On Sat, 21 May 2016 03:19:49 +, Dan Sommers wrote:
>> Is there something shorter and sweeter for the summation?
>
> from itertools import groupby
> from operator import itemgetter
>
> result = [(k,
>sum(map(itemgetter(2), v)),
>sum(map(itemgetter(3
On Wed, 18 May 2016 20:59:55 -0400, DFS wrote:
> Have aList = [
> ('x','Name1', 1, 85),
> ('x','Name2', 3, 219),
> ('x','Name2', 1, 21),
> ('x','Name3', 6, 169)
> ]
>
> want
>
> aList = [
> ('Name1', 1, 85),
> ('Name2', 4, 240),
> ('Name3', 6, 169)
> ]
[snip]
> Is there something shorter and
On 05/18/2016 09:53 PM, DFS wrote:
On 5/18/2016 10:58 PM, Larry Hudson wrote:
[snip...]
Why two loops? Put both summations in a single loop. Then you're only
scanning the alist once instead of twice.
groups1 = defaultdict(int)
groups2 = defaultdict(int)
for nm, matches, words in alist:
g
On 05/18/2016 05:59 PM, DFS wrote:
Have aList = [
('x','Name1', 1, 85),
('x','Name2', 3, 219),
('x','Name2', 1, 21),
('x','Name3', 6, 169)
]
want
aList = [
('Name1', 1, 85),
('Name2', 4, 240),
('Name3', 6, 169)
]
This drops the first element in each tuple:
alist = [(b,c,d) for a,b,c,d in ali
