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Closed 10 years ago.
Possible Duplicate:
Round a double to 2 significant figures after decimal point
i've got lat and long point like this,
x1: 11.955165229802363
y1: 79.8232913017273
i need to convert 4 decimal point
x1 = 11.9552
y1 = 79.8233
Try
double roundTwoDecimals(double d)
{
DecimalFormat twoDForm = new DecimalFormat("#.####");
return Double.valueOf(twoDForm.format(d));
}
Math.ceil(x1* 10000) / 10000
Replace 10000 with 10^N, where N is number of digits after dot. In case of 4 digits after dot, precision shouldn't be lost.
Try this
String.format("%.4f", 11.955165229802363)
Assuming you want to round/truncate the decimal, and speed is not a large consideration, you want to use BigDecimal(BigInteger unscaledVal, int scale) with scale set to 4.
DecimalFormat dtime = new DecimalFormat("#.####");
^^^^
x1= Double.valueOf(dtime.format(x1));
float round = Round(num,4);
System.out.println("Rounded data: " + round);
}
public float Round(float Rval, int Rpl) {
float p = (float)Math.pow(10,Rpl);
Rval = Rval * p;
float tmp = Math.round(Rval);
return (float)tmp/p;
}
If you only want to display the value like that, use a DecimalFormat to convert the value to a string, then display that.
If you really want to round it to four digits, you can achieve that by multiplying by 10000, rounding, then dividing again. However, I would advise against it, since not all decimal numbers can be properly represented in floating point format. Changes are you'll just get something like you already had.
If you really want four digits to work with as an internal state, use a BigDecimal instead. It is properly equipped to do what you want.
Related
This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 8 years ago.
If the value is 200.3456, it should be formatted to 200.34.
If it is 200, then it should be 200.00.
Here's an utility that rounds (instead of truncating) a double to specified number of decimal places.
For example:
round(200.3456, 2); // returns 200.35
Original version; watch out with this
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
long factor = (long) Math.pow(10, places);
value = value * factor;
long tmp = Math.round(value);
return (double) tmp / factor;
}
This breaks down badly in corner cases with either a very high number of decimal places (e.g. round(1000.0d, 17)) or large integer part (e.g. round(90080070060.1d, 9)). Thanks to Sloin for pointing this out.
I've been using the above to round "not-too-big" doubles to 2 or 3 decimal places happily for years (for example to clean up time in seconds for logging purposes: 27.987654321987 -> 27.99). But I guess it's best to avoid it, since more reliable ways are readily available, with cleaner code too.
So, use this instead
(Adapted from this answer by Louis Wasserman and this one by Sean Owen.)
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
BigDecimal bd = BigDecimal.valueOf(value);
bd = bd.setScale(places, RoundingMode.HALF_UP);
return bd.doubleValue();
}
Note that HALF_UP is the rounding mode "commonly taught at school". Peruse the RoundingMode documentation, if you suspect you need something else such as Bankers’ Rounding.
Of course, if you prefer, you can inline the above into a one-liner:
new BigDecimal(value).setScale(places, RoundingMode.HALF_UP).doubleValue()
And in every case
Always remember that floating point representations using float and double are inexact.
For example, consider these expressions:
999199.1231231235 == 999199.1231231236 // true
1.03 - 0.41 // 0.6200000000000001
For exactness, you want to use BigDecimal. And while at it, use the constructor that takes a String, never the one taking double. For instance, try executing this:
System.out.println(new BigDecimal(1.03).subtract(new BigDecimal(0.41)));
System.out.println(new BigDecimal("1.03").subtract(new BigDecimal("0.41")));
Some excellent further reading on the topic:
Item 48: "Avoid float and double if exact answers are required" in Effective Java (2nd ed) by Joshua Bloch
What Every Programmer Should Know About Floating-Point Arithmetic
If you wanted String formatting instead of (or in addition to) strictly rounding numbers, see the other answers.
Specifically, note that round(200, 0) returns 200.0. If you want to output "200.00", you should first round and then format the result for output (which is perfectly explained in Jesper's answer).
