Generate 8 unique random integers

Question

Hey guys, im currently doing a random integers programme. Here's what i got so far

    FirstVal=Random.Range(0,numOfPeeps);  
  SecondVal=Random.Range(0,numOfPeeps); 
  ThirdVal=Random.Range(0,numOfPeeps); 
  FourthVal=Random.Range(0,numOfPeeps); 
  FifthVal=Random.Range(0,numOfPeeps);
  SixthVal=Random.Range(0,numOfPeeps);
  SeventhVal=Random.Range(0,numOfPeeps);
  EightVal=Random.Range(0,numOfPeeps);
  
  while(FirstVal==SecondVal)
  {
  SecondVal=Random.Range(0,numOfPeeps);
  }
  
  if(numOfPeeps&gt;=3){
  while(ThirdVal==FirstVal||ThirdVal==SecondVal)
  {
  ThirdVal=Random.Range(0,numOfPeeps);
  }
  }
  
  if(numOfPeeps&gt;=4){
  while(FourthVal==FirstVal||FourthVal==SecondVal||FourthVal==ThirdVal)
  {
  FourthVal=Random.Range(0,numOfPeeps);
  }
  }
  
  if(numOfPeeps&gt;=5){
  while(FifthVal==FirstVal||FifthVal==SecondVal||FifthVal==ThirdVal||FifthVal==FourthVal)
  {
  FifthVal=Random.Range(0,numOfPeeps);
  }
  }
  
  if(numOfPeeps&gt;=6){
  while(SixthVal==FirstVal||SixthVal==SecondVal||SixthVal==ThirdVal||SixthVal==FourthVal||SixthVal==FifthVal)
  {
  SixthVal=Random.Range(0,numOfPeeps);
  }
  }
  
  if(numOfPeeps&gt;=7){
  while(SeventhVal==FirstVal||SeventhVal==SecondVal||SeventhVal==ThirdVal||SeventhVal==FourthVal||SeventhVal==FifthVal||SeventhVal==SixthVal)
  {
  SeventhVal=Random.Range(0,numOfPeeps);
  }
  }
  
  if(numOfPeeps&gt;=8){
  while(EightVal==FirstVal||EightVal==SecondVal||EightVal==ThirdVal||EightVal==FourthVal||EightVal==FifthVal||EightVal==SixthVal||EightVal==SeventhVal)
  {
  EightVal=Random.Range(0,numOfPeeps);
  }
  }

Answer 1

Best Answer

Answer by whydoidoit · Jul 24, 2012 at 06:58 PM

    using System.Linq;
    using System.Collections.Generic;

    var availableEntries = Enumerable.Range(1, numOfPeeps).ToList();

    firstVal = availableEntries[Random.Range(0, availableEntries.Count)];
    availableEntries.Remove(firstVal);

    secondVal = availableEntries[Random.Range(0, availableEntries.Count)];
    availableEntries.Remove(secondVal);

etcetera...

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whydoidoit · Jul 24, 2012 at 07:03 PM 1

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Or make the list in a random order like this:

    var availableEntries = Enumerable.Range(1, numOfPeeps).OrderBy(b=>Random.value).ToList();
    firstVal = availableEntries[0];
    secondVal = availableEntries[1];

DaveA · Jul 24, 2012 at 08:27 PM 0

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You want a 'deal' function, that's essentially what this OrderBy thing does. Create list of random numbers, sort them, keeping track of their indices. Resulting sorted list has 'random' indices, each, of course, unique.

Answer 2

Answer by glaiveorb · Jul 25, 2012 at 06:14 AM

Hey thanks for the feedback guys! Turns out it was an error with the numOfPeeps as i was not setting it properly!! Thanks btw!!

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Generate 8 unique random integers

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