Related
How do I generate a random int value in a specific range?
The following methods have bugs related to integer overflow:
randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum = minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
In Java 1.7 or later, the standard way to do this is as follows:
import java.util.concurrent.ThreadLocalRandom;
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);
See the relevant JavaDoc. This approach has the advantage of not needing to explicitly initialize a java.util.Random instance, which can be a source of confusion and error if used inappropriately.
However, conversely there is no way to explicitly set the seed so it can be difficult to reproduce results in situations where that is useful such as testing or saving game states or similar. In those situations, the pre-Java 1.7 technique shown below can be used.
Before Java 1.7, the standard way to do this is as follows:
import java.util.Random;
/**
* Returns a pseudo-random number between min and max, inclusive.
* The difference between min and max can be at most
* <code>Integer.MAX_VALUE - 1</code>.
*
* #param min Minimum value
* #param max Maximum value. Must be greater than min.
* #return Integer between min and max, inclusive.
* #see java.util.Random#nextInt(int)
*/
public static int randInt(int min, int max) {
// NOTE: This will (intentionally) not run as written so that folks
// copy-pasting have to think about how to initialize their
// Random instance. Initialization of the Random instance is outside
// the main scope of the question, but some decent options are to have
// a field that is initialized once and then re-used as needed or to
// use ThreadLocalRandom (if using at least Java 1.7).
//
// In particular, do NOT do 'Random rand = new Random()' here or you
// will get not very good / not very random results.
Random rand;
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
See the relevant JavaDoc. In practice, the java.util.Random class is often preferable to java.lang.Math.random().
In particular, there is no need to reinvent the random integer generation wheel when there is a straightforward API within the standard library to accomplish the task.
Note that this approach is more biased and less efficient than a nextInt approach, https://stackoverflow.com/a/738651/360211
One standard pattern for accomplishing this is:
Min + (int)(Math.random() * ((Max - Min) + 1))
The Java Math library function Math.random() generates a double value in the range [0,1). Notice this range does not include the 1.
In order to get a specific range of values first, you need to multiply by the magnitude of the range of values you want covered.
Math.random() * ( Max - Min )
This returns a value in the range [0,Max-Min), where 'Max-Min' is not included.
For example, if you want [5,10), you need to cover five integer values so you use
Math.random() * 5
This would return a value in the range [0,5), where 5 is not included.
Now you need to shift this range up to the range that you are targeting. You do this by adding the Min value.
Min + (Math.random() * (Max - Min))
You now will get a value in the range [Min,Max). Following our example, that means [5,10):
5 + (Math.random() * (10 - 5))
But, this still doesn't include Max and you are getting a double value. In order to get the Max value included, you need to add 1 to your range parameter (Max - Min) and then truncate the decimal part by casting to an int. This is accomplished via:
Min + (int)(Math.random() * ((Max - Min) + 1))
And there you have it. A random integer value in the range [Min,Max], or per the example [5,10]:
5 + (int)(Math.random() * ((10 - 5) + 1))
Use:
Random ran = new Random();
int x = ran.nextInt(6) + 5;
The integer x is now the random number that has a possible outcome of 5-10.
Use:
minValue + rn.nextInt(maxValue - minValue + 1)
With java-8 they introduced the method ints(int randomNumberOrigin, int randomNumberBound) in the Random class.
For example if you want to generate five random integers (or a single one) in the range [0, 10], just do:
Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();
The first parameter indicates just the size of the IntStream generated (which is the overloaded method of the one that produces an unlimited IntStream).
If you need to do multiple separate calls, you can create an infinite primitive iterator from the stream:
public final class IntRandomNumberGenerator {
private PrimitiveIterator.OfInt randomIterator;
/**
* Initialize a new random number generator that generates
* random numbers in the range [min, max]
* #param min - the min value (inclusive)
* #param max - the max value (inclusive)
*/
public IntRandomNumberGenerator(int min, int max) {
randomIterator = new Random().ints(min, max + 1).iterator();
}
/**
* Returns a random number in the range (min, max)
* #return a random number in the range (min, max)
*/
public int nextInt() {
return randomIterator.nextInt();
}
}
You can also do it for double and long values.
