Creating random numbers with no duplicates
The simplest way would be to create a list of the possible numbers (1..20 or whatever) and then shuffle them with Collections.shuffle
. Then just take however many elements you want. This is great if your range is equal to the number of elements you need in the end (e.g. for shuffling a deck of cards).
That doesn't work so well if you want (say) 10 random elements in the range 1..10,000 - you'd end up doing a lot of work unnecessarily. At that point, it's probably better to keep a set of values you've generated so far, and just keep generating numbers in a loop until the next one isn't already present:
if (max < numbersNeeded)
{
throw new IllegalArgumentException("Can't ask for more numbers than are available");
}
Random rng = new Random(); // Ideally just create one instance globally
// Note: use LinkedHashSet to maintain insertion order
Set<Integer> generated = new LinkedHashSet<Integer>();
while (generated.size() < numbersNeeded)
{
Integer next = rng.nextInt(max) + 1;
// As we're adding to a set, this will automatically do a containment check
generated.add(next);
}
Be careful with the set choice though - I've very deliberately used LinkedHashSet
as it maintains insertion order, which we care about here.
Yet another option is to always make progress, by reducing the range each time and compensating for existing values. So for example, suppose you wanted 3 values in the range 0..9. On the first iteration you'd generate any number in the range 0..9 - let's say you generate a 4.
On the second iteration you'd then generate a number in the range 0..8. If the generated number is less than 4, you'd keep it as is... otherwise you add one to it. That gets you a result range of 0..9 without 4. Suppose we get 7 that way.
On the third iteration you'd generate a number in the range 0..7. If the generated number is less than 4, you'd keep it as is. If it's 4 or 5, you'd add one. If it's 6 or 7, you'd add two. That way the result range is 0..9 without 4 or 6.
PHP - Generate random numbers with no duplicates
In your case, you need create range array of numbers, and then simply shuffle them
$days = range(1,24);
shuffle($days);
Thats all
Random number generator with no more than 2 duplicates
Have you tried checking if each generated index already appears twice in taslar
and if so, generating another one?
while (taslar.Count < 16)
{
int arrIndex = UnityEngine.Random.Range(0, tasSprites.Length);
tasSprite = tasSprites[arrIndex];
tasName = tasSprite.name;
int value = Int32.Parse(tasName);
if (taslar.Count(t => t == value) < 2)
{
taslar.Add(value);
}
}
How do I create a list of random numbers without duplicates
This will return a list of 10 numbers selected from the range 0 to 99, without duplicates.
import random
random.sample(range(100), 10)
With reference to your specific code example, you probably want to read all the lines from the file once and then select random lines from the saved list in memory. For example:
all_lines = f1.readlines()
for i in range(50):
lines = random.sample(all_lines, 40)
This way, you only need to actually read from the file once, before your loop. It's much more efficient to do this than to seek back to the start of the file and call f1.readlines()
again for each loop iteration.
Generating 10 random numbers without duplicate with fundamental techniques
You need to break
out of the for
loop if either of the conditions are met.
int[] number = new int[10];
int count=0;
int num;
Random r = new Random();
while(count<number.length){
num = r.nextInt(21);
boolean repeat=false;
do{
for(int i=0; i<number.length; i++){
if(num==number[i]){
repeat=true;
break;
}
else if(i==count){
number[count]=num;
count++;
repeat=true;
break;
}
}
}while(!repeat);
}
for(int j=0;j<number.length;j++){
System.out.print(number[j]+" ");
}
This will make YOUR code work but @gonzo proposed a better solution.
Random array generation with no duplicates
First of all rand()
is generatig random numbers but not wihout duplicates.
If you want to generate a random array without duplicates the rand()
method is not working at all.
Let say you want to generate an array of 1000 numbers. In the best case let say you generated the first 999 numbers without duplicates and last think to do is generating the last number. The probability of getting that number is 1/1000 so this is almost going to take forever to get generated. In practice only 10 numbers makes a big trouble.
The best method is to generate all your numbers by incrementation (or strictly monotonic sequence) is shuffle them. In this case there will be no duplicates
Here is an exemple on how to do it with 10 numbers. Even with 1000 numbers it's working.
Note: Suffle function from Jhon Leehey's answer.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void shuffle(int *arr, size_t n)
{
if (n > 1)
{
size_t i;
srand(time(NULL));
for (i = 0; i < n - 1; i++)
{
size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
int t = arr[j];
arr[j] = arr[i];
arr[i] = t;
}
}
}
int main()
{
int i;
int arr[10];
for (i=0; i<10; i++){
arr[i] = i;
}
shuffle(arr, 10);
for (i=0; i<10; i++){
printf("%d ", arr[i]);
}
}
Random number generator without duplication
Create a List
, and populate it with the elements in your range. Then shuffle()
the list, and the order is your random numbers. That is, the 0-th element is your first random number, the 1st element is your second random number, etc.
Random number generator without dupes in Javascript?
var nums = [1,2,3,4,5,6,7,8,9,10,11,12];
var gen_nums = [];
function in_array(array, el) {
for(var i = 0 ; i < array.length; i++)
if(array[i] == el) return true;
return false;
}
function get_rand(array) {
var rand = array[Math.floor(Math.random()*array.length)];
if(!in_array(gen_nums, rand)) {
gen_nums.push(rand);
return rand;
}
return get_rand(array);
}
for(var i = 0; i < 9; i++) {
console.log(get_rand(nums));
}
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