[Tutor] overlapping tuples
Narasimharao Nelluri
narasimha928 at gmail.com
Fri Feb 28 15:06:49 EST 2020
Hi peter,
Thanks a lot for your inputs, i will work your inputs and let you know. For
the last question if there is independent over lap for ex: in below case
python should return [(5,7),(6,8)]
tuple_overlap([(0, 2), (1, 3), (5, 7), (6, 8)])
I tried different combinations as of i didn't get perfect solution.
Thanks
Narasimha.
On Fri, Feb 28, 2020 at 11:40 AM Peter Otten <__peter__ at web.de> wrote:
> Narasimharao Nelluri wrote:
>
> > Hi David ,
> > Thanks for your feedback. Yes here i am solving Home work from one of my
> > assignments.
> > please check below i added your comments.
> >
> > NNELLURI-M-L13P:google_class nnelluri$ cat overlap.py
> > #!/usr/bin/env python3
> >
> >
> > def tuple_overlap(old_list,overlap = None):
> >
> > old_list.sort()
> > if overlap is None:
> > overlap = []
> > for fir,sec in zip(old_list,old_list[1:]):
> >
> > if fir[1] >= sec[0] and fir[1] <=sec[1]:=======>Here i am validating
> > 2nd element in first variablae is in-between seconds variable if it is
> > there is a overlap
> > overlap.append(fir)
> >
> > if sec[0] >= fir[0] and sec[1] <= fir[1]:=====> Here i am checking if
> > first element in second variable is in-between second variable , there is
> > a oberlap
> > overlap.append(sec)
> >
> > overlap = sorted(overlap)
> > return overlap
> >
> >
> > overlaps = tuple_overlap( [(1,10),(15,20),(101,110)] )
> > print( "Overlap 1 =", overlaps )
> > assert overlaps == []
> >
> > overlaps = tuple_overlap( [(1,20),(15,20),(101,110)])
> > print( "Overlap 2 =", overlaps )
> > assert overlaps ==[(1, 20), (15, 20)]
> >
> > overlaps = tuple_overlap( [(1,10),(15,20),(1,10),(1,10),(101,110)])
> > print( "Overlap 3 =", overlaps )
> > assert overlaps == [(1, 10), (1, 10),(1,10)]
> >
> >
> > NNELLURI-M-L13P:google_class nnelluri$
> >
> >
> > Please let know if you know correct solution.
>
> I would have to work it out myself ;) but I think I can give you a few
> hints:
>
> If you add some debugging information
>
> >>> tuple_overlap( [(1,10, "a"),(15,20, "b"),(1,10, "c"),(1,10, "d"),
> (101,110, "e")])
> [(1, 10, 'a'), (1, 10, 'c'), (1, 10, 'c'), (1, 10, 'd')]
>
> you can see that (1, 10, "c") occurs twice. Your code checks adjacent
> tuples, and if three tuples overlap the tuple in the middle may be added
> twice. In this situation you need to ensure that every tuple is added only
> once.
>
> But not only that:
>
> >>> tuple_overlap([(0, 10), (1, 3)])
> [(1, 3)]
>
> Hm, one overlapping tuple? with what? with itself?
>
> If tuple A overlaps with tuple B, tuple B overlaps with A, i. e. instead
> of
> two independent checks you can use a single one, and add both tuples if
> there is an overlap.
>
> That single check may be easier to implement if you ask yourself the
> opposite question: when do the tuples not overlap?
>
> Also consider [(0, 10), (1, 3), (5, 20)].
>
> Clearly you need to decide what to do if t1 overlaps with t2 and t3, but
> t2
> not with t2.
>
> Finally, does your task say something about independent overlaps, e. g.
> what
> should be returned by
>
> tuple_overlap([(0, 2), (1, 3), (5, 7), (6, 8)])
>
> ?
>
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