292
votes

Let's say I have a dictionary in which the keys map to integers like:

d = {'key1': 1,'key2': 14,'key3': 47}

Is there a syntactically minimalistic way to return the sum of the values in d—i.e. 62 in this case?

11
Just for fun: implement sum yourself in terms of reduce -- reduce is a more general form (e.g. sum, min and max can all be written in terms of reduce) and can solve other problems (e.g. product) easily.user166390
What about Guido's saying -- I think I remember this correctly -- that reduce is going away? I'm with you. Why remove it from the language?octopusgrabbus

11 Answers

577
votes

As you'd expect:

sum(d.values())
69
votes

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.

17
votes

Sure there is. Here is a way to sum the values of a dictionary.

>>> d = {'key1':1,'key2':14,'key3':47}
>>> sum(d.values())
62
6
votes
d = {'key1': 1,'key2': 14,'key3': 47}
sum1 = sum(d[item] for item in d)
print(sum1)

you can do it using the for loop

3
votes

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())
2
votes

sum(d.values()) - "d" -> Your dictionary Variable

1
votes

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.

enter image description here

0
votes

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

0
votes

USE sum() TO SUM THE VALUES IN A DICTIONARY.

Call dict.values() to return the values of a dictionary dict. Use sum(values) to return the sum of the values from the previous step.

d = {'key1':1,'key2':14,'key3':47}
values = d.values()
#Return values of a dictionary    
total = sum(values)
print(total)
0
votes

simplest/silliest solution:

https://trinket.io/python/a8a1f25353

d = {'key1': 1,'key2': 14,'key3': 47}
s = 0
for k in d:
    s += d[k]

print(s)

or if you want it fancier:

https://trinket.io/python/5fcd379536

import functools

d = {'key1': 1,'key2': 14,'key3': 47}
s = functools.reduce(lambda acc,k: acc+d[k], d, 0)

print(s)
-2
votes

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()))