I'm trying to randomly select 100 values from an array of 1000.
How do I do this?

Also, after selection, how can I add these 100 values into a new array.

Do I need to do something like this with a loop?

for (x = 0; x < 100; x++){
    smallArray = smallArray + (value1);
};

Thanks!

Recommended Answers

I'll give you a c++ equivalent :

for(i=0;i<100;i++){
array100[i]= array1000[random(1000)];
}

All you need now is to find an equivalent random function in java ;)

Jump to Post

I hope that's help for you:

for(i=0;i<100;i++){
   
          array100[i]= array1000[(int)(Math.random()*1000)];
   
      }

Sorry for english... I am Hungarian

Jump to Post

You can also use Random.nextInt(int) for the random int.

Keep in mind that the suggestions tendered so far will allow the chance that an element is selected multiple times. If you don't …

Jump to Post
for(int i=0;i<100;i++){
        keys_Array[i] = unsorted.get( random.nextInt(100) );
    }

Note that get() is used if 'unsorted' is an arraylist and not an array.

Jump to Post

All 14 Replies

I'll give you a c++ equivalent :

for(i=0;i<100;i++){
array100[i]= array1000[random(1000)];
}

All you need now is to find an equivalent random function in java ;)

I hope that's help for you:

for(i=0;i<100;i++){
   
          array100[i]= array1000[(int)(Math.random()*1000)];
   
      }

Sorry for english... I am Hungarian

You can also use Random.nextInt(int) for the random int.

Keep in mind that the suggestions tendered so far will allow the chance that an element is selected multiple times. If you don't want that to be possible then you'll need to put additional measures in place to prevent it.

You can also use Random.nextInt(int) for the random int.

Keep in mind that the suggestions tendered so far will allow the chance that an element is selected multiple times. If you don't want that to be possible then you'll need to put additional measures in place to prevent it.

Thanks for the help thusfar.

I'm a little confused.

for(int i=0;i<100;i++){
        keys_Array = random.nextInt(unsorted);
    }

keys_Array is size 100, and I want to have it select 100 elements from unsorted an arraylist filled with 1000 elements.

What do I need to change from this? I'm not so worried about multiple elements at the moment.

for(int i=0;i<100;i++){
        keys_Array[i] = unsorted.get( random.nextInt(100) );
    }

Note that get() is used if 'unsorted' is an arraylist and not an array.

for(int i=0;i<100;i++){
        keys_Array[i] = unsorted.get( random.nextInt(100) );
    }

Note that get() is used if 'unsorted' is an arraylist and not an array.

Thank you.
I changed to .get, says I have incompatible types.
Says found .Object, requires Int

Could you point out where I went wrong in this?
Appreciate it.

import java.util.*;

public class Lab5 {

    public static void main(String[] args) {
        Random random = new Random();
        ArrayList unsorted = new ArrayList(),
               sorted = new ArrayList();
        
        int[] keys_Array = new int[100];

        for(int count = 0; count < 1000; count++){
            int rand = random.nextInt(10000)+1;
            System.out.println(rand);

            unsorted.add(rand);
            sorted.add(rand);

        }
//    System.out.print("\n\nSize: " + unsorted.size());
 //   System.out.print("\n");
  //  System.out.print("Contents:" + unsorted);
        
        for(int i=0;i<100;i++){
            keys_Array[i] = unsorted.get(random.nextInt(100));
        }
    }
}

You have to cast the Object to Integer if you want to assign it to the array.

keys_Array[i] = (Integer)unsorted.get(random.nextInt(100));

Using generics to type your collections would avoid that.

Thanks a bunch!

If you don't want repeat selections, you can use this instead

keys_Array[i] = (Integer) unsorted.remove( random.nextInt(unsorted.size()) );

Can you explain the difference in that one? If I use it I would kind of like to know how it works.

It's actually removing the element from "unsorted". It grabs an element randomly between 0 and size()-1.

Alright, thats awesome. It can still get a repeat value though, based on what the first array contains right?

If "unsorted" contains any duplicates, yes. It just can't retrieve the same element from "unsorted" more than once. The first version could use unsorted.get(4) multiple times if 4 came up as the random number more than once.

Thanks for explaining in depth.
I understand this concept now :)

Be a part of the DaniWeb community

We're a friendly, industry-focused community of 1.19 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge.