A 64-bit double can represent integer +/- 253 exactly.
Given this fact, I choose to use a double type as a single type for all my types, since my largest integer is an unsigned 32-bit number.
But now I have to print these pseudo integers, but the problem is they are also mixed in with actual doubles.
So how do I print these doubles nicely in Java?
I have tried String.format("%f", value), which is close, except I get a lot of trailing zeros for small values.
Here's an example output of of %f
232.00000000
0.18000000000
1237875192.0
4.5800000000
0.00000000
1.23450000
What I want is:
232
0.18
1237875192
4.58
0
1.2345
Sure I can write a function to trim those zeros, but that's lot of performance loss due to string manipulation. Can I do better with other format code?
The answers by Tom E. and Jeremy S. are unacceptable as they both arbitrarily rounds to two decimal places. Please understand the problem before answering.
Please note that String.format(format, args...) is locale-dependent (see answers below).
If the idea is to print integers stored as doubles as if they are integers, and otherwise print the doubles with the minimum necessary precision:
public static String fmt(double d)
{
if(d == (long) d)
return String.format("%d",(long)d);
else
return String.format("%s",d);
}
Produces:
232
0.18
1237875192
4.58
0
1.2345
And does not rely on string manipulation.
String.format("%.2f", value);
In short:
If you want to get rid of trailing zeros and locale problems, then you should use:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); //340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
System.out.println(df.format(myValue)); //output: 0.00000021
Explanation:
Why other answers did not suit me:
Double.toString() or System.out.println or FloatingDecimal.toJavaFormatString uses scientific notations if double is less than 10^-3 or greater than or equal to 10^7
double myValue = 0.00000021d;
String.format("%s", myvalue); //output: 2.1E-7
by using %f, the default decimal precision is 6, otherwise you can hardcode it, but it results in extra zeros added if you have fewer decimals. Example:
double myValue = 0.00000021d;
String.format("%.12f", myvalue); // Output: 0.000000210000
by using setMaximumFractionDigits(0); or %.0f you remove any decimal precision, which is fine for integers/longs but not for double
double myValue = 0.00000021d;
System.out.println(String.format("%.0f", myvalue)); // Output: 0
DecimalFormat df = new DecimalFormat("0");
System.out.println(df.format(myValue)); // Output: 0
by using DecimalFormat, you are local dependent. In the French locale, the decimal separator is a comma, not a point:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0");
df.setMaximumFractionDigits(340);
System.out.println(df.format(myvalue)); // Output: 0,00000021
Using the ENGLISH locale makes sure you get a point for decimal separator, wherever your program will run.
Why using 340 then for setMaximumFractionDigits?
Two reasons:
setMaximumFractionDigits accepts an integer, but its implementation has a maximum digits allowed of DecimalFormat.DOUBLE_FRACTION_DIGITS which equals 340
Double.MIN_VALUE = 4.9E-324 so with 340 digits you are sure not to round your double and lose precision
Use:
if (d % 1.0 != 0)
return String.format("%s", d);
else
return String.format("%.0f", d);
This should work with the extreme values supported by Double. It yields:
0.12
12
12.144252
0
On my machine, the following function is roughly 7 times faster than the function provided by JasonD's answer, since it avoids String.format:
public static String prettyPrint(double d) {
int i = (int) d;
return d == i ? String.valueOf(i) : String.valueOf(d);
}
My two cents:
if(n % 1 == 0) {
return String.format(Locale.US, "%.0f", n));
} else {
return String.format(Locale.US, "%.1f", n));
}
if (d == Math.floor(d)) {
return String.format("%.0f", d); //Format is: 0 places after decimal point
} else {
return Double.toString(d);
}
More info: https://docs.oracle.com/javase/tutorial/java/data/numberformat.html
float price = 4.30;
DecimalFormat format = new DecimalFormat("0.##"); // Choose the number of decimal places to work with in case they are different than zero and zero value will be removed
format.setRoundingMode(RoundingMode.DOWN); // Choose your Rounding Mode
System.out.println(format.format(price));
This is the result of some tests:
4.30 => 4.3
4.39 => 4.39 // Choose format.setRoundingMode(RoundingMode.UP) to get 4.4
4.000000 => 4
4 => 4
Naw, never mind. The performance loss due to string manipulation is zero.
And here's the code to trim the end after %f:
private static String trimTrailingZeros(String number) {
if(!number.contains(".")) {
return number;
}
return number.replaceAll("\\.?0*$", "");
}
Use a DecimalFormat and setMinimumFractionDigits(0).
This one will gets the job done nicely:
public static String removeZero(double number) {
DecimalFormat format = new DecimalFormat("#.###########");
return format.format(number);
}
new DecimalFormat("00.#").format(20.236)
//out =20.2
new DecimalFormat("00.#").format(2.236)
//out =02.2
0 for minimum number of digits
Renders # digits
Please note that String.format(format, args...) is locale-dependent because it formats using the user's default locale, that is, probably with commas and even spaces inside like 123 456,789 or 123,456.789, which may be not exactly what you expect.
