I am trying to beautify a program by displaying 1.2 if it is 1.2 and 1 if it is 1 problem is I have stored the numbers into the arraylist as doubles. How can I check if a Number is a double or int?
Well, you can use:
if (x == Math.floor(x))
or even:
if (x == (long) x) // Performs truncation in the conversion
If the condition is true, i.e. the body of the if statement executes, then the value is an integer. Otherwise, it's not.
Note that this will view 1.00000000001 as still a double - if these are values which have been computed (and so may just be "very close" to integer values) you may want to add some tolerance. Also note that this will start failing for very large integers, as they can't be exactly represented in double anyway - you may want to consider using BigDecimal instead if you're dealing with a very wide range.
EDIT: There are better ways of approaching this - using DecimalFormat you should be able to get it to only optionally produce the decimal point. For example:
import java.text.*;
public class Test
{
public static void main(String[] args)
{
DecimalFormat df = new DecimalFormat("0.###");
double[] values = { 1.0, 3.5, 123.4567, 10.0 };
for (double value : values)
{
System.out.println(df.format(value));
}
}
}
Output:
1
3.5
123.457
10
Another simple & intuitive solution using the modulus operator (%)
if (x % 1 == 0) // true: it's an integer, false: it's not an integer
I am C# programmer so I tested this in .Net. This should work in Java too (other than the lines that use the Console class to display the output.
class Program
{
static void Main(string[] args)
{
double[] values = { 1.0, 3.5, 123.4567, 10.0, 1.0000000003 };
int num = 0;
for (int i = 0; i < values.Length; i++ )
{
num = (int) values[i];
// compare the difference against a very small number to handle
// issues due floating point processor
if (Math.Abs(values[i] - (double) num) < 0.00000000001)
{
Console.WriteLine(num);
}
else // print as double
{
Console.WriteLine(values[i]);
}
}
Console.Read();
}
}
Alternatively one can use this method too, I found it helpful.
double a = 1.99;
System.out.println(Math.floor(a) == Math.ceil(a));
You can use:
double x=4;
//To check if it is an integer.
return (int)x == x;
Related
When analysing data sets, such as data for human heights or for human weights, a common step is to adjust the data. This adjustment can be done by normalizing to values between 0 and 1, or throwing away outliers.
For this program, adjust the values by dividing all values by the largest value. The input begins with an integer indicating the number of floating-point values that follow. Assume that the list will always contain fewer than 20 floating-point values.
Output each floating-point value with two digits after the decimal point, which can be achieved as follows:
System.out.printf("%.2f", yourValue);
Ex: If the input is:
5 30.0 50.0 10.0 100.0 65.0
the output is:
0.30 0.50 0.10 1.00 0.65
The 5 indicates that there are five floating-point values in the list, namely 30.0, 50.0, 10.0, 100.0, and 65.0. 100.0 is the largest value in the list, so each value is divided by 100.0.
For coding simplicity, follow every output value by a space, including the last one.
This is my code so far:
import java.util.Scanner;
public class LabProgram {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
double numElements;
numElements = scnr.nextDouble();
double[] userList = new double[numElements];
int i;
double maxValue;
for (i = 0; i < userList.length; ++i) {
userList[i] = scnr.nextDouble();
}
maxValue = userList[i];
for (i = 0; i < userList.length; ++i) {
if (userList[i] > maxValue) {
maxValue = userList[i];
}
}
for (i = 0; i < userList.length; ++i) {
userList[i] = userList[i] / maxValue;
System.out.print(userList[i] + " ");
System.out.printf("%.2f", userList[i]);
}
}
}
I keep getting this output.
LabProgram.java:8: error: incompatible types: possible lossy conversion from double to int
double [] userList = new double [numElements];
^
1 error
I think my variable is messed up. I read through my book and could not find help. Can someone please help me on here. Thank you so much! This has been very stressful for me.
The specific error message is because the index and size of an element must be int. So declare and assign at once: int numElements = scnr.nextInt();
Better way of programming things:
skip manual input (aka Scanner and consorts). Makes you crazy and testing a 100'000'000 times slower
you can integrate the interactive part later, once the method is done. You already know how, your code already shows.
use an explicit method to do your work. Don't throw everything into the main method. This way you can run multiple examples/tests on the method, and you have a better implementation for later.
check for invalid input INSIDE the method that you implement. Once you can rely in such a method, you can keep on using it later on.
you could even move the example numbers to its own test method, so you can run multiple test methods. You will learn about Unit Testing later on.
