Fastest way to rank items with multiple values and weightings

Question

I have a collection of key value pairs like this:

{ 
   \'key1\': [value1_1, value2_1, value3_1, ...], 
   \'key2\': [value1_2, value2_2, value3_2, ...],
   ...
         


        

Answer 1

You can't get linear-time, but you can do it faster; this looks like a matrix-multiply to me, so I suggest you use numpy:

import numpy as np

keys = ['key1', 'key2', 'key3']

values = np.matrix([
    [1.1, 1.2, 1.3, 1.4],
    [2.1, 2.2, 2.3, 2.4],
    [3.1, 3.2, 3.3, 3.4]
])

weights = np.matrix([[10., 20., 30., 40.]]).transpose()

res = (values * weights).transpose().tolist()[0]

items = zip(res, keys)
items.sort(reverse=True)

which gives

[(330.0, 'key3'), (230.0, 'key2'), (130.0, 'key1')]

Edit: with thanks to @Ondro for np.dot and to @unutbu for np.argsort, here is an improved version entirely in numpy:

import numpy as np

# set up values
keys = np.array(['key1', 'key2', 'key3'])
values = np.array([
    [1.1, 1.2, 1.3, 1.4],    # values1_x
    [2.1, 2.2, 2.3, 2.4],    # values2_x
    [3.1, 3.2, 3.3, 3.4]     # values3_x
])
weights = np.array([10., 20., 30., 40.])

# crunch the numbers
res = np.dot(values, -weights)   # negative of weights!

order = res.argsort(axis=0)  # sorting on negative value gives
                             # same order as reverse-sort; there does
                             # not seem to be any way to reverse-sort
                             # directly
sortedkeys = keys[order].tolist()

which results in ['key3', 'key2', 'key1'].

Answer 2

Here's a normalization function, that will linearly transform your values into [0,1]

def normalize(val, ilow, ihigh, olow, ohigh):
    return ((val-ilow) * (ohigh-olow) / (ihigh - ilow)) + olow

Now, use normalize to compute a new dictionary with normalized values. Then, sort by the weighted sum:

def sort(d, weights, ranges):
    # ranges is a list of tuples containing the lower and upper bounds of the corresponding value

    newD = {k:[normalize(v,ilow, ihigh, 0, 1) for v,(ilow, ihigh) in zip(vals, ranges)] for k,val in d.iteritems()}  # d.items() in python3
    return sorted(newD, key=lambda k: sum(v*w for v,w in zip(newD[k], weights)))