You can edit your second code example to:
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
Just a small modification of your first solution would suffice.
Random rand = new Random();
randomNum = minimum + rand.nextInt((maximum - minimum) + 1);
See more here for implementation of Random
ThreadLocalRandom equivalent of class java.util.Random for multithreaded environment. Generating a random number is carried out locally in each of the threads. So we have a better performance by reducing the conflicts.
int rand = ThreadLocalRandom.current().nextInt(x,y);
x,y - intervals e.g. (1,10)
The Math.Random class in Java is 0-based. So, if you write something like this:
Random rand = new Random();
int x = rand.nextInt(10);
x will be between 0-9 inclusive.
So, given the following array of 25 items, the code to generate a random number between 0 (the base of the array) and array.length would be:
String[] i = new String[25];
Random rand = new Random();
int index = 0;
index = rand.nextInt( i.length );
Since i.length will return 25, the nextInt( i.length ) will return a number between the range of 0-24. The other option is going with Math.Random which works in the same way.
index = (int) Math.floor(Math.random() * i.length);
For a better understanding, check out forum post Random Intervals (archive.org).
It can be done by simply doing the statement:
Randomizer.generate(0, 10); // Minimum of zero and maximum of ten
Below is its source code.
File Randomizer.java
public class Randomizer {
public static int generate(int min, int max) {
return min + (int)(Math.random() * ((max - min) + 1));
}
}
It is just clean and simple.
Forgive me for being fastidious, but the solution suggested by the majority, i.e., min + rng.nextInt(max - min + 1)), seems perilous due to the fact that:
rng.nextInt(n) cannot reach Integer.MAX_VALUE.
(max - min) may cause overflow when min is negative.
A foolproof solution would return correct results for any min <= max within [Integer.MIN_VALUE, Integer.MAX_VALUE]. Consider the following naive implementation:
int nextIntInRange(int min, int max, Random rng) {
if (min > max) {
throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
}
int diff = max - min;
if (diff >= 0 && diff != Integer.MAX_VALUE) {
return (min + rng.nextInt(diff + 1));
}
int i;
do {
i = rng.nextInt();
} while (i < min || i > max);
return i;
}
Although inefficient, note that the probability of success in the while loop will always be 50% or higher.
I wonder if any of the random number generating methods provided by an Apache Commons Math library would fit the bill.
For example: RandomDataGenerator.nextInt or RandomDataGenerator.nextLong
I use this:
/**
* #param min - The minimum.
* #param max - The maximum.
* #return A random double between these numbers (inclusive the minimum and maximum).
*/
public static double getRandom(double min, double max) {
return (Math.random() * (max + 1 - min)) + min;
}
You can cast it to an Integer if you want.
As of Java 7, you should no longer use Random. For most uses, the
random number generator of choice is now
ThreadLocalRandom.For fork join pools and parallel
streams, use SplittableRandom.
Joshua Bloch. Effective Java. Third Edition.
Starting from Java 8
For fork join pools and parallel streams, use SplittableRandom that is usually faster, has a better statistical independence and uniformity properties in comparison with Random.
To generate a random int in the range [0, 1_000]:
int n = new SplittableRandom().nextInt(0, 1_001);
To generate a random int[100] array of values in the range [0, 1_000]:
int[] a = new SplittableRandom().ints(100, 0, 1_001).parallel().toArray();
To return a Stream of random values:
IntStream stream = new SplittableRandom().ints(100, 0, 1_001);
rand.nextInt((max+1) - min) + min;
Let us take an example.
Suppose I wish to generate a number between 5-10:
int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);
Let us understand this...
Initialize max with highest value and min with the lowest value.
Now, we need to determine how many possible values can be obtained. For this example, it would be:
5, 6, 7, 8, 9, 10
So, count of this would be max - min + 1.
i.e. 10 - 5 + 1 = 6
The random number will generate a number between 0-5.
i.e. 0, 1, 2, 3, 4, 5
Adding the min value to the random number would produce:
5, 6, 7, 8, 9, 10
Hence we obtain the desired range.
