Python - sum values in dictionary

Question

I have got pretty simple list:

example_list = [
    {'points': 400, 'gold': 2480},
    {'points': 100, 'gold': 610},
    {'points': 100, 'gold': 620},
    {'points': 100, 'gold': 620}
]

How can I sum all gold values? I'm looking for nice oneliner.

Now I'm using this code (but it's not the best solution):

total_gold = 0
for item in example_list:
    total_gold += example_list["gold"]

You shouldn't use `list` as the name of a local variable - doing so shadows the built-in `list` type and can cause problems. — g.d.d.c, Jul 27 '12 at 17:23
@sigo -- In this case, a nice one-liner exists, but generally, restricting answers to "nice one-liners" is probably a bad idea (as nicer multi-liners might exist). — mgilson, Jul 27 '12 at 17:28

g.d.d.c · Accepted Answer · 2020-11-06T14:12:57.320

221

sum(item['gold'] for item in myList)

edited Nov 06 '20 at 14:12

answered Jul 27 '12 at 17:22

g.d.d.c

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score 30 · Answer 2 · edited May 23 '17 at 12:26

30

If you're memory conscious:

sum(item['gold'] for item in example_list)

If you're extremely time conscious:

sum([item['gold'] for item in example_list])

In most cases just use the generator expression, as the performance increase is only noticeable on a very large dataset/very hot code path.

See this answer for an explanation of why you should avoid using map.

See this answer for some real-world timing comparisons of list comprehension vs generator expressions.

edited May 23 '17 at 12:26

Community

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answered Jul 27 '12 at 19:16

Ben Burns

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why does a similar `sum(item.gold for item in example_list)` not work? I get a `AttributeError: 'dict' object has no attribute 'gold'` – cryanbhu Nov 06 '20 at 12:08
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Because the dict type has no attribute (type member) named 'gold'. You're probably confusing Python for JavaScript. In JS, you can use either dot notation or index notation, as the dict-like object there is generally just a plain old JS object. In Python dicts are dicts, and you must use the index operator, so, long answer long: you want `item['gold']` instead of `item.gold`. – Ben Burns Nov 08 '20 at 04:27

mgilson · Answer 3 · 2012-07-27T17:33:55.300

8

If you prefer map, this works too:

 import operator
 total_gold = sum(map(operator.itemgetter('gold'),example_list))

But I think the generator posted by g.d.d.c is significantly better. This answer is really just to point out the existence of operator.itemgetter.

edited Jul 27 '12 at 17:33

answered Jul 27 '12 at 17:23

mgilson

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`_reduce = lambda fn, ar: fn( ar[0], _reduce(fn, ar[1:])) if len(ar) > 1 else ar[0]` `total_gold = _reduce((lambda x, y: x + y), (lambda fn, ar: [fn(ar[i]) for i in xrange(len(ar))])( (lambda x: x["gold"]), example_list))` I also think g.d.d.c's answer is good. This answer is really just to point out the existence of lambda :P – Moop Jul 27 '12 at 18:07
1

thanks -- comment formatting doesn't seem to respect newlines though, so it looks even worse than it should. – Moop Jul 27 '12 at 18:11
@Filipq -- Really, I posted the same answer as g.d.d.c did ~30 seconds after he did. I originally deleted it, but ever since I got the stupid moderator tools, I can still see my deleted answer. It was bugging me, so I thought to myself, there has to be another somewhat clean way to do this that illustrates something useful ... I probably should just delete it though ... I don't know. (I'm open to suggestions). – mgilson Jul 27 '12 at 18:14

score 5 · Answer 4 · edited Feb 25 '22 at 11:51

5

from collections import Counter
from functools import reduce
from operator import add

sum_dict = reduce(add, (map(Counter, example_list)))
# Counter({'points': 700, 'gold': 4330})
total_gold = sum_dict['gold']

edited Feb 25 '22 at 11:51

Bruno Mello

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answered Feb 24 '22 at 00:58

xmduhan

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nice, this allows for counting arbitrary items in the dict without having to know all keys (e.g. this would support counting all keys in the dict at once without specifying them all). – mdmjsh May 31 '22 at 10:55

score 0 · Answer 5 · answered Feb 25 '22 at 12:07

0

example_list = [
    {'points': 400, 'gold': 2480},
    {'points': 100, 'gold': 610},
    {'points': 100, 'gold': 620},
    {'points': 100, 'gold': 620}
]

result = np.sum([x['gold'] for x in example_list])


print(result)

output

answered Feb 25 '22 at 12:07

Golden Lion

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Is that different than using the non `np` sum? – General Grievance Feb 28 '22 at 02:54

Python - sum values in dictionary

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