Allocate an array of integers proportionally compensating for rounding errors

Question

I have an array of non-negative values. I want to build an array of values who\'s sum is 20 so that they are proportional to the first array.

This would be an easy probl

Answer 1

Your problem is similar to a proportional representation where you want to share N seats (in your case 20) among parties proportionnaly to the votes they obtain, in your case [3, 3, 3, 3, 3, 3, 18]

There are several methods used in different countries to handle the rounding problem. My code below uses the Hagenbach-Bischoff quota method used in Switzerland, which basically allocates the seats remaining after an integer division by (N+1) to parties which have the highest remainder:

def proportional(nseats,votes):
    """assign n seats proportionaly to votes using Hagenbach-Bischoff quota
    :param nseats: int number of seats to assign
    :param votes: iterable of int or float weighting each party
    :result: list of ints seats allocated to each party
    """
    quota=sum(votes)/(1.+nseats) #force float
    frac=[vote/quota for vote in votes]
    res=[int(f) for f in frac]
    n=nseats-sum(res) #number of seats remaining to allocate
    if n==0: return res #done
    if n<0: return [min(x,nseats) for x in res] # see siamii's comment
    #give the remaining seats to the n parties with the largest remainder
    remainders=[ai-bi for ai,bi in zip(frac,res)]
    limit=sorted(remainders,reverse=True)[n-1]
    #n parties with remainter larger than limit get an extra seat
    for i,r in enumerate(remainders):
        if r>=limit:
            res[i]+=1
            n-=1 # attempt to handle perfect equality
            if n==0: return res #done
    raise #should never happen

However this method doesn't always give the same number of seats to parties with perfect equality as in your case:

proportional(20,[3, 3, 3, 3, 3, 3, 18])
[2,2,2,2,1,1,10]