ansaurus

Question

Answer 1

A:

sum(i['count'] for i in d.values())

Ken 2009-12-17 02:10:25

Answer 2

+4 A:

>>> x = {0: {'count': 1000}, 1: {'count': 2000}}
>>> sum(v['count'] for v in x.values()) 
3000

Ned Deily 2009-12-17 02:10:33

Use .itervalues() to save some memory overhead, since .values() builds a new list for the values.

Matt Good 2009-12-17 07:59:26

Answer 3

+4 A:

A shorter one:

sum(d[k]['count'] for k in d)

dF 2009-12-17 02:18:59

It saves a few characters, but it's slower since you have to look up each key, instead of iterating over the values directly.

Matt Good 2009-12-17 07:58:24

Well technically, for the example it's 5% faster :) But you're right, for large dicts it's about 15% slower.

dF 2009-12-17 23:05:50

Answer 4

A:

How about using reduction in python?

reduce(lambda x,y: x+y, [v['count'] for v in a.values()])

Raymond Tay 2009-12-17 03:03:06

how to add all of these values in a python dictionary