Generate a random number for the difference of min and max by using the nextint(n) method and then add min number to the result:
Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
To generate a random number "in between two numbers", use the following code:
Random r = new Random();
int lowerBound = 1;
int upperBound = 11;
int result = r.nextInt(upperBound-lowerBound) + lowerBound;
This gives you a random number in between 1 (inclusive) and 11 (exclusive), so initialize the upperBound value by adding 1. For example, if you want to generate random number between 1 to 10 then initialize the upperBound number with 11 instead of 10.
Just use the Random class:
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
These methods might be convenient to use:
This method will return a random number between the provided minimum and maximum value:
public static int getRandomNumberBetween(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt(max - min) + min;
if (randomNumber == min) {
// Since the random number is between the min and max values, simply add 1
return min + 1;
} else {
return randomNumber;
}
}
and this method will return a random number from the provided minimum and maximum value (so the generated number could also be the minimum or maximum number):
public static int getRandomNumberFrom(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt((max + 1) - min) + min;
return randomNumber;
}
In case of rolling a dice it would be random number between 1 to 6 (not 0 to 6), so:
face = 1 + randomNumbers.nextInt(6);
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();
Or take a look to RandomUtils from Apache Commons.
You can achieve that concisely in Java 8:
Random random = new Random();
int max = 10;
int min = 5;
int totalNumber = 10;
IntStream stream = random.ints(totalNumber, min, max);
stream.forEach(System.out::println);
Here's a helpful class to generate random ints in a range with any combination of inclusive/exclusive bounds:
import java.util.Random;
public class RandomRange extends Random {
public int nextIncInc(int min, int max) {
return nextInt(max - min + 1) + min;
}
public int nextExcInc(int min, int max) {
return nextInt(max - min) + 1 + min;
}
public int nextExcExc(int min, int max) {
return nextInt(max - min - 1) + 1 + min;
}
public int nextIncExc(int min, int max) {
return nextInt(max - min) + min;
}
}
Another option is just using Apache Commons:
import org.apache.commons.math.random.RandomData;
import org.apache.commons.math.random.RandomDataImpl;
public void method() {
RandomData randomData = new RandomDataImpl();
int number = randomData.nextInt(5, 10);
// ...
}
I found this example Generate random numbers :
This example generates random integers in a specific range.
import java.util.Random;
/** Generate random integers in a certain range. */
public final class RandomRange {
public static final void main(String... aArgs){
log("Generating random integers in the range 1..10.");
int START = 1;
int END = 10;
Random random = new Random();
for (int idx = 1; idx <= 10; ++idx){
showRandomInteger(START, END, random);
}
log("Done.");
}
private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
if ( aStart > aEnd ) {
throw new IllegalArgumentException("Start cannot exceed End.");
}
//get the range, casting to long to avoid overflow problems
long range = (long)aEnd - (long)aStart + 1;
// compute a fraction of the range, 0 <= frac < range
long fraction = (long)(range * aRandom.nextDouble());
int randomNumber = (int)(fraction + aStart);
log("Generated : " + randomNumber);
}
private static void log(String aMessage){
System.out.println(aMessage);
}
}
An example run of this class :
Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.
public static Random RANDOM = new Random(System.nanoTime());
public static final float random(final float pMin, final float pMax) {
return pMin + RANDOM.nextFloat() * (pMax - pMin);
}
Here is a simple sample that shows how to generate random number from closed [min, max] range, while min <= max is true
You can reuse it as field in hole class, also having all Random.class methods in one place
Results example:
RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert
Sources:
import junit.framework.Assert;
import java.util.Random;
public class RandomUtils extends Random {
/**
* #param min generated value. Can't be > then max
* #param max generated value
* #return values in closed range [min, max].
*/
public int nextInt(int min, int max) {
Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
if (min == max) {
return max;
}
return nextInt(max - min + 1) + min;
}
}
It's better to use SecureRandom rather than just Random.
public static int generateRandomInteger(int min, int max) {
SecureRandom rand = new SecureRandom();
rand.setSeed(new Date().getTime());
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
rand.nextInt((max+1) - min) + min;
This is working fine.
