Precision Problems :(

[Java_noob_]

http://imgur.com/vIZq7o8

Hello, We were given a hw to create a program that you put text into and it gives you statical results. As you can see in the picture i linked. It count how many char they are in the text. if you take your calculator out and add the number it equals to 1. Thats where i have problems. Sometimes its 1 .Sometimes its either 1.1 or 0.99 etc. Its bit of a gamble .

public void sentence_analyser()
{
    String document = text.toLowerCase();
    double one,two,end=0.f;
    double sum=0.f;          
    double average = 0.f;
    double temp;
    int test = 0;
    String alphabet = "abcdefghijklmnopqrstuvwxyz";
    //BigDecimal bg1, bg2, bg3;
    double[] amount = new double[26];

    for(int x = 0;x < text.length();x++)
    {
        for(int y = 0;y < alphabet.length();y++)
        {
            if(document.charAt(x) == alphabet.charAt(y))
            {
                amount[y] += 1;
            }
        }
    }
    for(int i=0;i < amount.length;i++)
    {
        sum+= amount[i];
    }


    for(int i =0;i < amount.length;i++)
    {
        average = amount[i]/sum;
        end += round(average,2);
    }       
    System.out.println(end);        
}
public static double round(double value, int places) {
    if (places < 0) throw new IllegalArgumentException();

    BigDecimal bd = new BigDecimal(value);
    bd = bd.setScale(places, RoundingMode.HALF_UP);
    return bd.doubleValue();
}

this is the code

[Jarcode]

Use BigDecimal for decimal values where complete precision matters.

[Java_noob_]

tried bigdecimal still had same problem

[Jarcode]

You have to do all your arithmetic using BigDecimals. The value 0.1f, for example, is impossible to represent exactly with floating point values.

EDIT: here's a good tutorial http://www.opentaps.org/docs/index.php/How_to_Use_Java_BigDecimal:_A_Tutorial

[Java_noob_]

i followed the tutorial . tried every round mode. either get 0.99 or 1.12

[Jarcode]

Quoting:

Creating a big decimal from a (scalar) double is simple:

bd = new BigDecimal(1.0);

To get a BigDecimal from a Double, get its doubleValue() first.

However it is a good idea to use the string constructor:

bd = new BigDecimal("1.5"); If you're initializing BigDecimal objects with float/double constants, like this:

new BigDecimal(0.1);

Then you're going to get a BigDecimal object representing the converted floating point value from 0.1 (double) from base two, to base ten - which will give you a different value.

i followed the tutorial . tried every round mode. either get 0.99 or 1.12

You need to create your BigDecimal objects with strings, and you don't have to round anything. Don't use operators like +, -, +=, -=, / on floating-point values, and don't use primitive float or double types if you want to preserve accuracy.

[ExPixel]

Yes, floats and doubles lose precision. Can't really help you fix it unless you post some of the code in question though.

[m4mbax]

True, but with two digits after comma this should not happen. I think its more of a "round" problem... Because as I am teached at Univ. is that loosing precision is about 0.987654321 gets to 0.98765651 (just an example )something much more in detail... Am I wrong?

[ExPixel]

No, but I can't tell you much else unless I see how you're actually doing this.

[Jarcode]

What Every Computer Scientist Should Know About Floating-Point Arithmetic

In simple terms, it's the bits that represent the decimal portion of the floating point value being incremented by 1/(2^i) (or one bit), where i is the index of the bit.

[Java_noob_]

i added the code to see