Random sampling without replacement [closed]

Question

Create a function that will output a set of distinct random numbers drawn from a range. The order of the elements in the set is unimportant (they can even be sorted), but it must be possible for the contents of the set to be different each time the function is called.

The function will receive 3 parameters in whatever order you want:

1. Count of numbers in output set
2. Lower limit (inclusive)
3. Upper limit (inclusive)

Assume all numbers are integers in the range 0 (inclusive) to 2³¹ (exclusive). The output can be passed back any way you want (write to console, as an array, etc.)

Judging

Criteria includes the 3 R's

1. Run-time - tested on a quad-core Windows 7 machine with whatever compiler is freely or easily available (provide a link if necessary)
2. Robustness - does the function handle corner cases or will it fall into an infinite loop or produce invalid results - an exception or error on invalid input is valid
3. Randomness - it should produce random results that are not easily predictable with a random distribution. Using the built in random number generator is fine. But there should be no obvious biases or obvious predictable patterns. Needs to be better than that random number generator used by the Accounting Department in Dilbert

If it is robust and random then it comes down to run-time. Failing to be robust or random greatly hurts its standings.

Answer 1

Python

import random

def sample(n, lower, upper):
    result = []
    pool = {}
    for _ in xrange(n):
        i = random.randint(lower, upper)
        x = pool.get(i, i)
        pool[i] = pool.get(lower, lower)
        lower += 1
        result.append(x)
    return result

I probably just re-invented some well-known algorithm, but the idea is to (conceptually) perform a partial Fisher-Yates shuffle of the range lower..upper to get the length n prefix of a uniformly shuffled range.

Of course, storing the whole range would be rather expensive, so I only store the locations where the elements have been swapped.

This way, the algorithm should perform well both in the case where you're sampling numbers from a tight range (e.g. 1000 numbers in the range 1..1000), as well as the case where you're sampling numbers from a large range.

I'm not sure about the quality of randomness from the built-in generator in Python, but it's relatively simple to swap in any generator that can generate integers uniformly from some range.

Answer 2

python 2.7

import random
print(lambda x,y,z:random.sample(xrange(y,z),x))(input(),input(),input())

not sure what your standing is on using builtin random methods, but here you go anyways. nice and short

edit: just noticed that range() doesn't like to make big lists. results in a memory error. will see if there is any other way to do this...

edit2: range was the wrong function, xrange works. The maximum integer is actually 2**31-1 for python

test:

python sample.py
10
0
2**31-1
[786475923, 2087214992, 951609341, 1894308203, 173531663, 211170399, 426989602, 1909298419, 1424337410, 2090382873]

Answer 3

C

Returns an array containing x unique random ints between min and max. (caller must free)

#include <stdlib.h>
#include <stdint.h>
#define MAX_ALLOC ((uint32_t)0x40000000)  //max allocated bytes, fix per platform
#define MAX_SAMPLES (MAX_ALLOC/sizeof(uint32_t))

int* randsamp(uint32_t x, uint32_t min, uint32_t max)
{
   uint32_t r,i=x,*a;
   if (!x||x>MAX_SAMPLES||x>(max-min+1)) return NULL;
   a=malloc(x*sizeof(uint32_t));
   while (i--) {
      r= (max-min+1-i);
      a[i]=min+=(r ? rand()%r : 0);
      min++;
   }
   while (x>1) {
      r=a[i=rand()%x--];
      a[i]=a[x];
      a[x]=r;
   }
   return a;
}

Works by generating x sequential random integers in the range, then shuffling them. Add a seed(time) somewhere in caller if you don't want the same results every run.

Answer 4

Ruby >= 1.8.7

def pick(num, min, max)
  (min..max).to_a.sample(num)
end

p pick(5, 10, 20) #=>[12, 18, 13, 11, 10]

Answer 5

R

s <- function(n, lower, upper) sample(lower:upper,n); s(10,0,2^31-2)

Answer 6

The question is not correct. Do you need uniform sampling or not? In the case uniform sampling is needed I have the following code in R, which has average complexity O(s log s), where s is the sample size.

