[Tutor] Flatten a list in tuples and remove doubles

eryksun eryksun at gmail.com
Sat Jul 28 20:41:21 CEST 2012


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')]


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