You may prefer to use String.format((Locale)null, format, args...).
For example,
double f = 123456.789d;
System.out.println(String.format(Locale.FRANCE,"%f",f));
System.out.println(String.format(Locale.GERMANY,"%f",f));
System.out.println(String.format(Locale.US,"%f",f));
prints
123456,789000
123456,789000
123456.789000
and this is what will String.format(format, args...) do in different countries.
EDIT Ok, since there has been a discussion about formalities:
res += stripFpZeroes(String.format((Locale) null, (nDigits!=0 ? "%."+nDigits+"f" : "%f"), value));
...
protected static String stripFpZeroes(String fpnumber) {
int n = fpnumber.indexOf('.');
if (n == -1) {
return fpnumber;
}
if (n < 2) {
n = 2;
}
String s = fpnumber;
while (s.length() > n && s.endsWith("0")) {
s = s.substring(0, s.length()-1);
}
return s;
}
I made a DoubleFormatter to efficiently convert a great numbers of double values to a nice/presentable string:
double horribleNumber = 3598945.141658554548844;
DoubleFormatter df = new DoubleFormatter(4, 6); // 4 = MaxInteger, 6 = MaxDecimal
String beautyDisplay = df.format(horribleNumber);
If the integer part of V has more than MaxInteger => display V in scientific format (1.2345E+30). Otherwise, display in normal format (124.45678).
the MaxDecimal decide numbers of decimal digits (trim with bankers' rounding)
Here the code:
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.text.DecimalFormatSymbols;
import java.text.NumberFormat;
import java.util.Locale;
import com.google.common.base.Preconditions;
import com.google.common.base.Strings;
/**
* Convert a double to a beautiful String (US-local):
*
* double horribleNumber = 3598945.141658554548844;
* DoubleFormatter df = new DoubleFormatter(4,6);
* String beautyDisplay = df.format(horribleNumber);
* String beautyLabel = df.formatHtml(horribleNumber);
*
* Manipulate 3 instances of NumberFormat to efficiently format a great number of double values.
* (avoid to create an object NumberFormat each call of format()).
*
* 3 instances of NumberFormat will be reused to format a value v:
*
* if v < EXP_DOWN, uses nfBelow
* if EXP_DOWN <= v <= EXP_UP, uses nfNormal
* if EXP_UP < v, uses nfAbove
*
* nfBelow, nfNormal and nfAbove will be generated base on the precision_ parameter.
*
* #author: DUONG Phu-Hiep
*/
public class DoubleFormatter
{
private static final double EXP_DOWN = 1.e-3;
private double EXP_UP; // always = 10^maxInteger
private int maxInteger_;
private int maxFraction_;
private NumberFormat nfBelow_;
private NumberFormat nfNormal_;
private NumberFormat nfAbove_;
private enum NumberFormatKind {Below, Normal, Above}
public DoubleFormatter(int maxInteger, int maxFraction){
setPrecision(maxInteger, maxFraction);
}
public void setPrecision(int maxInteger, int maxFraction){
Preconditions.checkArgument(maxFraction>=0);
Preconditions.checkArgument(maxInteger>0 && maxInteger<17);
if (maxFraction == maxFraction_ && maxInteger_ == maxInteger) {
return;
}
maxFraction_ = maxFraction;
maxInteger_ = maxInteger;
EXP_UP = Math.pow(10, maxInteger);
nfBelow_ = createNumberFormat(NumberFormatKind.Below);
nfNormal_ = createNumberFormat(NumberFormatKind.Normal);
nfAbove_ = createNumberFormat(NumberFormatKind.Above);
}
private NumberFormat createNumberFormat(NumberFormatKind kind) {
// If you do not use the Guava library, replace it with createSharp(precision);
final String sharpByPrecision = Strings.repeat("#", maxFraction_);
NumberFormat f = NumberFormat.getInstance(Locale.US);
// Apply bankers' rounding: this is the rounding mode that
// statistically minimizes cumulative error when applied
// repeatedly over a sequence of calculations
f.setRoundingMode(RoundingMode.HALF_EVEN);
if (f instanceof DecimalFormat) {
DecimalFormat df = (DecimalFormat) f;
DecimalFormatSymbols dfs = df.getDecimalFormatSymbols();
// Set group separator to space instead of comma
//dfs.setGroupingSeparator(' ');
// Set Exponent symbol to minus 'e' instead of 'E'
if (kind == NumberFormatKind.Above) {
dfs.setExponentSeparator("e+"); //force to display the positive sign in the exponent part
} else {
dfs.setExponentSeparator("e");
}
df.setDecimalFormatSymbols(dfs);
// Use exponent format if v is outside of [EXP_DOWN,EXP_UP]
if (kind == NumberFormatKind.Normal) {
if (maxFraction_ == 0) {
df.applyPattern("#,##0");
} else {
df.applyPattern("#,##0."+sharpByPrecision);
}
} else {
if (maxFraction_ == 0) {
df.applyPattern("0E0");
} else {
df.applyPattern("0."+sharpByPrecision+"E0");
}
}
}
return f;
}
public String format(double v) {
if (Double.isNaN(v)) {
return "-";
}
if (v==0) {
return "0";
}
final double absv = Math.abs(v);
if (absv<EXP_DOWN) {
return nfBelow_.format(v);
}
if (absv>EXP_UP) {
return nfAbove_.format(v);
}
return nfNormal_.format(v);
}
/**
* Format and higlight the important part (integer part & exponent part)
*/
public String formatHtml(double v) {
if (Double.isNaN(v)) {
return "-";
}
return htmlize(format(v));
}
/**
* This is the base alogrithm: create a instance of NumberFormat for the value, then format it. It should
* not be used to format a great numbers of value
*
* We will never use this methode, it is here only to understanding the Algo principal:
*
* format v to string. precision_ is numbers of digits after decimal.
* if EXP_DOWN <= abs(v) <= EXP_UP, display the normal format: 124.45678
* otherwise display scientist format with: 1.2345e+30
*
* pre-condition: precision >= 1
*/
#Deprecated
public String formatInefficient(double v) {
// If you do not use Guava library, replace with createSharp(precision);
final String sharpByPrecision = Strings.repeat("#", maxFraction_);
final double absv = Math.abs(v);
NumberFormat f = NumberFormat.getInstance(Locale.US);
// Apply bankers' rounding: this is the rounding mode that
// statistically minimizes cumulative error when applied
// repeatedly over a sequence of calculations
f.setRoundingMode(RoundingMode.HALF_EVEN);
if (f instanceof DecimalFormat) {
DecimalFormat df = (DecimalFormat) f;
DecimalFormatSymbols dfs = df.getDecimalFormatSymbols();
// Set group separator to space instead of comma
dfs.setGroupingSeparator(' ');
// Set Exponent symbol to minus 'e' instead of 'E'
if (absv>EXP_UP) {
dfs.setExponentSeparator("e+"); //force to display the positive sign in the exponent part
} else {
dfs.setExponentSeparator("e");
}
df.setDecimalFormatSymbols(dfs);
//use exponent format if v is out side of [EXP_DOWN,EXP_UP]
if (absv<EXP_DOWN || absv>EXP_UP) {
df.applyPattern("0."+sharpByPrecision+"E0");
} else {
df.applyPattern("#,##0."+sharpByPrecision);
}
}
return f.format(v);
}
/**
* Convert "3.1416e+12" to "<b>3</b>.1416e<b>+12</b>"
* It is a html format of a number which highlight the integer and exponent part
*/
private static String htmlize(String s) {
StringBuilder resu = new StringBuilder("<b>");
int p1 = s.indexOf('.');
if (p1>0) {
resu.append(s.substring(0, p1));
resu.append("</b>");
} else {
p1 = 0;
}
int p2 = s.lastIndexOf('e');
if (p2>0) {
resu.append(s.substring(p1, p2));
resu.append("<b>");
resu.append(s.substring(p2, s.length()));
resu.append("</b>");
} else {
resu.append(s.substring(p1, s.length()));
if (p1==0){
resu.append("</b>");
}
}
return resu.toString();
}
}
Note: I used two functions from the Guava library. If you don't use Guava, code it yourself:
/**
* Equivalent to Strings.repeat("#", n) of the Guava library:
*/
private static String createSharp(int n) {
StringBuilder sb = new StringBuilder();
for (int i=0; i<n; i++) {
sb.append('#');
}
return sb.toString();
}
String s = String.valueof("your int variable");
while (g.endsWith("0") && g.contains(".")) {
g = g.substring(0, g.length() - 1);
if (g.endsWith("."))
{
g = g.substring(0, g.length() - 1);
}
}
You said you choose to store your numbers with the double type. I think this could be the root of the problem, because it forces you to store integers into doubles (and therefore losing the initial information about the value's nature). What about storing your numbers in instances of the Number class (superclass of both Double and Integer) and rely on polymorphism to determine the correct format of each number?
I know it may not be acceptable to refactor a whole part of your code due to that, but it could produce the desired output without extra code/casting/parsing.
