Question

I have a string like: "2E6 3.34e-5 3 4.6" and I want to use replaceAll to replace tokens like:

"((\\-)?[0-9]+(\\.([0-9])+)?)(E|e)((\\-)?[0-9]+(\\.([0-9])+)?)"

(i.e. two numbers with e or E between them) into the equivalent normal number format (i.e. replace "2E6" with "2000000" and "3.34e-5" with "0.0000334")

I wrote:

value.replaceAll("((\\-)?[0-9]+(\\.([0-9])+)?)(E|e)((\\-)?[0-9]+(\\.([0-9])+)?)", "($1)*10^($6)");

but I would like to actually multiply the 1st argument by 10 to the power of the 2nd argument, not just writing it that way .. Any ideas?

UPDATE

I did the following based on your suggesions:

Pattern p = Pattern.compile("((\\-)?[0-9]+(\\.([0-9])+)?)(E|e)((\\-)?[0-9]+(\\.([0-9])+)?)");
Matcher m = p.matcher("2E6 3.34e-5 3 4.6");
StringBuffer sb = new StringBuffer();
while (m.find()) {
    m.appendReplacement(sb, "WHAT HERE??"); // What HERE ??
}
m.appendTail(sb);
System.out.println(sb.toString());

UPDATE

Finally, this is what I reached:

// 32 #'s because this is the highest precision I need in my application
private static NumberFormat formatter = new DecimalFormat("#.################################");

private static String fix(String values) {
    String[] values_array = values.split(" ");
    StringBuilder result = new StringBuilder();
    for(String value:values_array){
        try{
            result.append(formatter.format(new Double(value))).append(" ");
        }catch(NumberFormatException e){ //If not a valid double, copy it as is
            result.append(value).append(" ");
        }
    }
    return result.toString().substring(0, result.toString().length()-1);
}

Answer 1

If you need to convert scientific number notation into normal formal you can use DecimalFormat

public static void main(String[] args) {
    NumberFormat formatter = new DecimalFormat();

    double num1 = 2E6;
    formatter = new DecimalFormat("##########");
    System.out.println(formatter.format(num1)); 

    double num2 = 3.3e-5;
    formatter = new DecimalFormat("#.##########");
    System.out.println(formatter.format(num2));
}

Just add the logic to split the intial string over spaces and apply the above logic.

You can check more about the symbols like #(in this case) at the javadoc for DecimalFormat.

Symbol Location     Localized?  Meaning   
------------------------------------------------------------
#      Number       Yes         Digit, zero shows as absent