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 do I generate random integers within a specific range in Java?
(72 answers)
Closed 11 months ago.
I have to do this in Java:
Write a method fillArray() that takes three integers: (s, min, max),
and returns an array of size s having random integers with values
between min and max.
This is the code I wrote
public static void fillArray(int s, int min, int max) {
int[] random = new int[s];
for (int i = 0; i < s; i++) {
int n = (int) (Math.random()*100 %s);
if (n > min && n < max) {
random[i] = n;
}
}
System.out.printf("Here's an array of size %d, whose elements vary between %d and %d: \n", s, min, max);
System.out.print(Arrays.toString(random));
}
The problem is, when I implement my method in the main, fillArray(10, 10, 20), it gives me arrays of size 10, with elements at 0.
I tried playing around with this specific expression in the code
int n = (int) (Math.random()*100 %s);
and changing what I do after *100.
Sometimes it works for most elements, but I still get some elements which are 0, which is wrong since the minimum is 10.
Any idea how I could fix that?
Math.random()
Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0.
So since the random numbers you generate are always between [0 - 1) (Zero inclusive & 1 exclusiv you need to multiply them by your desired max number to generate random numbers which are greter than one. Example to generate random numbers between [0 - 15) you can do
Math.random() * 15
to get values between [0.0 - 14.999..].
To include 15 in your values you need to multiply with (15 + 1) = 16
Math.random() * (15 + 1) or generaly
Math.random() * (max + 1)
Since there is also a min value in your requierment, you need to exclude values between 0 - min. To do so you could add simply the min value to the result of the above:
(Math.random() * (max + 1)) + min
But wait, let see an example for min = 5 and max = 15. Since
Math.random() * (max + 1)
returns values between [0.0 - 15.999..] adding min = 5 to each value:
(Math.random() * (max + 1)) + min
will result in values in range [5.0 - 20.999..] which is not desired. To change that you will need to substract min from max while multiplying:
(Math.random() * (max - min + 1)) + min
which will result in the correct range [5.0 - 15.999..] where you need to apply casting (int) to get your random integers instead of dobles. So your method could look like
public static void fillArray(int s, int min, int max) {
int[] random = new int[s];
for (int i = 0; i < s; i++) {
int n = (int) (Math.random() * (max - min + 1)) + min;
random[i] = n;
}
System.out.printf("Here's an array of size %d, whose elements vary between %d and %d: \n", s, min, max);
System.out.print(Arrays.toString(random));
}
So basically I need to generate a number between a min and max (could be negative as well) and there should be a possibility the number could end with a decimal 0.5.
So, for example, the numbers could be 5, 6, -8.5, 3.5.
I have this to generate my number (example):
rand.nextInt(20 + 1 + 10) - 10;
You could generate an integer with twice the range you want, and then halve it.
Something like this:
int min = -5;
int max = 15;
double r = min + 0.5 * rand.nextInt(2 * (max - min));
If you want max included in the random range, then:
double r = min + 0.5 * rand.nextInt(2 * (max + 1 - min));
Math.round((min + (rand.nextDouble() * Math.abs(max - min))) / 0.5) * 0.5;
should do the trick by creating a correct value in range [min, max] (including both min and max as possible values!)
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.
I want to make a method in java where I can input the maximum and minimum numbers and it will give me a random number, but with a higher chance of a bigger number than a smaller one. How would I do this?
There is lots of ways to do this depending on what you mean by higher. You can skew your distribution with a function.
int num = (int) (func(Math.random()) * (max - min)) + min;
Your func could be Math.sqrt or Math.pow(x, n) where n < 1 to favour higher numbers.
I would like it to be 10% more likely to have an answer over ten.
If this is your requirement, you actually have two distributions.
private static final Random rand = new Random();
public static int randBetween(int min, int max) {
return rand.nextInt(max - min + 1) + min;
}
int next = rand.nextInt(100) < 10 ? // a 10% chance
randBetween(10, max) : // random of at least 10
randBetween(min, max); // otherwise any number.