Log base 2 in Java for doubles

Question

I'm trying to calculate the entropy of English using the following Java function

public static void calculateEntropy()
    {
        for(int i = 0; i < letterFrequencies[i]; i++)
        {
            entropy += letterFrequencies[i] * (Math.log(letterFrequencies[i])/Math.log(2));
        }
        entropy *= -1;
    }

The formula I'm using requires log base 2 but Java only has natural log and log base 10. I'm trying to use the change of base formula to get the log base 2 of letterFrequencies[i]. I do not know if I am implementing it correctly because I'm expecting an answer close to 4.18 but instead getting roughly .028

@turbo OP's using this formula: http://en.wikipedia.org/wiki/Logarithm#Change_of_base — Luiggi Mendoza, Feb 09 '15 at 16:12
Probably not a dup but related to your case: http://stackoverflow.com/q/3305059/1065197 — Luiggi Mendoza, Feb 09 '15 at 16:12
Make sure all the other pieces actually work as expected. You could be trying to solve the wrong problem. I'd remove the log part to check if values are what is expected. — Voicu, Feb 09 '15 at 16:27
`i < letterFrequencies[i]` Is this correct? Isn't it `i < letterFrequencies.length`? — fps, Feb 09 '15 at 16:59
It's supposed to be letterFrequencies.length. That was my problem — Nick Gilbert, Feb 09 '15 at 17:20
You could also use the for-each loop that is available for all collection types in Java. Something like `for(double freq : letterFrequencies) {entropy +=freq*Math.log(freq); } entropy /= -Math.log(2);`. — Lutz Lehmann, Feb 09 '15 at 17:24

score 2 · Answer 1 · answered Feb 09 '15 at 17:42

2

The problem is in the for's stop condition:

i < letterFrequencies[i] should be i < letterFrequencies.length.

Furthermore, I'd use Guava's DoubleMath.log2() method, which is optimized as @LutzL suggested.

answered Feb 09 '15 at 17:42

fps

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score -3 · Answer 2 · edited May 23 '17 at 12:20

-3

Mathematically that implementation is correct, but it doesn't work out in code. You could instead write your own implementation, witch runs significantly faster:

public static int log2(int n){
    if(n <= 0) throw new IllegalArgumentException();
    return 31 - Integer.numberOfLeadingZeros(n);
}

Source: How do you calculate log base 2 in Java for integers?

edited May 23 '17 at 12:20

Community

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answered Feb 09 '15 at 16:12

flaghacker

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*Yes, that is a correct implementation of log2* mathematically it's correct, but programatically it isn't. – Luiggi Mendoza Feb 09 '15 at 16:13
Yes. That's also explained in the question from where you obtained the piece of code you posted. – Luiggi Mendoza Feb 09 '15 at 16:16
Is it ok to edit my answer like this? (I'm new to stackoverflow) – flaghacker Feb 09 '15 at 16:18
return value gives you Math.floor(Math.log(n)/Math.log(2)) – Yossarian42 Jan 24 '21 at 06:40

Log base 2 in Java for doubles

2 Answers2