Example:
import java.util.ArrayList;
import java.util.List;
public class UseMixedNumbers {
public static void main(String[] args) {
List<Number> listNumbers = new ArrayList<Number>();
listNumbers.add(232);
listNumbers.add(0.18);
listNumbers.add(1237875192);
listNumbers.add(4.58);
listNumbers.add(0);
listNumbers.add(1.2345);
for (Number number : listNumbers) {
System.out.println(number);
}
}
}
Will produce the following output:
232
0.18
1237875192
4.58
0
1.2345
For Kotlin you can use an extension like:
fun Double.toPrettyString() =
if(this - this.toLong() == 0.0)
String.format("%d", this.toLong())
else
String.format("%s", this)
This is what I came up with:
private static String format(final double dbl) {
return dbl % 1 != 0 ? String.valueOf(dbl) : String.valueOf((int) dbl);
}
It is a simple one-liner and only casts to int if it really needs to.
Format price with grouping, rounding, and no unnecessary zeroes (in double).
Rules:
No zeroes at the end (2.0000 = 2; 1.0100000 = 1.01)
Two digits maximum after a point (2.010 = 2.01; 0.20 = 0.2)
Rounding after the 2nd digit after a point (1.994 = 1.99; 1.995 = 2; 1.006 = 1.01; 0.0006 -> 0)
Returns 0 (null/-0 = 0)
Adds $ (= $56/-$56)
Grouping (101101.02 = $101,101.02)
More examples:
-99.985 = -$99.99
10 = $10
10.00 = $10
20.01000089 = $20.01
It is written in Kotlin as a fun extension of Double (because it is used in Android), but it can be converted to Java easily, because Java classes were used.
/**
* 23.0 -> $23
*
* 23.1 -> $23.1
*
* 23.01 -> $23.01
*
* 23.99 -> $23.99
*
* 23.999 -> $24
*
* -0.0 -> $0
*
* -5.00 -> -$5
*
* -5.019 -> -$5.02
*/
fun Double?.formatUserAsSum(): String {
return when {
this == null || this == 0.0 -> "$0"
this % 1 == 0.0 -> DecimalFormat("$#,##0;-$#,##0").format(this)
else -> DecimalFormat("$#,##0.##;-$#,##0.##").format(this)
}
}
How to use:
var yourDouble: Double? = -20.00
println(yourDouble.formatUserAsSum()) // will print -$20
yourDouble = null
println(yourDouble.formatUserAsSum()) // will print $0
About DecimalFormat: https://docs.oracle.com/javase/6/docs/api/java/text/DecimalFormat.html
A simple solution with locale in mind:
double d = 123.45;
NumberFormat numberFormat = NumberFormat.getInstance(Locale.GERMANY);
System.out.println(numberFormat.format(d));
Since comma is used as decimal separator in Germany, the above will print:
123,45
Here's another answer that has an option to append decimal ONLY IF decimal was not zero.
/**
* Example: (isDecimalRequired = true)
* d = 12345
* returns 12,345.00
*
* d = 12345.12345
* returns 12,345.12
*
* ==================================================
* Example: (isDecimalRequired = false)
* d = 12345
* returns 12,345 (notice that there's no decimal since it's zero)
*
* d = 12345.12345
* returns 12,345.12
*
* #param d float to format
* #param zeroCount number decimal places
* #param isDecimalRequired true if it will put decimal even zero,
* false will remove the last decimal(s) if zero.
*/
fun formatDecimal(d: Float? = 0f, zeroCount: Int, isDecimalRequired: Boolean = true): String {
val zeros = StringBuilder()
for (i in 0 until zeroCount) {
zeros.append("0")
}
var pattern = "#,##0"
if (zeros.isNotEmpty()) {
pattern += ".$zeros"
}
val numberFormat = DecimalFormat(pattern)
var formattedNumber = if (d != null) numberFormat.format(d) else "0"
if (!isDecimalRequired) {
for (i in formattedNumber.length downTo formattedNumber.length - zeroCount) {
val number = formattedNumber[i - 1]
if (number == '0' || number == '.') {
formattedNumber = formattedNumber.substring(0, formattedNumber.length - 1)
} else {
break
}
}
}
return formattedNumber
}
Here are two ways to achieve it. First, the shorter (and probably better) way:
public static String formatFloatToString(final float f)
{
final int i = (int)f;
if(f == i)
return Integer.toString(i);
return Float.toString(f);
}
And here's the longer and probably worse way:
public static String formatFloatToString(final float f)
{
final String s = Float.toString(f);
int dotPos = -1;
for(int i=0; i<s.length(); ++i)
if(s.charAt(i) == '.')
{
dotPos = i;
break;
}
if(dotPos == -1)
return s;
int end = dotPos;
for(int i = dotPos + 1; i<s.length(); ++i)
{
final char c = s.charAt(i);
if(c != '0')
end = i + 1;
}
final String result = s.substring(0, end);
return result;
}
public static String fmt(double d) {
String val = Double.toString(d);
String[] valArray = val.split("\\.");
long valLong = 0;
if(valArray.length == 2) {
valLong = Long.parseLong(valArray[1]);
}
if (valLong == 0)
return String.format("%d", (long) d);
else
return String.format("%s", d);
}
I had to use this because d == (long)d was giving me violation in a SonarQube report.