# The Tree growing algorithm for uniform sampling without replacement
# by Pavel Ruzankin 
quicksample = function (n,size)
# n - the number of items to choose from
# size - the sample size
{
  s=as.integer(size)
  if (s>n) {
    stop("Sample size is greater than the number of items to choose from")
  }
  # upv=integer(s) #level up edge is pointing to
  leftv=integer(s) #left edge is poiting to; must be filled with zeros
  rightv=integer(s) #right edge is pointig to; must be filled with zeros
  samp=integer(s) #the sample
  ordn=integer(s) #relative ordinal number

  ordn[1L]=1L #initial value for the root vertex
  samp[1L]=sample(n,1L) 
  if (s > 1L) for (j in 2L:s) {
    curn=sample(n-j+1L,1L) #current number sampled
    curordn=0L #currend ordinal number
    v=1L #current vertice
    from=1L #how have come here: 0 - by left edge, 1 - by right edge
    repeat {
      curordn=curordn+ordn[v]
      if (curn+curordn>samp[v]) { #going down by the right edge
        if (from == 0L) {
          ordn[v]=ordn[v]-1L
        }
        if (rightv[v]!=0L) {
          v=rightv[v]
          from=1L
        } else { #creating a new vertex
          samp[j]=curn+curordn
          ordn[j]=1L
          # upv[j]=v
          rightv[v]=j
          break
        }
      } else { #going down by the left edge
        if (from==1L) {
          ordn[v]=ordn[v]+1L
        }
        if (leftv[v]!=0L) {
          v=leftv[v]
          from=0L
        } else { #creating a new vertex
          samp[j]=curn+curordn-1L
          ordn[j]=-1L
          # upv[j]=v
          leftv[v]=j
          break
        }
      }
    }
  }
  return(samp)  
}

Of course, one may rewrite it in C for better performance. The complexity of this algorithm is discussed in: Rouzankin, P. S.; Voytishek, A. V. On the cost of algorithms for random selection. Monte Carlo Methods Appl. 5 (1999), no. 1, 39-54. http://dx.doi.org/10.1515/mcma.1999.5.1.39

You may look through this paper for another algorithm with the same average complexity.

But if you do not need uniform sampling, only requiring that all sampled numbers be different, then the situation changes dramatically. It is not hard to write an algorithm that has average complexity O(s).

See also for uniform sampling: P. Gupta, G. P. Bhattacharjee. (1984) An efficient algorithm for random sampling without replacement. International Journal of Computer Mathematics 16:4, pages 201-209. DOI: 10.1080/00207168408803438

Teuhola, J. and Nevalainen, O. 1982. Two efficient algorithms for random sampling without replacement. /IJCM/, 11(2): 127–140. DOI: 10.1080/00207168208803304

In the last paper the authors use hash tables and claim that their algorithms have O(s) complexity. There is one more fast hash table algorithm, which will soon be implemented in pqR (pretty quick R): https://stat.ethz.ch/pipermail/r-devel/2017-October/075012.html

Answer 7

APL, 18 22 bytes

{⍵[0]+(1↑⍺)?⍵[1]-⍵[0]}

Declares an anonymous function that takes two arguments ⍺ and ⍵. ⍺ is the number of random numbers you want, ⍵ is a vector containing the lower and upper bounds, in that order.

a?b picks a random numbers between 0-b without replacement. By taking ⍵[1]-⍵[0] we get the range size. Then we pick ⍺ numbers (see below) from that range and add the lower bound. In C, this would be

lower + rand() * (upper - lower)

⍺ times without replacement. Parentheses not needed because APL operates right-to-left.

Assuming I've understood the conditions correctly, this fails the 'robustness' criteria because the function will fail if given improper arguments (e.g. passing a vector instead of a scalar as ⍺).

In the event that ⍺ is a vector rather than a scalar, 1↑⍺ takes the first element of ⍺. For a scalar, this is the scalar itself. For a vector, it's the first element. This should make the function meet the 'robustness' criteria.

