How to use dictionary values to calculate sum of weighted values without explicitly defined weights?

Question

As an example I have a dictionary as follows:

mydict = {A:['asdasd',2,3], B:['asdasd',4,5], C:['rgerg',9,10]}

How can I get just one number that is essentially the sum of the weighted values of all the keys in the dict (this example: A, B, C ) weighted by the last number in the list ( list[-1] )?

So for instance, this would be the sum of:

(2/15)*3 + (4/15)*5 + (9/15)*10 = 7.73

Where 15 is the sum of 2,4,9.

At the moment I'm creating too many variables that just iterate through the dictionary. I'm sure there's a quick efficient Python way to do it.

Answer 1

Using the fact that (2/15)*3 + (4/15)*5 + (9/15)*10 is just the same as (2*3 + 4*5 + 9*10)/15 , you don't have to pre-calculate the total, so you could do this in one pass of the dict with reduce , but perhaps it isn't as readable.

Turning the dict into a series of tuples:

>>> d = {A:['asdasd',2,3], B:['asdasd',4,5], C:['rgerg',9,10]}
>>> from functools import reduce    # Py3
>>> x,y = reduce(lambda t, e: (t[0]+e[0], t[1]+e[1]), ((a*b, a) for _,a,b in d.values()))
>>> x/y   # float(y) in Py2
7.733333333333333

Or without the intermediate generator and using an initial value (probably the fastest):

>>> x,y = reduce(lambda t, e: (t[0]+e[1]*e[2], t[1]+e[1]), d.values(), (0,0))
>>> x/y   # float(y) in Py2
7.733333333333333

Or you could zip up the results and sum:

>>> x,y = (sum(x) for x in zip(*((a*b, a) for _,a,b in d.values())))
>>> x/y   # float(y) in Py2
7.733333333333333

Or you could do what @ranlot suggests though I would avoid the intermediate lists:

>>> sum(a*b for _,a,b in d.values())/sum(a for _,a,b in d.values())
7.733333333333333

Though this seems to be fractionally slower than either of the ones above.

Answer 2

myDict = {'A':['asdasd', 2, 3], 'B':['asdasd', 4, 5], 'C':['rgerg', 9, 10]}
normConst = sum([dict[x][-2] for x in myDict])
sum([dict[x][-1]*dict[x][-2] / float(normConst) for x in myDict])