Rounding an arrays of values to 100%

Question

Hi all,

I have an dictionary of lists of objects as shown below:

IDictionary<string, IList> MyItemDictionary

I work out the percentages by doing a for each through the dictionary with the code below:

IList<double> percentages = new List<double>();    

foreach(KeyValuePair<string, IList> pair in MyItemDictionary)
{
    double percentage = (100d/totalItemsCount)*pair.Value.Count;
    percentages.Add(percentage);
}

Basically I need to process the percentages list and round each percentage to a whole number BUT have them add up to 100. Accuracy is not of the highest importance but a sub 1 percentage i.e. 0.45 needs to be rounded to 1.

Does anyone know how to do this?

Thanks for your time and I look forward to your responses!

Adam

Answer 1

If your percentage list ends up containing 1.5 + 1.5 + .. + 1.5 = 100.0 and you were to round them into 2 + 2 + .. + 2, you would end up with 134 as the total sum (~67 entries), not 100. The way to fix this, is to distribute the error (134-100=34) among the existing percentages. In this case, you would subtract 1 from 34 of the percentages so you end up with a series of 1 + 2 + 1 + 2 + .. + 2 = 100.0.

To find what "every other" means you simply do int(numberOfPercentages / theError) and that should give you the interval.

Also, you must take care not to subtract anything from your sub 1 percentages.

Oh, and in case all your percentages are sub 1, the problem cannot be solved :-(

Answer 2

Can you have an "Other" or "Misc" category? If you have many small percentages, and you round them all up or down, you will accumulate a huge error as wic pointed out. You might need a minmimum threshold beyond which you lump everything together as miscellaneous stuff.

Otherwise, I second wic's idea about summing up the percentages to find the error and then distributing the error over the largest percentages.

Answer 3

Take a look at how seats in a parliament are assigned based on votes for proportional representation. I would suggest the largest remainder method, because it uses time proportional to the number of items in your list.