This question already has answers here:
How to concatenate int values in java?
(22 answers)
Closed 3 years ago.
I am writing some code that gives me a random string of numbers, they need to list under on integer but each number needs to be under a different math.random. For Instance, if two separate number are listed like 5 and 7, I don't want it to print 12, I would like it to print 57. But i don't want to use the System.out.println(Number1+Number2); way.
I have tried using the "&" Sign multiple ways but none seem to work.
int Number1 = 1 + (int)(Math.random() * ((5) + 1));
int Number2 = 1 + (int)(Math.random() * ((5) + 1));
int Number3 = 1 + (int)(Math.random() * ((5) + 1));
int finalcode=Number1&Number2&Number3;
System.out.println("Promo Code Genorator:");
System.out.println(" ");
System.out.println("Your Promo Code Is: "+finalcode);
Instead, what happens is it picks the lowest number from there and prints them. Any Ideas?
It is suggested to use String if you want to combine a variety of numbers together.
You can write it like this:
int Number1 = 1 + (int)(Math.random() * ((5) + 1));
int Number2 = 1 + (int)(Math.random() * ((5) + 1));
int Number3 = 1 + (int)(Math.random() * ((5) + 1));
String finalcode = String.valueOf(Number1) + String.valueOf(Number2) + String.valueOf(Number3);
System.out.println("Promo Code Genorator:");
System.out.println(" ");
System.out.println("Your Promo Code Is: "+finalcode);
If you really need your final code to be a integer, you can use
int finalcode = Integer.parseInt(String.valueOf(Number1) + String.valueOf(Number2) + String.valueOf(Number3));
in which Integer.parseInt(String string) takes in a string and return a integer.
FYI, if you want to convert it to long instead of integer, use Long.parseLong(String string).
Hope this helps!
This seems like a nice problem to solve using streams and functional programming.
import java.util.stream.Stream;
import java.util.stream.Collectors;
public class LazyNumbersTest {
public static void main(String[] args) {
int result = Integer.parseInt(
Stream.generate(() -> (int)(Math.random() * ((5) + 1)))
.peek(System.out::println)
.limit(3)
.map(Object::toString)
.collect(Collectors.joining(""))
);
System.out.println("Promo Code Generator:");
System.out.println(" ");
System.out.println("Your Promo Code Is: " + result);
}
}
It's an infinite stream of random numbers, from which we pluck the first 3 and convert them to string, then join them together and convert to an integer.
If you want more or less numbers in the calculation, just change the number in limit. It's a shame I couldn't include the parseInt in the stream of fluent operations (without making it really ugly).
I am a beginner java programmer, studying off basic youtube videos and overflow forums, and i came across a question on an textbook sheet that asked me to use a for loop program and print this table out as well as fill in each blank
Number Square Cube (n+1)/(2n-18)
______________________________________
1 # # #
2 # # #
3 # # #
4 # # #
5 # # #
I thought I should try it out to test myself. I came up with the following program that works perfectly for the Number, Square and Cube part of the table, but I don't understand how to generate the numbers using the given formula. The formula I initialized as a variable (double) doesn't print the actual results, and I honestly don't have a clue as to what to do. I'd rather a simple explanation than a complex one and simple changes to the code rather than complex. As I said I am a beginner, and many different methods may go right over my head. Thank you so much in advance (also an extra task asks that I print out the sums of each column. I don't know how to do that at all, and would like an explanation if possible, but wouldn't mind if I don't receive one)
int number;
int maxValue;
Scanner keyboard = new Scanner(System.in);
System.out.println("how many numbers do you want the max value to be");
maxValue = keyboard.nextInt();
System.out.println("Number\tSquare\tCube (n+1)/(2n-18)");
System.out.println("--------------------------------");
for (number = 1; number <= maxValue; number++) {
double formula = (number + 1) / (number * 2);
System.out.println(
number + "\t\t\t" + number * number + "\t\t\t" +
number * number * number + "\t\t\t" + formula);
}
Your formula should be:
double formula = (double)(number + 1) / (number * 2 - 18);
Two issues:
missing -18
the / is doing an integer division unless you cast at least one operand into a double
Oh! one more thing: when number==9, there is a division by zero. A double division gives you "Infinity" whereas an integer division throws an exception.
