Как найти сумму значений словаря питон

asked Feb 2, 2011 at 23:01

As you’d expect:

sum(d.values())

answered Feb 2, 2011 at 23:02

In Python 2 you can avoid making a temporary copy of all the values by using the itervalues() dictionary method, which returns an iterator of the dictionary’s keys:

sum(d.itervalues())

In Python 3 you can just use d.values() because that method was changed to do that (and itervalues() was removed since it was no longer needed).

To make it easier to write version independent code which always iterates over the values of the dictionary’s keys, a utility function can be helpful:

import sys

def itervalues(d):
    return iter(getattr(d, ('itervalues', 'values')[sys.version_info[0]>2])())

sum(itervalues(d))

This is essentially what Benjamin Peterson’s six module does.

answered Feb 2, 2011 at 23:18

answered Feb 2, 2011 at 23:04

answered Mar 8, 2019 at 8:35

I feel sum(d.values()) is the most efficient way to get the sum.

You can also try the reduce function to calculate the sum along with a lambda expression:

reduce(lambda x,y:x+y,d.values())

phihag’s answer (and similar ones) won’t work in python3.

For python 3:

d = {'key1': 1,'key2': 14,'key3': 47}
sum(list(d.values()))

Update!
There are complains that it doesn’t work! I just attach a screenshot from my terminal. Could be some mismatch in versions etc.

answered Dec 13, 2019 at 10:17

You could consider ‘for loop’ for this:

  d = {'data': 100, 'data2': 200, 'data3': 500}
  total = 0
  for i in d.values():
        total += i

total = 800

answered Apr 9, 2020 at 3:51

nemonemo

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You can get a generator of all the values in the dictionary, then cast it to a list and use the sum() function to get the sum of all the values.

Example:

c={"a":123,"b":4,"d":4,"c":-1001,"x":2002,"y":1001}

sum(list(c.values()))

Mateusz JagiełłoMateusz Jagiełło

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

answered Jul 27, 2012 at 17:22

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.

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.

answered Jul 27, 2012 at 17:23

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']

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

 4330

answered Feb 25, 2022 at 12:07