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Algo for dividing a number into (almost) equal whole numbers

I have a situation where I have invoice spreadsheets incoming with single rows that span multiple months with a quantity column containing the summation of the quantity for all months spanned.

In order to run month-by-month analytics, we need to split the total quantity into equal(ish) quantities across n rows where n is the number of months spanned.

These numbers can be off by one or two, but the smaller the difference between each element the better.

I have a rough mockup I did in python but I feel there's a better way to do this somehow. Note: Please excuse... everything:

from __future__ import division
import math
def evenDivide(num, div):
    splits = []
    sNum = str(num/div)
    remainder = float(sNum[sNum.index('.'):])
    #print "Remainder is " + str(remainder)
    integer = math.floor(num/div)
    #print "Integer is " + str(integer)
    totRemainder = round(remainder * div, 2)
    #print "Total Remainder is " + str(totRemainder)
    for index in range(div):
        if (totRemainder > 0):
            totRemainder -= 1 if (index%2 == 0) else 0
            if (index % 2 == 0):
                splits.append(int(integer + 1)) 
            else:
                splits.append(int(integer))
        else:
            splits.append(int(integer))
    for index in range(div):
        if(totRemainder > 0):
            if (index % 2 == 1):
                splits[index] += 1
                totRemainder -= 1

    return splits

def EvalSolution(splits):
    total = 0
    for index in range(len(splits)):
        total += splits[index]
    return total

def testEvenDivide():
    for index in range(20000):
        for jndex in range(3, 200):
            if (EvalSolution(evenDivide(index, jndex)) != index):
                print "Error for " + str(index) + ", " + str(jndex)
like image 390
Sed Avatar asked Dec 03 '13 10:12

Sed


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How do you divide a number into equal parts?

If the number is being split into exactly 'N' parts then every part will have the value X/N and the remaining X%N part can be distributed among any X%N numbers. Thus, if X % N == 0 then the minimum difference will always be '0' and the sequence will contain all equal numbers i.e. x/n.


1 Answers

If space is an issue, this one-liner may help:

num, div = 15, 4
print ([num // div + (1 if x < num % div else 0)  for x in range (div)])
# result: [4, 4, 4, 3]
like image 112
Poe Dator Avatar answered Oct 14 '22 04:10

Poe Dator