how to group by function if one of the group has relationship with another one in the group?

[Ho Yeung Lee]

actually i used in this application
if same color is neighbor like connected then group them

i use for segmentation of words in screen capture

https://stackoverflow.com/questions/45294829/how-to-group-by-function-if-any-one-of-the-group-members-has-neighbor-relationsh

i asked here too, but i do not know how to use partial and do not know what center is.


On Tuesday, July 25, 2017 at 5:00:25 PM UTC+8, Peter Otten wrote:
> Ho Yeung Lee wrote:
> 
> > from itertools import groupby
> > 
> > testing1 = [(1,1),(2,3),(2,4),(3,5),(3,6),(4,6)]
> > def isneighborlocation(lo1, lo2):
> >     if abs(lo1[0] - lo2[0]) == 1  or lo1[1] == lo2[1]:
> >         return 1
> >     elif abs(lo1[1] - lo2[1]) == 1  or lo1[0] == lo2[0]:
> >         return 1
> >     else:
> >         return 0
> > 
> > groupda = groupby(testing1, isneighborlocation)
> > for key, group1 in groupda:
> >     print key
> >     for thing in group1:
> >         print thing
> > 
> > expect output 3 group
> > group1 [(1,1)]
> > group2 [(2,3),(2,4]
> > group3 [(3,5),(3,6),(4,6)]
> 
> groupby() calculates the key value from the current item only, so there's no 
> "natural" way to apply it to your problem.
> 
> Possible workarounds are to feed it pairs of neighbouring items (think 
> zip()) or a stateful key function. Below is an example of the latter:
> 
> $ cat sequential_group_class.py
> from itertools import groupby
> 
> missing = object()
> 
> class PairKey:
>     def __init__(self, continued):
>         self.prev = missing
>         self.continued = continued
>         self.key = False
> 
>     def __call__(self, item):
>         if self.prev is not missing and not self.continued(self.prev, item):
>             self.key = not self.key
>         self.prev = item
>         return self.key
> 
> def isneighborlocation(lo1, lo2):
>     x1, y1 = lo1
>     x2, y2 = lo2
>     dx = x1 - x2
>     dy = y1 - y2
>     return dx*dx + dy*dy <= 1
> 
> items = [(1,1),(2,3),(2,4),(3,5),(3,6),(4,6)]
> 
> for key, group in groupby(items, key=PairKey(isneighborlocation)):
>     print key, list(group)
> 
> $ python sequential_group_class.py 
> False [(1, 1)]
> True [(2, 3), (2, 4)]
> False [(3, 5), (3, 6), (4, 6)]