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

[Piet van Oostrum]

Peter Otten <__peter__ at web.de> writes:

> 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)]

That only works if
(a) The elements in the list are already clustered on group (i.e. all
elements of a group are adjacent)
(b) In a group the order is such that adjacent elements are direct
neigbours, i.e. their distance is at most 1.

So 'groupby' is not a natural solution for this problem.
-- 
Piet van Oostrum <piet-l at vanoostrum.org>
WWW: http://piet.vanoostrum.org/
PGP key: [8DAE142BE17999C4]