Example:

Input: 100 {⍵[0]+⍺?⍵[1]-⍵[0]} 0 100
Output: 34 10 85 2 46 56 32 8 36 79 77 24 90 70 99 61 0 21 86 50 83 5 23 27 26 98 88 66 58 54 76 20 91 72 71 65 63 15 33 11 96 60 43 55 30 48 73 75 31 13 19 3 45 44 95 57 97 37 68 78 89 14 51 47 74 9 67 18 12 92 6 49 41 4 80 29 82 16 94 52 59 28 17 87 25 84 35 22 38 1 93 81 42 40 69 53 7 39 64 62

Answer 8

Scala

object RandSet {
  val random = util.Random 

  def rand (count: Int, lower: Int, upper: Int, sofar: Set[Int] = Set.empty): Set[Int] =
    if (count == sofar.size) sofar else 
    rand (count, lower, upper, sofar + (random.nextInt (upper-lower) + lower)) 
}

object RandSetRunner {

  def main (args: Array [String]) : Unit = {
    if (args.length == 4) 
      (0 until args (0).toInt).foreach { unused => 
      println (RandSet.rand (args (1).toInt, args (2).toInt, args (3).toInt).mkString (" "))
    }
    else Console.err.println ("usage: scala RandSetRunner OUTERCOUNT COUNT MIN MAX")
  }
}

compile and run:

scalac RandSetRunner.scala 
scala RandSetRunner 200 15 0 100

The second line will run 200 tests with 15 values from 0 to 100, because Scala produces fast bytecode but needs some startup time. So 200 starts with 15 values from 0 to 100 would consume more time.

Sample on a 2 Ghz Single Core:

time scala RandSetRunner 100000 10 0 1000000 > /dev/null

real    0m2.728s
user    0m2.416s
sys     0m0.168s

Logic:

Using the build-in random and recursively picking numbers in the range (max-min), adding min and checking, if the size of the set is the expected size.

Critique:

• It will be fast for small samples of big ranges, but if the task is to pick nearly all elements of a sample (999 numbers out of 1000) it will repeatedly pick numbers, already in the set.
• From the question, I'm unsure, whether I have to sanitize against unfulfillable requests like Take 10 distinct numbers from 4 to 8. This will now lead to an endless loop, but can easily be avoided with a prior check which I will append if requested.

Answer 9

Scheme

Not sure why you need 3 parameters passed nor why i need to assume any range...

(import srfi-1) ;; for iota
(import srfi-27) ;; randomness
(import srfi-43) ;; for vector-swap!

(define rand (random-source-make-integers
               default-random-source))

;; n: length, i: lower limit
(define (random-range n i)
  (let ([v (list->vector (iota n i))])
    (let f ([n n])
      (let* ([i (rand n)] [n (- n 1)])
        (if (zero? n) v
            (begin (vector-swap! v n i) (f n)))))))

Answer 10

R

random <- function(count, from, to) {
  rand.range <- to - from

  vec <- c()

  for (i in 1:count) {
    t <- sample(rand.range, 1) + from
    while(i %in% vec) {
      t <- sample(rand.range, 1) + from
    }
    vec <- c(vec, t)
  }

  return(vec)
}

Answer 11

C++

This code is best when drawing many samples from the range.

#include <exception>
#include <stdexcept>
#include <cstdlib>

template<typename OutputIterator>
 void sample(OutputIterator out, int n, int min, int max)
{
  if (n < 0)
    throw std::runtime_error("negative sample size");
  if (max < min)
    throw std::runtime_error("invalid range");
  if (n > max-min+1)
    throw std::runtime_error("sample size larger than range");

  while (n>0)
  {
    double r = std::rand()/(RAND_MAX+1.0);
    if (r*(max-min+1) < n)
    {
      *out++ = min;
      --n;
    }
    ++min;
  }
}

Answer 12

Q (19 characters)

f:{(neg x)?y+til z}

Then use f[x;y;z] as [count of numbers in output set;starting point;size of range]

e.g. f[5;10;10] will output 5 distinct random numbers between 10 and 19 inclusive.

q)\ts do[100000;f[100;1;10000]]
2418 131456j

Results above show performance at 100,000 iterations of picking 100 random numbers between 1 & 10,000.

