I have this data:
self.data = [(1, 1, 5.0),
(1, 2, 3.0),
(1, 3, 4.0),
(2, 1, 4.0),
(2, 2, 2.0)]
>
Below "fixes" several annoyances with Python's itertools.groupby
.
def groupby2(l, key=lambda x:x, val=lambda x:x, agg=lambda x:x, sort=True):
if sort:
l = sorted(l, key=key)
return ((k, agg((val(x) for x in v))) \
for k,v in itertools.groupby(l, key=key))
Specifically,
key
as named parameter only.tuple(key, grouped_values)
where values are specified by 3rd parameter.Example Usage
import itertools
from operator import itemgetter
from statistics import *
t = [('a',1), ('b',2), ('a',3)]
for k,v in groupby2(t, itemgetter(0), itemgetter(1), sum):
print(k, v)
This prints,
a 4
b 2
Play with this code