Example code:
public class LabProgram {
public static void main(final String[] args) {
final double[] initialValues = new double[] { 30.0, 50.0, 10.0, 100.0, 65.0 };
final double[] adjustedValues = normalizeValuesByHighest(initialValues);
System.out.println("Adjusted values:");
for (final double d : adjustedValues) {
System.out.printf("%.2f ", Double.valueOf(d));
}
// expected otuput is 0.30 0.50 0.10 1.00 0.65
System.out.println();
System.out.println("All done.");
}
static public double[] normalizeValuesByHighest(final double[] pInitialValues) {
if (pInitialValues == null) throw new IllegalArgumentException("Invalid double[] given!");
if (pInitialValues.length < 1) throw new IllegalArgumentException("double[] given contains no elements!");
// detect valid max value
double tempMaxValue = -Double.MAX_VALUE;
boolean hasValues = false;
for (final double d : pInitialValues) {
if (Double.isNaN(d)) continue;
tempMaxValue = Math.max(tempMaxValue, d);
hasValues = true;
}
if (!hasValues) throw new IllegalArgumentException("double[] given contains no valid elements, only NaNs!");
// create return array
final double maxValue = tempMaxValue; // final from here on
final double[] ret = new double[pInitialValues.length];
for (int i = 0; i < pInitialValues.length; i++) {
ret[i] = pInitialValues[i] / maxValue; // NaN will stay NaN
}
return ret;
}
}
Output:
Adjusted values:
0,30 0,50 0,10 1,00 0,65
All done.
Edit: This has to do with how computers handle floating point operations, a fact that every programmer faces once in a lifetime. I didn't understand this correctly when I asked the question.
I know the simplest way to start dealing with this would be:
val floatNumber: Float = 123.456f
val decimalPart = floatNumber - floatNumber.toInt() //This would be 0.456 (I don't care about precision as this is not the main objective of my question)
Now in a real world with a pen and a piece of paper, if I want to "convert" the decimal part 0.456 to integer, I just need to multiply 0.456 * 1000, and I get the desired result, which is 456 (an integer number).
Many proposed solutions suggest splitting the number as string and extracting the decimal part this way, but I need the solution to be obtained mathematically, not using strings.
Given a number, with an unknown number of decimals (convert to string and counting chars after . or , is not acceptable), I need to "extract" it's decimal part as an integer using only math.
Read questions like this with no luck:
How to get the decimal part of a float?
How to extract fractional digits of double/BigDecimal
If someone knows a kotlin language solution, it would be great. I will post this question also on the math platform just in case.
How do I get whole and fractional parts from double in JSP/Java?
Update:
Is there a "mathematical" way to "calculate" how many decimals a number has? (It is obvious when you convert to string and count the chars, but I need to avoid using strings) It would be great cause calculating: decimal (0.456) * 10 * number of decimals(3) will produce the desired result.
Update 2
This is not my use-case, but I guess it will clarify the idea:
Suppose you want to calculate a constant(such as PI), and want to return an integer with at most 50 digits of the decimal part of the constant. The constant doesn't have to be necessarily infinite (can be for example 0.5, in which case "5" will be returned)
I would just multiply the fractional number by 10 (or move the decimal point to the right) until it has no fractional part left:
public static long fractionalDigitsLong(BigDecimal value) {
BigDecimal fractional = value.remainder(BigDecimal.ONE);
long digits;
do {
fractional = fractional.movePointRight(1); // or multiply(BigDecimal.TEN)
digits = fractional.longValue();
} while (fractional.compareTo(BigDecimal.valueOf(digits)) != 0);
return digits;
}
Note 1: using BigDecimal to avoid floating point precision problems
Note 2: using compareTo since equals also compares the scale ("0.0" not equals "0.00")
(sure the BigDecimal already knows the size of the fractional part, just the value returned by scale())
Complement:
If using BigDecimal the whole problem can be compressed to:
public static BigInteger fractionalDigits(BigDecimal value) {
return value.remainder(BigDecimal.ONE).stripTrailingZeros().unscaledValue();
}
stripping zeros can be suppressed if desired
I am not sure if it counts against you on this specific problem if you use some String converters with a method(). That is one way to get the proper answer. I know that you stated you couldn't use String, but would you be able to use Strings within a Custom made method? That could get you the answer that you need with precision. Here is the class that could help us convert the number:
class NumConvert{
String theNum;
public NumConvert(String theNum) {
this.theNum = theNum;
}
public int convert() {
String a = String.valueOf(theNum);
String[] b = a.split("\\.");
String b2 = b[1];
int zeros = b2.length();
String num = "1";
for(int x = 0; x < zeros; x++) {
num += "0";
}
float c = Float.parseFloat(theNum);
int multiply = Integer.parseInt(num);
float answer = c - (int)c;
int integerForm = (int)(answer * multiply);
return integerForm;
}
}
Then within your main class:
public class ChapterOneBasics {
public static void main(String[] args) throws java.io.IOException{
NumConvert n = new NumConvert("123.456");
NumConvert q = new NumConvert("123.45600128");
System.out.println(q.convert());
System.out.println(n.convert());
}
}
output:
45600128
456
Float or Double are imprecise, just an approximation - without precision. Hence 12.345 is somewhere between 12.3449... and 12.3450... .
