I have big problem with reading big numbers from excel in java. When i read 71674705 i get 7.1674705E which is not ok.
Example:
double num =
cell.getNumericCellValue();
how can i prevent conversion between numbers that number will stay like 71674705.
The number itself is not changing, only the representation when you convert it to a String. A double variable like you are using does not have an explicit format defined.
You can use java.text.NumberFormat (javadoc) to format the number any way you would like to see it.
num is just a double, i.e. a binary floating-point number. That formatting (exponential notation) is only an issue for printing. So if you're just debugging, you shouldn't have to worry. For production output, you can use a NumberFormat such as DecimalFormat.
If you are sure you are dealing with an integer and you want it as an integer value you can always try to do:
Integer.valueOf(num)
Edit: not valid for big numbers of course :-)
Related
I am suprise to see the below program please advise how it is behaving like this as i aam very much concerened with poitn to precison after decimal point , below is the progrmam
double fixedRate = 0.997500000000; //**output --->0.9975
// BigDecimal fixedRate = new BigDecimal("0.997500000000");
double fixedRate1 = 0.1234567890123456789;
System.out.println(fixedRate);
System.out.println(fixedRate1);
and the output is
0.9975
0.12345678901234568
now please advise for the first the ouput is 0.9975 but late on for next it is not truncating after decimal points but why for first then.
The precision is not lost. It is just not printed because you do not need more digits to distinguish the printed value from any other double value.
If you do want to force a certain number of fractional digits, take a look at System.out.printf() or String.format().
See https://docs.oracle.com/javase/7/docs/api/java/util/Formatter.html for more details and possible formats.
The result may look like this:
System.out.printf("%.19f%n",fixedRate);
System.out.printf("%.19f%n", fixedRate1);
This is a printing problem.Please use this format to print your values and tell me the result:
double fixedRate = 0.997500000000;
double fixedRate1 = 0.1234567890123456789;
System.out.println(String.format("%.19f", fixedRate ));
System.out.println(String.format("%.19f", fixedRate1 ));
Good Luck !
According to your question
for the first the ouput is 0.9975 but late on for next it is not
truncating after decimal points but why for first then
Since double is numeric datatype and hence cannot hold the leading and trailing zeros.
A double doesn't care about formatting - it's about storage only. When you print it, it is converted to a String (using Double's static toString method).
In simple terms the value 0.9975 is not different from 0.997500000000or it is same as 0.997500000000 as zeros after a number will not have any value.
But consider if you had value like this 0.9975000000001 then all the numbers will be printed. Check it here.
If you want to format the value then you can see this question : How to properly display a price up to two decimals (cents) including trailing zeros in Java?
This might seem really naive but I am not sure what is the decimal representation of an integer?
Can you provide some code in Java to convert an integer to its decimal representation?
Thanks!
My answer would have been Integer.toString(value).
Done! I have now "calculated" the decimal representation of an integer (int value).
As the javadoc says:
The argument is converted to signed decimal representation and returned as a string.
Of course, they probably wanted you to write the code that implements that method, rather than using the library method.
You can use things like this :
Integer i = 10;
Double iDecimal = new Double(i);
System.out.println(iDecimal); //would print 10.0 i.e. decimal
How do I parse ANY Number type from a String in Java?
I'm aware of the methods like Integer.parseInt(myString) and Double.parseDouble(myString) but what I optimally want is a method like Number.parseNumber(myString) which doesn't exist. How can I achieve that behaviour in another way? (I want the parsed Number to reflect the String "exactly" in terms of for example number of decimals).
Example:
"0" => Number (internally of Integer subclass)
"0.0" => Number (internally of Double subclass)
Also, no ugliness like checking for decimal separators etc.
Number number = NumberFormat.getInstance().parse(myString);
Seems to do the trick...
Better way is to use BigDecimal class.
BigDecimal n = new BigDecimal("1.0");
Methods for needed values
n.byteValue();
n.intValue();
n.shortValue();
n.longValue();
n.floatValue();
n.doubleValue();
You are probably looking for BigDecimal.
Please remember that a number can be represented in many forms as a string. e.g. 1 is the same number as 1.0000000.
I have an api which takes a number as a String input and i need to get the Float value of the number. I currently use the Float.ParseFloat method to get the float value of my String number.
According the java documentation of Float.ParseFloat, it doesn't mention anything about the input being greater than the Float.MAX_VALUE.
One of the ways I was thinking of doing this was by checking the length of the input String is greater than the length of the Float.MAX_VALUE.
Pls suggest how I can go about handling this.
Although the javadoc doesn't make it clear, when I tested it, parseFloat of a String too large simply produced a Float of 'infinity'. You could use the isInfinite() method after creation to check the value.
Using something like BigDecimal would probably be a safer option here, especially if you'll be performing any arithmetic on your value.
You can use greater precision. Try double or BigDecimal. There are also arbitrary precision libraries which are open.
Here you can find how much each IEEE 754 format can hold: http://en.wikipedia.org/wiki/IEEE_754-2008 . Float would be near 1.234567*10^38
If you can't parse it properly (e.g. if there are too many significant digits or the exponent is too big: 1.23456789012345e5000) you won't be able either to hold it in a single precision float.
If the number is too big the result is set to Float.POSITIVE_INFINITY, as the rules of IEEE FP arithmetic require, and as a 10-second test shows.
The exponent clips to the maximum exponent value. See the source, line 1197.
Perhaps check for some maximum useful value for your application?
I'm having trouble with (what I suspect is) a rounding error.
I have a string, 0.686357E-01, which I'm trying to convert to a double. I've been able to split it up using the Pattern.split() function, and I'm capturing the base and the exponent values just fine. However, once I try to multiply them appropriately, I get this as a result: 0.06863570000000001.
Here's my relevant code:
pattern = Pattern.compile("E\\+?");
String[] number = pattern.split(string);
double base = Double.parseDouble(number[0]);
int exponent = Integer.parseInt(number[1]);
number= base*Math.pow(10, exponent);
So, how do I avoid the rounding error? (There are ways that I can work around it, but if it's possible to do, then I'd like to know how to fix the issue)
Thanks.
You don't need to split it, Double.parseDouble can handle those kinds of numbers just fine.
double number = Double.parseDouble("0.686357E-01");
See? It works!
0.0686357 is not exactly representable as a double-precision value.
Two solutions:
Use e.g. BigDecimal.
Limit the displayed precision to a certain number of significant figures when converting back to human-readable.
Floating point numbers do not have perfect precision. If that is an issue, use BigDecimal:
String string = "0.686357E-01";
BigDecimal number = new BigDecimal(string);
System.out.println(number);
Double will print always like that, but the value will remain correct. You'll need to format the output to get the correct value. See DecimalFormat class.