Your formula does not match the textbook. Try this:
System.out.println("Number\tSquare\tCube (n+1)/(2n-18)");
System.out.println("--------------------------------");
for (int number=1; number <= maxValue; number++) {
double square = Math.pow(number, 2);
double cube = Math.pow(number, 3);
double formula = (number + 1) / (number * 2 - 18);
System.out.println(number + "\t\t\t" + square + "\t\t\t" + cube + "\t\t\t" + formula);
}
Note: As pointed out by #MauricePerry an input of 9 would cause a divide by zero. Rather than try to catch this unchecked exception, I think you should control your input values so that this does not happen.
Your formula is not correct and you are missing few \t.
int number;
int maxValue;
Scanner keyboard = new Scanner(System.in);
System.out.println("how many numbers do you want the max value to be");
maxValue = keyboard.nextInt();
System.out.println("Number\t\tSquare\t\tCube \t\t(n+1)/(2n-18)");
System.out.println("-------------------------------------------------------");
for (number = 1; number <= maxValue; number++) {
double formula = (number + 1) / (number * 2 - 18);
System.out
.println(number + "\t\t" + number * number + "\t\t" + number * number * number + "\t\t" + formula);
}
I'm not sure which outputs are you getting, but it seems to me you're using integer division when trying to get the result of (n+1)/(2n-18).
Try using:
double decimalNumber = (double) number;
double formula = (decimalNumber + 1) / (decimalNumber * 2 - 18);
How do I generate a random int value in a specific range?
The following methods have bugs related to integer overflow:
randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum = minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
In Java 1.7 or later, the standard way to do this is as follows:
import java.util.concurrent.ThreadLocalRandom;
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);
See the relevant JavaDoc. This approach has the advantage of not needing to explicitly initialize a java.util.Random instance, which can be a source of confusion and error if used inappropriately.
However, conversely there is no way to explicitly set the seed so it can be difficult to reproduce results in situations where that is useful such as testing or saving game states or similar. In those situations, the pre-Java 1.7 technique shown below can be used.
Before Java 1.7, the standard way to do this is as follows:
import java.util.Random;
/**
* Returns a pseudo-random number between min and max, inclusive.
* The difference between min and max can be at most
* <code>Integer.MAX_VALUE - 1</code>.
*
* #param min Minimum value
* #param max Maximum value. Must be greater than min.
* #return Integer between min and max, inclusive.
* #see java.util.Random#nextInt(int)
*/
public static int randInt(int min, int max) {
// NOTE: This will (intentionally) not run as written so that folks
// copy-pasting have to think about how to initialize their
// Random instance. Initialization of the Random instance is outside
// the main scope of the question, but some decent options are to have
// a field that is initialized once and then re-used as needed or to
// use ThreadLocalRandom (if using at least Java 1.7).
//
// In particular, do NOT do 'Random rand = new Random()' here or you
// will get not very good / not very random results.
Random rand;
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
See the relevant JavaDoc. In practice, the java.util.Random class is often preferable to java.lang.Math.random().
In particular, there is no need to reinvent the random integer generation wheel when there is a straightforward API within the standard library to accomplish the task.
Note that this approach is more biased and less efficient than a nextInt approach, https://stackoverflow.com/a/738651/360211
One standard pattern for accomplishing this is:
Min + (int)(Math.random() * ((Max - Min) + 1))
The Java Math library function Math.random() generates a double value in the range [0,1). Notice this range does not include the 1.
In order to get a specific range of values first, you need to multiply by the magnitude of the range of values you want covered.
Math.random() * ( Max - Min )
This returns a value in the range [0,Max-Min), where 'Max-Min' is not included.