This means that 12.340 cannot be distinghuished from 12.34. The "decimal part" would be 34 divided by 100.
Also 12.01 would have a "decimal part" 1 divided by 100, and too 12.1 would have 1 divided by 10.
So a complete algorith would be (using java):
int[] decimalsAndDivider(double x) {
int decimalPart = 0;
int divider = 1;
final double EPS = 0.001;
for (;;) {
double error = x - (int)x;
if (-EPS < error && error < EPS) {
break;
}
x *= 10;
decimalPart = 10 * decimalPart + ((int)(x + EPS) % 10);
divider *= 10;
}
return new int[] { decimalPart, divider };
}
I posted the below solution yesterday after testing it for a while, and later found that it does not always work due to problems regarding precision of floats, doubles and bigdecimals. My conclusion is that this problem is unsolvable if you want infinite precision:
So I re-post the code just for reference:
fun getDecimalCounter(d: Double): Int {
var temp = d
var tempInt = Math.floor(d)
var counter = 0
while ((temp - tempInt) > 0.0 ) {
temp *= 10
tempInt = Math.floor(temp)
counter++
}
return counter
}
fun main(args: Array <String> ) {
var d = 3.14159
if (d < 0) d = -d
val decimalCounter = getDecimalCounter(d)
val decimalPart = (d - Math.floor(d))
var decimalPartInt = Math.round(decimalPart * 10.0.pow(decimalCounter))
while (decimalPartInt % 10 == 0L) {
decimalPartInt /= 10
}
println(decimalPartInt)
}
I dropped floats because of lesser precision and used doubles.
The final rounding is also necessary due to precision.
A simple comparison of two double values in Java creates some problems. Let's consider the following simple code snippet in Java.
package doublecomparision;
final public class DoubleComparision
{
public static void main(String[] args)
{
double a = 1.000001;
double b = 0.000001;
System.out.println("\n"+((a-b)==1.0));
}
}
The above code appears to return true, the evaluation of the expression ((a-b)==1.0) but it doesn't. It returns false instead because the evaluation of this expression is 0.9999999999999999 which was actually expected to be 1.0 which is not equal to 1.0 hence, the condition evaluates to boolean false. What is the best and suggested way to overcome such a situation?
Basically you shouldn't do exact comparisons, you should do something like this:
double a = 1.000001;
double b = 0.000001;
double c = a-b;
if (Math.abs(c-1.0) <= 0.000001) {...}
Instead of using doubles for decimal arithemetic, please use java.math.BigDecimal. It would produce the expected results.
For reference take a look at this stackoverflow question
You can use Double.compare; It compares the two specified double values.
int mid = 10;
for (double j = 2 * mid; j >= 0; j = j - 0.1) {
if (j == mid) {
System.out.println("Never happens"); // is NOT printed
}
if (Double.compare(j, mid) == 0) {
System.out.println("No way!"); // is NOT printed
}
if (Math.abs(j - mid) < 1e-6) {
System.out.println("Ha!"); // printed
}
}
System.out.println("Gotcha!");
Consider this line of code:
Math.abs(firstDouble - secondDouble) < Double.MIN_NORMAL
It returns whether firstDouble is equal to secondDouble. I'm unsure as to whether or not this would work in your exact case (as Kevin pointed out, performing any math on floating points can lead to imprecise results) however I was having difficulties with comparing two double which were, indeed, equal, and yet using the 'compareTo' method didn't return 0.