I am using this for formatting numbers without trailing zeroes in our JSF application. The original built-in formatters required you to specify max numbers of fractional digits which could be useful here also in case you have too many fractional digits.
/**
* Formats the given Number as with as many fractional digits as precision
* available.<br>
* This is a convenient method in case all fractional digits shall be
* rendered and no custom format / pattern needs to be provided.<br>
* <br>
* This serves as a workaround for {#link NumberFormat#getNumberInstance()}
* which by default only renders up to three fractional digits.
*
* #param number
* #param locale
* #param groupingUsed <code>true</code> if grouping shall be used
*
* #return
*/
public static String formatNumberFraction(final Number number, final Locale locale, final boolean groupingUsed)
{
if (number == null)
return null;
final BigDecimal bDNumber = MathUtils.getBigDecimal(number);
final NumberFormat numberFormat = NumberFormat.getNumberInstance(locale);
numberFormat.setMaximumFractionDigits(Math.max(0, bDNumber.scale()));
numberFormat.setGroupingUsed(groupingUsed);
// Convert back for locale percent formatter
return numberFormat.format(bDNumber);
}
/**
* Formats the given Number as percent with as many fractional digits as
* precision available.<br>
* This is a convenient method in case all fractional digits shall be
* rendered and no custom format / pattern needs to be provided.<br>
* <br>
* This serves as a workaround for {#link NumberFormat#getPercentInstance()}
* which does not renders fractional digits.
*
* #param number Number in range of [0-1]
* #param locale
*
* #return
*/
public static String formatPercentFraction(final Number number, final Locale locale)
{
if (number == null)
return null;
final BigDecimal bDNumber = MathUtils.getBigDecimal(number).multiply(new BigDecimal(100));
final NumberFormat percentScaleFormat = NumberFormat.getPercentInstance(locale);
percentScaleFormat.setMaximumFractionDigits(Math.max(0, bDNumber.scale() - 2));
final BigDecimal bDNumberPercent = bDNumber.multiply(new BigDecimal(0.01));
// Convert back for locale percent formatter
final String strPercent = percentScaleFormat.format(bDNumberPercent);
return strPercent;
}
work with given decimal length...
public static String getLocaleFloatValueDecimalWithLength(Locale loc, float value, int length) {
//make string from float value
return String.format(loc, (value % 1 == 0 ? "%.0f" : "%."+length+"f"), value);
}
0.0 -> 0%
1.0 -> 100%
0.1 -> 10%
0.11 -> 11%
0.01 -> 1%
0.111 -> 11.1%
0.001 -> 0.1%
0.1111 -> 11.11%
0.0001 -> 0.01%
".replace()" is added because I was always getting wrong separator
import java.text.NumberFormat
fun Double.formating(): String {
val defaultFormat: NumberFormat = NumberFormat.getPercentInstance()
defaultFormat.minimumFractionDigits = 0
defaultFormat.maximumFractionDigits = 2
return defaultFormat.format(this).replace(",", ".")
}
Here is an answer that actually works (combination of different answers here)
public static String removeTrailingZeros(double f)
{
if(f == (int)f) {
return String.format("%d", (int)f);
}
return String.format("%f", f).replaceAll("0*$", "");
}
The best way to do this is as below:
public class Test {
public static void main(String args[]){
System.out.println(String.format("%s something", new Double(3.456)));
System.out.println(String.format("%s something", new Double(3.456234523452)));
System.out.println(String.format("%s something", new Double(3.45)));
System.out.println(String.format("%s something", new Double(3)));
}
}
Output:
3.456 something
3.456234523452 something
3.45 something
3.0 something
The only issue is the last one where .0 doesn't get removed. But if you are able to live with that then this works best. %.2f will round it to the last two decimal digits. So will DecimalFormat. If you need all the decimal places, but not the trailing zeros then this works best.
String s = "1.210000";
while (s.endsWith("0")){
s = (s.substring(0, s.length() - 1));
}
This will make the string to drop the tailing 0-s.
Part of my program tests if two numbers are equal. Because certain operations take only doubles and others take only ints, I am comparing ints and doubles. The programs is returning that the two are equal even when they only round to being equal (i.e. 7.5=7). I only want the program to return true if the two actually are equal. I've tried the solutions listed here: How to test if a double is an integer
to determine if my doubles are ints.