For example, if you want [5,10), you need to cover five integer values so you use
Math.random() * 5
This would return a value in the range [0,5), where 5 is not included.
Now you need to shift this range up to the range that you are targeting. You do this by adding the Min value.
Min + (Math.random() * (Max - Min))
You now will get a value in the range [Min,Max). Following our example, that means [5,10):
5 + (Math.random() * (10 - 5))
But, this still doesn't include Max and you are getting a double value. In order to get the Max value included, you need to add 1 to your range parameter (Max - Min) and then truncate the decimal part by casting to an int. This is accomplished via:
Min + (int)(Math.random() * ((Max - Min) + 1))
And there you have it. A random integer value in the range [Min,Max], or per the example [5,10]:
5 + (int)(Math.random() * ((10 - 5) + 1))
Use:
Random ran = new Random();
int x = ran.nextInt(6) + 5;
The integer x is now the random number that has a possible outcome of 5-10.
Use:
minValue + rn.nextInt(maxValue - minValue + 1)
With java-8 they introduced the method ints(int randomNumberOrigin, int randomNumberBound) in the Random class.
For example if you want to generate five random integers (or a single one) in the range [0, 10], just do:
Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();
The first parameter indicates just the size of the IntStream generated (which is the overloaded method of the one that produces an unlimited IntStream).
If you need to do multiple separate calls, you can create an infinite primitive iterator from the stream:
public final class IntRandomNumberGenerator {
private PrimitiveIterator.OfInt randomIterator;
/**
* Initialize a new random number generator that generates
* random numbers in the range [min, max]
* #param min - the min value (inclusive)
* #param max - the max value (inclusive)
*/
public IntRandomNumberGenerator(int min, int max) {
randomIterator = new Random().ints(min, max + 1).iterator();
}
/**
* Returns a random number in the range (min, max)
* #return a random number in the range (min, max)
*/
public int nextInt() {
return randomIterator.nextInt();
}
}
You can also do it for double and long values.
You can edit your second code example to:
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
Just a small modification of your first solution would suffice.
Random rand = new Random();
randomNum = minimum + rand.nextInt((maximum - minimum) + 1);
See more here for implementation of Random
ThreadLocalRandom equivalent of class java.util.Random for multithreaded environment. Generating a random number is carried out locally in each of the threads. So we have a better performance by reducing the conflicts.
int rand = ThreadLocalRandom.current().nextInt(x,y);
x,y - intervals e.g. (1,10)
The Math.Random class in Java is 0-based. So, if you write something like this:
Random rand = new Random();
int x = rand.nextInt(10);
x will be between 0-9 inclusive.
So, given the following array of 25 items, the code to generate a random number between 0 (the base of the array) and array.length would be:
String[] i = new String[25];
Random rand = new Random();
int index = 0;
index = rand.nextInt( i.length );
Since i.length will return 25, the nextInt( i.length ) will return a number between the range of 0-24. The other option is going with Math.Random which works in the same way.
index = (int) Math.floor(Math.random() * i.length);
For a better understanding, check out forum post Random Intervals (archive.org).
It can be done by simply doing the statement:
Randomizer.generate(0, 10); // Minimum of zero and maximum of ten
Below is its source code.
File Randomizer.java
public class Randomizer {
public static int generate(int min, int max) {
return min + (int)(Math.random() * ((max - min) + 1));
}
}
It is just clean and simple.
Forgive me for being fastidious, but the solution suggested by the majority, i.e., min + rng.nextInt(max - min + 1)), seems perilous due to the fact that:
rng.nextInt(n) cannot reach Integer.MAX_VALUE.
(max - min) may cause overflow when min is negative.
A foolproof solution would return correct results for any min <= max within [Integer.MIN_VALUE, Integer.MAX_VALUE]. Consider the following naive implementation:
int nextIntInRange(int min, int max, Random rng) {
if (min > max) {
throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
}
int diff = max - min;
if (diff >= 0 && diff != Integer.MAX_VALUE) {
return (min + rng.nextInt(diff + 1));
}
int i;
do {
i = rng.nextInt();
} while (i < min || i > max);
return i;
}
Although inefficient, note that the probability of success in the while loop will always be 50% or higher.