I'm just leaving this there in case anyone needs to compare to check if they are indeed equal, and not just similar.
Just use Double.compare() method to compare double values.
Double.compare((d1,d2) == 0)
double d1 = 0.0;
double d2 = 0.0;
System.out.println(Double.compare((d1,d2) == 0)) // true
double a = 1.000001;
double b = 0.000001;
System.out.println( a.compareTo(b) );
Returns:
-1 : 'a' is numerically less than 'b'.
0 : 'a' is equal to 'b'.
1 : 'a' is greater than 'b'.
I want to find the closest fraction equal to 16/76. Whenever i run this i get 1 no matter what. I am doing this for Java class in school.
public class ClassOne {
public static double limit = 16/76;
public static double difference = 1;
public static double numer = 1;
public static double denom = 1;
public static void main(String[] args)
{
for(int i = 1;i<=100;i++)
{
for(int x = 1;x<=100;x++)
{
double temp = limit-(double)(x/i);
System.out.println((x/i));
if(Math.abs(temp) < difference && x/i != 16/76){difference = temp;numer = x; denom = i;
System.out.println("hi");}
}
}
System.out.println(numer + " " + denom);
}
}
A few problems
Use (double)x/(double)i where needed - probably assign to a temp
difference=temp should be difference=Math.abs(temp) or put the abs on the original temp computation
x/i = 16/76 is subject to floating point errors and so may not be hit when you want. May want to use something like 16*i != 76*x which can be computed in integers. I get 21 100.
You're encountering integer division here:
limit - (double)(x/i)
...and here:
public static double limit = 16/76;
Both x and i are int. The cast here will take effect after the division operation has taken effect.
Change your cast so that it applies immediately to one of the variables instead:
limit - ((double)x)/i
Also, consider wherever else you're doing any quotient that isn't with a floating point number (i.e. has a decimal after it or is explicitly cast to a double) - if you need it to be a floating point number, then use the appropriate cast.
A simple comparison of two double values in Java creates some problems. Let's consider the following simple code snippet in Java.
package doublecomparision;
final public class DoubleComparision
{
public static void main(String[] args)
{
double a = 1.000001;
double b = 0.000001;
System.out.println("\n"+((a-b)==1.0));
}
}
The above code appears to return true, the evaluation of the expression ((a-b)==1.0) but it doesn't. It returns false instead because the evaluation of this expression is 0.9999999999999999 which was actually expected to be 1.0 which is not equal to 1.0 hence, the condition evaluates to boolean false. What is the best and suggested way to overcome such a situation?
Basically you shouldn't do exact comparisons, you should do something like this:
double a = 1.000001;
double b = 0.000001;
double c = a-b;
if (Math.abs(c-1.0) <= 0.000001) {...}
Instead of using doubles for decimal arithemetic, please use java.math.BigDecimal. It would produce the expected results.
For reference take a look at this stackoverflow question
You can use Double.compare; It compares the two specified double values.
int mid = 10;
for (double j = 2 * mid; j >= 0; j = j - 0.1) {
if (j == mid) {
System.out.println("Never happens"); // is NOT printed
}
if (Double.compare(j, mid) == 0) {
System.out.println("No way!"); // is NOT printed
}
if (Math.abs(j - mid) < 1e-6) {
System.out.println("Ha!"); // printed
}
}
System.out.println("Gotcha!");
Consider this line of code:
Math.abs(firstDouble - secondDouble) < Double.MIN_NORMAL
It returns whether firstDouble is equal to secondDouble. I'm unsure as to whether or not this would work in your exact case (as Kevin pointed out, performing any math on floating points can lead to imprecise results) however I was having difficulties with comparing two double which were, indeed, equal, and yet using the 'compareTo' method didn't return 0.
I'm just leaving this there in case anyone needs to compare to check if they are indeed equal, and not just similar.
Just use Double.compare() method to compare double values.
Double.compare((d1,d2) == 0)
double d1 = 0.0;
double d2 = 0.0;
System.out.println(Double.compare((d1,d2) == 0)) // true
double a = 1.000001;
double b = 0.000001;
System.out.println( a.compareTo(b) );
Returns:
-1 : 'a' is numerically less than 'b'.
0 : 'a' is equal to 'b'.
1 : 'a' is greater than 'b'.