All of them appear to work - they compile, but the program still returns 7=7.5
I've tried going the other direction also - converting my ints to doubles - same result. How do I get my program to acknowledge the difference? With the most current suggestion:
import acm.program.ConsoleProgram;
import java.awt.Color;
import acm.io.IODialog;
import java.text.*;
import static java.lang.Math.*;
import java.util.*;
/** Tests to see if user color matches sample colors */
public class ColorMatch extends ConsoleProgram
{
//defining sample colors
Color[] dmc =
{
new Color(255,255,255),
new Color(43,57,41),
new Color(213,39,86),
new Color(0,160,130),
new Color(0,0,0),
};
public void run()
{
average();
}
//averages three colors, then tests for match to given color
public void average()
{
//asks for stitch color
IODialog dialog = new IODialog();
int stitchRed = dialog.readInt("Enter red value: ");
int stitchGreen = dialog.readInt("Enter green value: ");
int stitchBlue = dialog.readInt("Enter blue value: ");
Color stitchColor = new Color(stitchRed,stitchGreen,stitchBlue);
//gets averages for dmc colors
for (Color i:dmc)
{
for (Color j:dmc)
{
for (Color k:dmc)
{
int indexI = Arrays.asList(dmc).indexOf(i);
int indexJ = Arrays.asList(dmc).indexOf(j);
int indexK = Arrays.asList(dmc).indexOf(k);
if (indexI <= indexJ && indexJ <= indexK)
{
int iRed = i.getRed();
int jRed = j.getRed();
int kRed = k.getRed();
int iGreen = i.getGreen();
int jGreen = j.getGreen();
int kGreen = k.getGreen();
int iBlue = i.getBlue();
int jBlue = j.getBlue();
int kBlue = k.getBlue();
double redAverage = (iRed+jRed+kRed)/3;
double greenAverage = (iGreen+jGreen+kGreen)/3;
double blueAverage = (iBlue+jBlue+kBlue)/3;
if (redAverage == (int)redAverage && greenAverage == (int)greenAverage && blueAverage == (int)blueAverage)
{
int rAverage = (int)redAverage;
int gAverage = (int)greenAverage;
int bAverage = (int)blueAverage;
Color colorAverage = new Color(rAverage,gAverage,bAverage);
//tests to see if any average equals the stitch color
if (colorAverage.equals(stitchColor))
{
println("The color match is: " + i + ", " + j + ", " + k);
}
}
}
}
}
}
I plugged in 85s as my test numbers.
The only result should be (0,0,0)+(0,0,0)+(255,255,255), but it is also yielding (43,57,41)+(213,39,86)+(0,160,130) . (41+86+130)/3=85.7!=85.
I think the problem is that you are making a comparison of the 'int' and the 'double' values after casting the double value to int which truncates the decimal part.
For example (7==7.5) is false, but (7==(int)7.5) is true because (int)7.5 = 7.
So if you want strict comparison between an int and a double, you can compare them straight forward without casting. If you want to know more on how casting double to int works, please refer How does double to int cast work in Java.
I am compiling a Java program using for loop to find out the biggest value of long. However, nothing was printed when I run the program. Why?
Here's my code:
class LongMaxMin {
public static void main(String args[]) {
long i = 0L;
long result = 0L;
for (; ; ) {
result = i++;
if (i<0)
break;
}
System.out.println("The biggest integer:" + result);
}
Mostly because of time.
A long will have a max of about ~9.22 quintillion. You're starting at zero and incrementing up. That means you need to go through 9 quintillion loops before it wraps over and breaks. I just tried to run 2 billion operations in my javascript console and times out for a couple of minutes before I force quit.
If you sit there and let it run long enough, you'll get your output. Alternatively, start i at something close to the max already, like 9,223,372,036,854,700,000, and see if it still gives you the same issues. In Java 8, adding underscore to numeric literals is allowed. Initializing i to something like 9_223_372_036_854_700_000L will give you something in a more timely manner.
The max long is significantly high, at 9.223372e+18. For specifics, 9,223,372,036,854,775,807 is the number in question. This also contributes to that whole "this works, it'll just take WAY too long" theory.
I was curious how long it would take so I wrote a class to do the same thing. Wrote it with a separate thread to update results to the console every 1 second.
"int" results
1,343,211,433 37.4518434691484288634492200 % left
Max Value: 2,147,483,647
Time Taken (seconds): **1.588**
"long" results
1,220,167,357 99.9999999867709190074470400 % left
2,519,937,368 99.9999999726787843108699600 % left
3,881,970,343 99.9999999579115932059510100 % left
5,210,983,861 99.9999999435023997711689800 % left
6,562,562,290 99.9999999288485570811055300 % left
7,853,387,353 99.9999999148534037050721500 % left
9,137,607,100 99.9999999009298653086103000 % left
10,467,975,104 99.9999998865059865071902600 % left
11,813,910,300 99.9999998719133278719112300 % left
13,183,196,499 99.9999998570674971548090400 % left
...it continues on and on...