I wonder if any of the random number generating methods provided by an Apache Commons Math library would fit the bill.
For example: RandomDataGenerator.nextInt or RandomDataGenerator.nextLong
I use this:
/**
* #param min - The minimum.
* #param max - The maximum.
* #return A random double between these numbers (inclusive the minimum and maximum).
*/
public static double getRandom(double min, double max) {
return (Math.random() * (max + 1 - min)) + min;
}
You can cast it to an Integer if you want.
As of Java 7, you should no longer use Random. For most uses, the
random number generator of choice is now
ThreadLocalRandom.For fork join pools and parallel
streams, use SplittableRandom.
Joshua Bloch. Effective Java. Third Edition.
Starting from Java 8
For fork join pools and parallel streams, use SplittableRandom that is usually faster, has a better statistical independence and uniformity properties in comparison with Random.
To generate a random int in the range [0, 1_000]:
int n = new SplittableRandom().nextInt(0, 1_001);
To generate a random int[100] array of values in the range [0, 1_000]:
int[] a = new SplittableRandom().ints(100, 0, 1_001).parallel().toArray();
To return a Stream of random values:
IntStream stream = new SplittableRandom().ints(100, 0, 1_001);
rand.nextInt((max+1) - min) + min;
Let us take an example.
Suppose I wish to generate a number between 5-10:
int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);
Let us understand this...
Initialize max with highest value and min with the lowest value.
Now, we need to determine how many possible values can be obtained. For this example, it would be:
5, 6, 7, 8, 9, 10
So, count of this would be max - min + 1.
i.e. 10 - 5 + 1 = 6
The random number will generate a number between 0-5.
i.e. 0, 1, 2, 3, 4, 5
Adding the min value to the random number would produce:
5, 6, 7, 8, 9, 10
Hence we obtain the desired range.
Generate a random number for the difference of min and max by using the nextint(n) method and then add min number to the result:
Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
To generate a random number "in between two numbers", use the following code:
Random r = new Random();
int lowerBound = 1;
int upperBound = 11;
int result = r.nextInt(upperBound-lowerBound) + lowerBound;
This gives you a random number in between 1 (inclusive) and 11 (exclusive), so initialize the upperBound value by adding 1. For example, if you want to generate random number between 1 to 10 then initialize the upperBound number with 11 instead of 10.
Just use the Random class:
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
These methods might be convenient to use:
This method will return a random number between the provided minimum and maximum value:
public static int getRandomNumberBetween(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt(max - min) + min;
if (randomNumber == min) {
// Since the random number is between the min and max values, simply add 1
return min + 1;
} else {
return randomNumber;
}
}
and this method will return a random number from the provided minimum and maximum value (so the generated number could also be the minimum or maximum number):
public static int getRandomNumberFrom(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt((max + 1) - min) + min;
return randomNumber;
}
In case of rolling a dice it would be random number between 1 to 6 (not 0 to 6), so:
face = 1 + randomNumbers.nextInt(6);
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();
Or take a look to RandomUtils from Apache Commons.
You can achieve that concisely in Java 8:
Random random = new Random();
int max = 10;
int min = 5;
int totalNumber = 10;
IntStream stream = random.ints(totalNumber, min, max);
stream.forEach(System.out::println);
Here's a helpful class to generate random ints in a range with any combination of inclusive/exclusive bounds:
import java.util.Random;
public class RandomRange extends Random {
public int nextIncInc(int min, int max) {
return nextInt(max - min + 1) + min;
}
public int nextExcInc(int min, int max) {
return nextInt(max - min) + 1 + min;
}
public int nextExcExc(int min, int max) {
return nextInt(max - min - 1) + 1 + min;
}
public int nextIncExc(int min, int max) {
return nextInt(max - min) + min;
}
}
Another option is just using Apache Commons:
import org.apache.commons.math.random.RandomData;
import org.apache.commons.math.random.RandomDataImpl;
public void method() {
RandomData randomData = new RandomDataImpl();
int number = randomData.nextInt(5, 10);
// ...
