" % field5,
>> field6
> This looks like a good problem for itertools.groupby. My solution
> below needs error checking and testing, but this is where I'd start:
>
> data = [
>    (0, '3eA', 'Dupont', 'Juliette', 0, 11.0, 10.0),
>    (0, '3eA', 'Dupont', 'Juliette', 1, 4.0, 5.0),
>    (0, '3eA', 'Dupont', 'Juliette', 2, 17.5, 30.0),
>    (0, '3eA', 'Dupont', 'Juliette', 3, 3.0, 5.0),
>    (0, '3eA', 'Dupont', 'Juliette', 4, 4.5, 10.0),
>    (0, '3eA', 'Dupont', 'Juliette', 5, 35.5, 60.0),
>    (1, '3eA', 'Pop', 'Iggy', 0, 12.0, 10.0),
>    (1, '3eA', 'Pop', 'Iggy', 1, 3.5, 5.0),
>    (1, '3eA', 'Pop', 'Iggy', 2, 11.5, 30.0),
>    (1, '3eA', 'Pop', 'Iggy', 3, 4.0, 5.0),
>    (1, '3eA', 'Pop', 'Iggy', 4, 5.5, 10.0),
>    (1, '3eA', 'Pop', 'Iggy', 5, 40.5, 60.0),
> ]
>
> from operator import itemgetter
> from itertools import groupby
>
> #first sort by keyfunc, then group by it
> keyfunc = itemgetter(0,1,2,3)
> groups = groupby(sorted(data, key=keyfunc), keyfunc)
>
> result = []
> for group, records in groups:
>      temp = tuple('%s/%s' % r[5:] for r in sorted(records, key=itemgetter(4)))
>      result.append(group + temp)
>
>>>> result
> [(0, '3eA', 'Dupont', 'Juliette', '11.0/10.0', '4.0/5.0', '17.5/30.0',
> '3.0/5.0', '4.5/10.0', '35.5/60.0'), (1, '3eA', 'Pop', 'Iggy',
> '12.0/10.0', '3.5/5.0', '11.5/30.0', '4.0/5.0', '5.5/10.0',
> '40.5/60.0')]
>
Hmmmm... it happened again. I spend some time and effort to solve a 
problem, I feel like I'm almost a Real Python Programmer... and someone 
spoils my pride showing me some standard module whose name I barely 
remember, that can solve that same problem in a few lines...

Hey! That's a GREAT solution, Eryksun! Nothing wrong with you, really!

Every time this happens, I have to admit that I'm a newbie and I've 
still got a lot to learn about Python. Especially about its wonderful 
standard library. Better than Apple's App Store: for anything you can 
think, there's a Module. Problem is, I still can't readily recall which 
to use for a given problem.
My bad, now I'll RTFM again and I will study very carefully the operator 
and itertools modules. Who knows, maybe in a few decades I'll be able to 
say "This looks like a good problem for Module X" too.

Francesco