Answer 13

R, 31 or 40 bytes (depending on the meaning of the word “range”)

If the input has 3 numbers, a[1], a[2], a[3], and by “range” you mean “an integer sequence from a[2] to a[3]”, then you have this:

a=scan();sample(a[2]:a[3],a[1])

If you have an array n from which you are about to resample, but under the restriction of the lower and upper limits, like “resample values of the given array n from the range a[1]...a[2]”, then use this:

a=scan();sample(n[n>=a[2]&n<=a[3]],a[1])

I am quite surprised why the previous result was not golfed considering the built-in sample with replacement facilities! We create a vector that satisfies the range condition and re-sample it.

• Robustness: the corner cases (sequences of the same length as the range to sample from) are handled by default.
• Run-time: extremely fast because it is built in.
• Randomness: the seed is automatically changed every time the RNG is invoked.

Answer 14

Javascript (using external library) (64 bytes / 104 bytes??)

(a,b,n)=>_.Range(0,n).Select(x=>Math.random()*(b-a)+a).ToArray()

Link to lib: https://github.com/mvegh1/Enumerable/

Code explanation: Lambda expression accepts min,max,count as args. Create a collection of size n, and map each element to a random number fitting the min/max criteria. Convert to native JS array and return it. I ran this also on an input of size 5,000,000, and after applying a distinct transform still showed 5,000,000 elements. If it is agreed upon that this isn't safe enough of a guarantee for distinctness I will update the answer

I included some statistics in the image below...

EDIT: The below image shows code / performance that guarantees every element will be distinct. It's much much slower (6.65 seconds for 50,000 elements) vs the original code above for same args (0.012 seconds)

Answer 15

K (oK), 14 bytes

Solution:

{y+(-x)?1+z-y}

Try it online!

Example:

> {y+(-x)?1+z-y}. 10 10 20      / note: there are two ways to provide input, dot or
13 20 16 17 19 10 14 12 11 18
> {y+(-x)?1+z-y}[10;10;20]      / explicitly with [x;y;z]
12 11 13 19 15 17 18 20 14 10

Explanation:

Takes 3 implicit inputs per spec:

• x, count of numbers in output set,
• y, lower limit (inclusive)
• z, upper limit (inclusive)

---

{y+(-x)?1+z-y} / the solution
{            } / lambda function with x, y and z as implicit inputs
          z-y  / subtract lower limit from upper limit
        1+     / add 1
   (-x)?       / take x many distinct items from 0..(1+z=y)
 y+            / add lower limit

Notes:

Also a polyglot in q/kdb+ with an extra set of brackets: {y+((-)x)?1+z-y} (16 bytes).

Answer 16

Axiom + its library

f(n:PI,a:INT,b:INT):List INT==
    r:List INT:=[]
    a>b or n>99999999 =>r
    d:=1+b-a
    for i in 1..n repeat
          r:=concat(r,a+random(d)$INT)
    r

The above f() function return as error the empty list, in the case f(n,a,b) with a>b. In other cases of invalid input, it not run with one error message in Axiom window, because argument will be not of the right type. Examples

(6) -> f(1,1,5)
   (6)  [2]
                                                       Type: List Integer
(7) -> f(1,1,1)
   (7)  [1]
                                                       Type: List Integer
(10) -> f(10,1,1)
   (10)  [1,1,1,1,1,1,1,1,1,1]
                                                       Type: List Integer
(11) -> f(10,-20,-1)
   (11)  [- 10,- 4,- 18,- 5,- 5,- 11,- 15,- 1,- 20,- 1]
                                                       Type: List Integer
(12) -> f(10,-20,-1)
   (12)  [- 4,- 5,- 3,- 4,- 18,- 1,- 2,- 14,- 19,- 8]
                                                       Type: List Integer
(13) -> f(10,-20,-1)
   (13)  [- 18,- 12,- 12,- 19,- 19,- 15,- 5,- 17,- 19,- 4]
                                                       Type: List Integer
(14) -> f(10,-20,-1)
   (14)  [- 8,- 11,- 20,- 10,- 4,- 8,- 11,- 3,- 10,- 16]
                                                       Type: List Integer
(15) -> f(10,9,-1)
   (15)  []
                                                       Type: List Integer
(16) -> f(10,0,100)
   (16)  [72,83,41,35,27,0,33,18,60,38]
                                                       Type: List Integer