}
I found this example Generate random numbers :
This example generates random integers in a specific range.
import java.util.Random;
/** Generate random integers in a certain range. */
public final class RandomRange {
public static final void main(String... aArgs){
log("Generating random integers in the range 1..10.");
int START = 1;
int END = 10;
Random random = new Random();
for (int idx = 1; idx <= 10; ++idx){
showRandomInteger(START, END, random);
}
log("Done.");
}
private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
if ( aStart > aEnd ) {
throw new IllegalArgumentException("Start cannot exceed End.");
}
//get the range, casting to long to avoid overflow problems
long range = (long)aEnd - (long)aStart + 1;
// compute a fraction of the range, 0 <= frac < range
long fraction = (long)(range * aRandom.nextDouble());
int randomNumber = (int)(fraction + aStart);
log("Generated : " + randomNumber);
}
private static void log(String aMessage){
System.out.println(aMessage);
}
}
An example run of this class :
Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.
public static Random RANDOM = new Random(System.nanoTime());
public static final float random(final float pMin, final float pMax) {
return pMin + RANDOM.nextFloat() * (pMax - pMin);
}
Here is a simple sample that shows how to generate random number from closed [min, max] range, while min <= max is true
You can reuse it as field in hole class, also having all Random.class methods in one place
Results example:
RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert
Sources:
import junit.framework.Assert;
import java.util.Random;
public class RandomUtils extends Random {
/**
* #param min generated value. Can't be > then max
* #param max generated value
* #return values in closed range [min, max].
*/
public int nextInt(int min, int max) {
Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
if (min == max) {
return max;
}
return nextInt(max - min + 1) + min;
}
}
It's better to use SecureRandom rather than just Random.
public static int generateRandomInteger(int min, int max) {
SecureRandom rand = new SecureRandom();
rand.setSeed(new Date().getTime());
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
rand.nextInt((max+1) - min) + min;
This is working fine.
Say i have five variables a,b,c,d and e which are all assigned a random number from five different ranges corresponding to to the five variables a to e.
Is there a way to assign a sixth variable say x, that chooses a random number from a range with upper value equal to the total of a+b+c+d+e?
for example, say :-
a=5 (range 0-10)
b=1043 (range 0-2000)
c=37 (range 0-38)
d=2 (range 0-100)
e=20 (range 5-30)
then x = (random number) (range 5- 1107)
Thanks for any assistance
You can use
Min + (int)(Math.random() * ((Max - Min) + 1))
Where Min = a and Max = a+b+c+d+e
From java.util.Random:
Random rand = new Random();
rand.nextInt(X);
//returns an `int` in range: [0,X)
So, if you have a, b, ..., n, to get a number in between 0 and the sum of the variables, we do:
rand.nextInt(a + b + ... + n + 1); //returns an `int` in range [0, sum(a,b,...,n)]
As other users have suggested, you can also use Math.random(); however, Math.random() returns a double in range: [0,1). So, if you have min = 0, max = 9, and want a number between [0,9], then you need:
Min + (int)(Math.round((Math.random() * (Max - Min))))
//need to cast to int since Math.round(double returns a long)
or you could do:
Min + (int)(Math.random() * (Max - Min + 1))
Edit: I missunderstood your problem, but you just need to sum each var and do the same thing. Some answers already showed that.
It is just a algorithm problem.
int lra = 0; // lower range of a
int ura = 10; // upper range of a
int lrb = 0;
int urb = 2000;
// a couple of ranges for each var...
int a = lra + ((int) (Math.random() * (ura - lra)));
// the same for each var...
int lrs = lra + lrb + ...; // sum of lower ranges...
int urs = ura + urb + ...; // sum of upper ranges
int x = lrs + ((int) (Math.random() * (urs - lrs)));
Something like this should do the job
Random generator = new Random();
generator.setSeed(System.currentTimeMillis());
long range = a+b+c+d+e;
long fraction = (long)(range * generator.nextDouble());
randomNumber = (int)(fraction);