1,362,032,97 - difference between the 2nd and 3rd values (1 second)
6,771,768,529 seconds - how many seconds it would take to reach long's max value (Long.MAX_VALUE / 2nd3rdDifference)
6,771,768,529 seconds = 214.73 years (per conversion by google search)
So if my calculations are correct...you'd be dead of old age by the time an average computer calculated the max value of long via incrementing and checking if it's overflowed. Your children would be dead to. Your grandchildren, they might be around when it finished...
Code for Max Value Calculation
import java.math.BigDecimal;
import java.math.RoundingMode;
import java.text.NumberFormat;
public class MainLongMaxTest {
// /*
public static final long MAX_VALUE = Long.MAX_VALUE;
public static long value = 0;
public static long previousValue = 0;
// */
/*
public static final int MAX_VALUE = Integer.MAX_VALUE;
public static int value = 0;
public static int previousValue = 0;
*/
public static boolean done;
public static BigDecimal startTime;
public static BigDecimal endTime;
public static void main(String[] args) {
Runnable task = new StatusPrinterRunnable();
new Thread(task).start(); // code waits 1 second before result printing loop
done = false;
startTime = new BigDecimal(System.currentTimeMillis());
while(value >= 0) {
previousValue = value;
value += 1;
}
endTime = new BigDecimal(System.currentTimeMillis());
done = true;
}
}
class StatusPrinterRunnable implements Runnable {
public static final NumberFormat numberFormat = NumberFormat.getNumberInstance();
private static long SLEEP_TIME = 1000;
#Override
public void run() {
try { Thread.sleep(SLEEP_TIME); } catch (InterruptedException e) { throw new RuntimeException(e); }
while(!MainLongMaxTest.done) {
long value = MainLongMaxTest.value;
//long valuesLeft = MAX_VALUE - value;
BigDecimal maxValueBd = new BigDecimal(MainLongMaxTest.MAX_VALUE);
BigDecimal valueBd = new BigDecimal(value);
BigDecimal differenceBd = maxValueBd.subtract(valueBd);
BigDecimal percentLeftBd = differenceBd.divide(maxValueBd, 25, RoundingMode.HALF_DOWN);
percentLeftBd = percentLeftBd.multiply(new BigDecimal(100));
String numberAsString = numberFormat.format(value);
String percentLeftAsString = percentLeftBd.toString();
String message = "" + numberAsString + "\t" + percentLeftAsString + " % left";
System.out.println(message);
try { Thread.sleep(SLEEP_TIME); } catch (InterruptedException e) { throw new RuntimeException(e); }
}
BigDecimal msTaken = MainLongMaxTest.endTime.subtract(MainLongMaxTest.startTime);
BigDecimal secondsTaken = msTaken.divide(new BigDecimal("1000"));
System.out.println();
System.out.println("Max Value: " + numberFormat.format(MainLongMaxTest.previousValue));
System.out.println("Time Taken (seconds): " + secondsTaken);
}
}
I think your logic is correct just it will take a lot of time to reach that value.
the maximum Long value can hold is Long.MAX_value which is 9223372036854775807L
to speed up the logic, I modified the program as below and got the expected result.
public static void main(String args[]) {
long i = 9223372036854775806L;
long result = 0L;
for (; ; ) {
result = i++;
if (i<0) {
System.out.println("result"+result);
System.out.println("i"+i);
break;
}
}
System.out.println("The biggest integer: is" + result);
}
Output:
result9223372036854775807
i-9223372036854775808
The biggest integer: is9223372036854775807
result has the maximum value it can hold after that it changes to its minimum value.
You can get the result in one step if you take advantage of binary algebra by:
result = -1L >>> 1;
How could I convert from float to string or string to float?
In my case I need to make the assertion between 2 values string (value that I have got from table) and float value that I have calculated.
String valueFromTable = "25";
Float valueCalculated =25.0;
I tried from float to string:
String sSelectivityRate = String.valueOf(valueCalculated);
but the assertion fails
Using Java’s Float class.
float f = Float.parseFloat("25");
String s = Float.toString(25.0f);
To compare it's always better to convert the string to float and compare as two floats. This is because for one float number there are multiple string representations, which are different when compared as strings (e.g. "25" != "25.0" != "25.00" etc.)