If you just want to print a double with two digits after the decimal point, use something like this:
double value = 200.3456;
System.out.printf("Value: %.2f", value);
If you want to have the result in a String instead of being printed to the console, use String.format() with the same arguments:
String result = String.format("%.2f", value);
Or use class DecimalFormat:
DecimalFormat df = new DecimalFormat("####0.00");
System.out.println("Value: " + df.format(value));
I think this is easier:
double time = 200.3456;
DecimalFormat df = new DecimalFormat("#.##");
time = Double.valueOf(df.format(time));
System.out.println(time); // 200.35
Note that this will actually do the rounding for you, not just formatting.
The easiest way, would be to do a trick like this;
double val = ....;
val = val*100;
val = Math.round(val);
val = val /100;
if val starts at 200.3456 then it goes to 20034.56 then it gets rounded to 20035 then we divide it to get 200.34.
if you wanted to always round down we could always truncate by casting to an int:
double val = ....;
val = val*100;
val = (double)((int) val);
val = val /100;
This technique will work for most cases because for very large doubles (positive or negative) it may overflow. but if you know that your values will be in an appropriate range then this should work for you.
Please use Apache commons math:
Precision.round(10.4567, 2)
function Double round2(Double val) {
return new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue();
}
Note the toString()!!!!
This is because BigDecimal converts the exact binary form of the double!!!
These are the various suggested methods and their fail cases.
// Always Good!
new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue()
Double val = 260.775d; //EXPECTED 260.78
260.77 - WRONG - new BigDecimal(val).setScale(2,RoundingMode.HALF_UP).doubleValue()
Double val = 260.775d; //EXPECTED 260.78
260.77 - TRY AGAIN - Math.round(val * 100.d) / 100.0d
Double val = 256.025d; //EXPECTED 256.03d
256.02 - OOPS - new DecimalFormat("0.00").format(val)
// By default use half even, works if you change mode to half_up
Double val = 256.025d; //EXPECTED 256.03d
256.02 - FAIL - (int)(val * 100 + 0.5) / 100.0;
double value= 200.3456;
DecimalFormat df = new DecimalFormat("0.00");
System.out.println(df.format(value));
If you really want the same double, but rounded in the way you want you can use BigDecimal, for example
new BigDecimal(myValue).setScale(2, RoundingMode.HALF_UP).doubleValue();
double d = 28786.079999999998;
String str = String.format("%1.2f", d);
d = Double.valueOf(str);
For two rounding digits. Very simple and you are basically updating the variable instead of just display purposes which DecimalFormat does.
x = Math.floor(x * 100) / 100;
Rounding a double is usually not what one wants. Instead, use String.format() to represent it in the desired format.
In your question, it seems that you want to avoid rounding the numbers as well? I think .format() will round the numbers using half-up, afaik?
so if you want to round, 200.3456 should be 200.35 for a precision of 2. but in your case, if you just want the first 2 and then discard the rest?
You could multiply it by 100 and then cast to an int (or taking the floor of the number), before dividing by 100 again.
200.3456 * 100 = 20034.56;
(int) 20034.56 = 20034;
20034/100.0 = 200.34;
You might have issues with really really big numbers close to the boundary though. In which case converting to a string and substring'ing it would work just as easily.
value = (int)(value * 100 + 0.5) / 100.0;
This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 4 years ago.
I have this float 98.01645 and I'm getting 98.02 with this function:
String lastCredit = String.format("%.2f", AppSingleton.getInstance().credit);
That I want is get 98.01 (only two decimals or not rounded number). I'm trying but I can't get the way to do it work.
Doing it manually:
String lastCredit = String.format("%.2f", java.lang.Math.floor(100.0*AppSingleton.getInstance().credit)*0.01);
Multiplying by 100.0 to move the decimal point two to the right, then rounding down, then moving the decimal point two to the left by multiplying with 0.01.
You can try this
String lastCredit = String.format("%.2f",Math.floor((98.01645 * 100)) / 100);
System.out.println(lastCredit);
You basically multiply the value by 100 because you need 2 numbers after the decimal and round down that value. After you have the result you divide it by 100 again.
