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c#javaarraysprobability
David James Ball
David James Ball

Reputation: 913

'Normalizing' Probability Array

I have an array of doubles representing probabilities of certain events happening, [25,25,25,10,15] for event A,B..E. The numbers add up to 100.

Through my analysis I want to be able to cross off the possibility of a certain event happening or not.

So, If I find that event E is impossible, then I set that index to 0. How do I re-normalize the array so that the total adds up to 100, and the relative probability of each event is maintained?

I will use C# or Java.

Upvotes: 0

Views: 1214

Answers (4)

Peter Lawrey
Peter Lawrey

Reputation: 533530

A more general solution is to normalise an array to give it a specific total.

public static void setTotal(double total, double[] values) {
    double sum = 0;
    for (double value : values) sum += value;
    double factor = total / sum;
    for (int i = 0; i < values.length; i++)
        values[i] *= factor;
}

This allows you to alter any value and ensure the total is whatever you want (provided the sum is not zero)

Upvotes: 0

ddmps
ddmps

Reputation: 4380

I'm assuming the probabilities will stay the same (i.e the first double's new probably should be 25/85 in your example). Then what you need to do is save the double you've removed, subtract 100 by it and divide each double by it, and multiply by 100.

Java code:

double[] doubleArray = new double[] { 25, 25 ,25, 10, 15 }; //or any other array
double temp = 100-doubleArray[4]; //or any other specific value in the array
doubleArray[4] = 0;
for (int i; i<doubleArray.length; i++) {
    doubleArray[i] = doubleArray[i]/temp*100;
}

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726639

You can re-distribute the probability of the removed event in many ways, for example

  • Proportionally - Replace each number with the value p/sum(p), where sum(p) is the sum of remaining probabilities.
  • Equally - Divide the removed value by the number of remaining items, and add the result to each of the remaining probability.

The "correct" answer depends on the specifics of your problem.

Upvotes: 2

Itay Karo
Itay Karo

Reputation: 18286

sum = 100 - P(E)
for each event e:
  if e is E:
    P(E) = 0
  else:
    P(e) = P(e)*100/sum

Upvotes: 0

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