Float to string - String.valueOf()
float amount=100.00f;
String strAmount=String.valueOf(amount);
// or Float.toString(float)
String to Float - Float.parseFloat()
String strAmount="100.20";
float amount=Float.parseFloat(strAmount)
// or Float.valueOf(string)
You can try this sample of code:
public class StringToFloat
{
public static void main (String[] args)
{
// String s = "fred"; // do this if you want an exception
String s = "100.00";
try
{
float f = Float.valueOf(s.trim()).floatValue();
System.out.println("float f = " + f);
}
catch (NumberFormatException nfe)
{
System.out.println("NumberFormatException: " + nfe.getMessage());
}
}
}
found here
I believe the following code will help:
float f1 = 1.23f;
String f1Str = Float.toString(f1);
float f2 = Float.parseFloat(f1Str);
This is a possible answer, this will also give the precise data, just need to change the decimal point in the required form.
public class TestStandAlone {
/**
* This method is to main
* #param args void
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
try {
Float f1=152.32f;
BigDecimal roundfinalPrice = new BigDecimal(f1.floatValue()).setScale(2,BigDecimal.ROUND_HALF_UP);
System.out.println("f1 --> "+f1);
String s1=roundfinalPrice.toPlainString();
System.out.println("s1 "+s1);
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
Output will be
f1 --> 152.32
s1 152.32
If you're looking for, say two decimal places..
Float f = (float)12.34;
String s = new DecimalFormat ("#.00").format (f);
well this method is not a good one, but easy and not suggested. Maybe i should say this is the least effective method and the worse coding practice but, fun to use,
float val=10.0;
String str=val+"";
the empty quotes, add a null string to the variable str, upcasting 'val' to the string type.
There are three ways to convert float to String.
"" + f
Float.toString(f)
String.valueOf(f)
There are two ways Convert String to float
Float.valueOf(str)
Float.parseFloat(str);
Example:-
public class Test {
public static void main(String[] args) {
System.out.println("convert FloatToString " + convertFloatToString(34.0f));
System.out.println("convert FloatToStr Using Float Method " + convertFloatToStrUsingFloatMethod(23.0f));
System.out.println("convert FloatToStr Using String Method " + convertFloatToStrUsingFloatMethod(233.0f));
float f = Float.valueOf("23.00");
}
public static String convertFloatToString(float f) {
return "" + f;
}
public static String convertFloatToStrUsingFloatMethod(float f) {
return Float.toString(f);
}
public static String convertFloatToStrUsingStringMethod(float f) {
return String.valueOf(f);
}
}
String str = "1234.56";
float num = 0.0f;
int digits = str.length()- str.indexOf('.') - 1;
float factor = 1f;
for(int i=0;i<digits;i++) factor /= 10;
for(int i=str.length()-1;i>=0;i--){
if(str.charAt(i) == '.'){
factor = 1;
System.out.println("Reset, value="+num);
continue;
}
num += (str.charAt(i) - '0') * factor;
factor *= 10;
}
System.out.println(num);
To go the full manual route: This method converts doubles to strings by shifting the number's decimal point around and using floor (to long) and modulus to extract the digits. Also, it uses counting by base division to figure out the place where the decimal point belongs. It can also "delete" higher parts of the number once it reaches the places after the decimal point, to avoid losing precision with ultra-large doubles. See commented code at the end. In my testing, it is never less precise than the Java float representations themselves, when they actually show these imprecise lower decimal places.
/**
* Convert the given double to a full string representation, i.e. no scientific notation
* and always twelve digits after the decimal point.
* #param d The double to be converted
* #return A full string representation
*/
public static String fullDoubleToString(final double d) {
// treat 0 separately, it will cause problems on the below algorithm
if (d == 0) {
return "0.000000000000";
}
// find the number of digits above the decimal point
double testD = Math.abs(d);
int digitsBeforePoint = 0;
while (testD >= 1) {
// doesn't matter that this loses precision on the lower end
testD /= 10d;
++digitsBeforePoint;
}
// create the decimal digits
StringBuilder repr = new StringBuilder();
// 10^ exponent to determine divisor and current decimal place
int digitIndex = digitsBeforePoint;
double dabs = Math.abs(d);
while (digitIndex > 0) {
// Recieves digit at current power of ten (= place in decimal number)
long digit = (long)Math.floor(dabs / Math.pow(10, digitIndex-1)) % 10;
repr.append(digit);
--digitIndex;
}
// insert decimal point
if (digitIndex == 0) {
repr.append(".");
}
// remove any parts above the decimal point, they create accuracy problems
long digit = 0;
dabs -= (long)Math.floor(dabs);
// Because of inaccuracy, move to entirely new system of computing digits after decimal place.
while (digitIndex > -12) {
// Shift decimal point one step to the right
dabs *= 10d;
final var oldDigit = digit;
digit = (long)Math.floor(dabs) % 10;
repr.append(digit);
// This may avoid float inaccuracy at the very last decimal places.
// However, in practice, inaccuracy is still as high as even Java itself reports.
// dabs -= oldDigit * 10l;
--digitIndex;
}
return repr.insert(0, d < 0 ? "-" : "").toString();
}
Note that while StringBuilder is used for speed, this method can easily be rewritten to use arrays and therefore also work in other languages.