I would not call this elegant, but you can just use String.format to get 3 decimal places and remove the last using substring:
String lastCredit = String.format("%.3f", AppSingleton.getInstance().credit);
lastCredit = lastCredit.substring(0, lastCredit.length() - 1);
Use below code instead of String lastCredit = String.format("%.2f", AppSingleton.getInstance().credit);
DecimalFormat decimalFormat = new DecimalFormat("#.##");
decimalFormat.setRoundingMode(RoundingMode.FLOOR);
String lastCredit = decimalFormat.format(f);
This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 7 years ago.
I'm beginner in android developing. How can I make a double datatype i.e 56.32534 into 56.33? But I want to take it in a double variable. I know a way but it takes in String.
Double value = 56.32534;
DecimalFormat df = new DecimalFormat("0.##");
String newvalue = df.format(value);
Here, newvalueis a string. But I want to take it in a double datatype.
I need to use it's numeric value not just in display purpose only.
You can use Math.round():
double value = 56.32534;
double rounded = Math.round(100 * value) / 100.0;
The multiplication and then division by 100 is necessary because Math.round() rounds to the nearest long value.
If you want to generalize this to a variable number of digits, you can use something like this:
public double round(double value, int digits) {
double scale = Math.pow(10, digits);
return Math.round(value * scale) / scale;
}
This will even work if digits is not positive; it will round to the nearest integer if digits is 0, to the nearest 10 if digits is -1, to the nearest 100 if digits is -2, etc.
We need a little bit more precision on what you want to do. Do you want to display it that way or just use its numerical value that way ?
If it's the display your solution is ok. You could always get back the float number from the string afterwards.
On the other hand if you want to use it as a numerical value what you could do is
double rounded = (double) Math.round(myfloat * 100) /100;
I think it works. Can't test it as I'm on my mobile though
I have to write a program with the following requirements:
I have a variable of type float, say float a = 3333.333f;
I have a variable of type int, say int b = 9999;
When I perform a*b in calculator, the result will be 33329996.667
After rounding up the decimals to 2 places, I want to print the value as 33329996.67 in java. I tried with long, double, float, big decimal, But couldnt succeed.
Can anyone please help me solving this?
float only has 7 digits of precision, so its not a good choice for a result with more than 7 digits. double has up to 16 digits of accuracy and is a better choice.
double a = 3333.333;
int b = 9999;
System.out.printf("%.2f", a * b);
prints
33329996.67
To determine the number of digits after the comma you have to apply a little trick:
First shift the comma to the right for the amount of digits you want to have, then cut the whole number e.g. with Math.ceil(float f) and then shift the comma back to the left.
That will illustrate that:
float f = 33329996.667;
float f2 = Math.ceil((f * 100)) / 100;
f2 now has the value 33329996.67.
Hope this helps.
EDIT: For formatting have a look here
This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 8 years ago.
If the value is 200.3456, it should be formatted to 200.34.
If it is 200, then it should be 200.00.
Here's an utility that rounds (instead of truncating) a double to specified number of decimal places.
For example:
round(200.3456, 2); // returns 200.35
Original version; watch out with this
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
long factor = (long) Math.pow(10, places);
value = value * factor;
long tmp = Math.round(value);
return (double) tmp / factor;
}
This breaks down badly in corner cases with either a very high number of decimal places (e.g. round(1000.0d, 17)) or large integer part (e.g. round(90080070060.1d, 9)). Thanks to Sloin for pointing this out.
I've been using the above to round "not-too-big" doubles to 2 or 3 decimal places happily for years (for example to clean up time in seconds for logging purposes: 27.987654321987 -> 27.99). But I guess it's best to avoid it, since more reliable ways are readily available, with cleaner code too.
So, use this instead
(Adapted from this answer by Louis Wasserman and this one by Sean Owen.)
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
BigDecimal bd = BigDecimal.valueOf(value);
bd = bd.setScale(places, RoundingMode.HALF_UP);
return bd.doubleValue();
}
Note that HALF_UP is the rounding mode "commonly taught at school". Peruse the RoundingMode documentation, if you suspect you need something else such as Bankers’ Rounding.
Of course, if you prefer, you can inline the above into a one-liner:
new BigDecimal(value).setScale(places, RoundingMode.HALF_UP).doubleValue()
And in every case
Always remember that floating point representations using float and double are inexact.
