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 alist]
Slicing may be more efficient (perhaps, perhaps not -- don't know). And probably effectively no
difference, but slightly shorter to write.
alist = [t[1:] for t in alist]
And this summation creates 2 lists that then have to be combined:
groups1 = defaultdict(int)
for nm, matches, words in alist:
groups1[nm] += matches
groups2 = defaultdict(int)
for nm, matches, words in alist:
groups2[nm] += words
Is there something shorter and sweeter for the summation?
Thanks
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:
groups1[nm] += matches
groups2[nm] += words
-=- Larry -=-
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