For example, consider these expressions:
999199.1231231235 == 999199.1231231236 // true
1.03 - 0.41 // 0.6200000000000001
For exactness, you want to use BigDecimal. And while at it, use the constructor that takes a String, never the one taking double. For instance, try executing this:
System.out.println(new BigDecimal(1.03).subtract(new BigDecimal(0.41)));
System.out.println(new BigDecimal("1.03").subtract(new BigDecimal("0.41")));
Some excellent further reading on the topic:
Item 48: "Avoid float and double if exact answers are required" in Effective Java (2nd ed) by Joshua Bloch
What Every Programmer Should Know About Floating-Point Arithmetic
If you wanted String formatting instead of (or in addition to) strictly rounding numbers, see the other answers.
Specifically, note that round(200, 0) returns 200.0. If you want to output "200.00", you should first round and then format the result for output (which is perfectly explained in Jesper's answer).
If you just want to print a double with two digits after the decimal point, use something like this:
double value = 200.3456;
System.out.printf("Value: %.2f", value);
If you want to have the result in a String instead of being printed to the console, use String.format() with the same arguments:
String result = String.format("%.2f", value);
Or use class DecimalFormat:
DecimalFormat df = new DecimalFormat("####0.00");
System.out.println("Value: " + df.format(value));
I think this is easier:
double time = 200.3456;
DecimalFormat df = new DecimalFormat("#.##");
time = Double.valueOf(df.format(time));
System.out.println(time); // 200.35
Note that this will actually do the rounding for you, not just formatting.
The easiest way, would be to do a trick like this;
double val = ....;
val = val*100;
val = Math.round(val);
val = val /100;
if val starts at 200.3456 then it goes to 20034.56 then it gets rounded to 20035 then we divide it to get 200.34.
if you wanted to always round down we could always truncate by casting to an int:
double val = ....;
val = val*100;
val = (double)((int) val);
val = val /100;
This technique will work for most cases because for very large doubles (positive or negative) it may overflow. but if you know that your values will be in an appropriate range then this should work for you.
Please use Apache commons math:
Precision.round(10.4567, 2)
function Double round2(Double val) {
return new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue();
}
Note the toString()!!!!
This is because BigDecimal converts the exact binary form of the double!!!
These are the various suggested methods and their fail cases.
// Always Good!
new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue()
Double val = 260.775d; //EXPECTED 260.78
260.77 - WRONG - new BigDecimal(val).setScale(2,RoundingMode.HALF_UP).doubleValue()
Double val = 260.775d; //EXPECTED 260.78
260.77 - TRY AGAIN - Math.round(val * 100.d) / 100.0d
Double val = 256.025d; //EXPECTED 256.03d
256.02 - OOPS - new DecimalFormat("0.00").format(val)
// By default use half even, works if you change mode to half_up
Double val = 256.025d; //EXPECTED 256.03d
256.02 - FAIL - (int)(val * 100 + 0.5) / 100.0;
double value= 200.3456;
DecimalFormat df = new DecimalFormat("0.00");
System.out.println(df.format(value));
If you really want the same double, but rounded in the way you want you can use BigDecimal, for example
new BigDecimal(myValue).setScale(2, RoundingMode.HALF_UP).doubleValue();
double d = 28786.079999999998;
String str = String.format("%1.2f", d);
d = Double.valueOf(str);
For two rounding digits. Very simple and you are basically updating the variable instead of just display purposes which DecimalFormat does.
x = Math.floor(x * 100) / 100;
Rounding a double is usually not what one wants. Instead, use String.format() to represent it in the desired format.
In your question, it seems that you want to avoid rounding the numbers as well? I think .format() will round the numbers using half-up, afaik?
so if you want to round, 200.3456 should be 200.35 for a precision of 2. but in your case, if you just want the first 2 and then discard the rest?
You could multiply it by 100 and then cast to an int (or taking the floor of the number), before dividing by 100 again.
200.3456 * 100 = 20034.56;
(int) 20034.56 = 20034;
20034/100.0 = 200.34;
You might have issues with really really big numbers close to the boundary though. In which case converting to a string and substring'ing it would work just as easily.
value = (int)(value * 100 + 0.5) / 100.0;