Randomly choosing an item from a Swift array without repeating - ios

Question

This code picks a random color from a array of pre-set colors. How do I make it so the same color doesn't get picked more than once?
var colorArray = [(UIColor.redColor(), "red"), (UIColor.greenColor(), "green"), (UIColor.blueColor(), "blue"), (UIColor.yellowColor(), "yellow"), (UIColor.orangeColor(), "orange"), (UIColor.lightGrayColor(), "grey")]
var random = { () -> Int in
return Int(arc4random_uniform(UInt32(colorArray.count)))
} // makes random number, you can make it more reusable
var (sourceColor, sourceName) = (colorArray[random()])

Answer 1

Create an array of indexes. Remove one of the indexes from the array and then use that to fetch a color.
Something like this:
var colorArray = [
(UIColor.redColor(), "red"),
(UIColor.greenColor(), "green"),
(UIColor.blueColor(), "blue"),
(UIColor.yellowColor(), "yellow"),
(UIColor.orangeColor(), "orange"),
(UIColor.lightGrayColor(), "grey")]
var indexes = [Int]();
func randomItem() -> UIColor
{
if indexes.count == 0
{
print("Filling indexes array")
indexes = Array(0..< colorArray.count)
}
let randomIndex = Int(arc4random_uniform(UInt32(indexes.count)))
let anIndex = indexes.removeAtIndex(randomIndex)
return colorArray[anIndex].0;
}
The code above creates an array indexes. The function randomItem looks to see if indexes is empty. if it is, it populates it with index values ranging from 0 to colorArray.count - 1.
It then picks a random index in the indexes array, removes the value at that index in the indexes array, and uses it to fetch and return an object from your colorArray. (It doesn't remove objects from the colorArray. It uses indirection, and removes objects from the indexesArray, which initially contains an index value for each entry in your colorArray.
The one flaw in the above is that after you fetch the last item from indexArray, you populate it with a full set of indexes, and it's possible that the next color you get from the newly repopulated array will be the same as the last one you got.
It's possible to add extra logic to prevent this.

Answer 2

based on the fact, that arc4random_uniform generate not only random, but also uniformly distributed numbers
import Foundation // arc4random_uniform
class Random {
var r:UInt32
let max: UInt32
init(max: UInt32) {
self.max = max
r = arc4random_uniform(max)
}
var next: UInt32 {
var ret: UInt32
repeat {
ret = arc4random_uniform(max)
} while r == ret
r = ret
return r
}
}
// usage example
let r = Random(max: 5)
for i in 0..<10 {
print(r.r, r.next) // there will never be a pair of the same numbers in the
// generated stream
}
/*
2 4
4 0
0 3
3 0
0 3
3 4
4 1
1 3
3 4
4 3
*/
simple test for different k and stream length of one milion
class Random {
var r:UInt32
let max: UInt32
init(max: UInt32) {
self.max = max
r = arc4random_uniform(max)
}
var next: (UInt32, Int) {
var i = 0
var ret: UInt32
repeat {
ret = arc4random_uniform(max)
i += 1
} while r == ret
r = ret
return (r,i)
}
}
for k in 3..<16 {
let r = Random(max: UInt32(k))
var repetition = 0
var sum = 0
for i in 0..<1000000 {
let j = r.next
repetition = max(repetition, j.1)
sum += j.1
}
print("maximum of while repetition for k:", k, "is", repetition, "with average of", Double(sum) / Double(1000000) )
}
prints
maximum of while repetition for k: 3 is 15 with average of 1.499832
maximum of while repetition for k: 4 is 12 with average of 1.334008
maximum of while repetition for k: 5 is 9 with average of 1.250487
maximum of while repetition for k: 6 is 8 with average of 1.199631
maximum of while repetition for k: 7 is 8 with average of 1.167501
maximum of while repetition for k: 8 is 7 with average of 1.142799
maximum of while repetition for k: 9 is 8 with average of 1.124096
maximum of while repetition for k: 10 is 6 with average of 1.111178
maximum of while repetition for k: 11 is 7 with average of 1.099815
maximum of while repetition for k: 12 is 7 with average of 1.091041
maximum of while repetition for k: 13 is 6 with average of 1.083582
maximum of while repetition for k: 14 is 6 with average of 1.076595
maximum of while repetition for k: 15 is 6 with average of 1.071965
finaly, here is more Swifty and functional approach based on the same idea
import Foundation
func random(max: Int)->()->Int {
let max = UInt32(max)
var last = arc4random_uniform(max)
return {
var r = arc4random_uniform(max)
while r == last {
r = arc4random_uniform(max)
}
last = r
return Int(last)
}
}
let r0 = random(8)
let r1 = random(4)
for i in 0..<20 {
print(r0(), terminator: " ")
}
print("")
for i in 0..<20 {
print(r1(), terminator: " ")
}
/*
4 5 4 3 4 0 5 6 7 3 6 7 5 4 7 4 7 2 1 6
0 3 0 1 0 2 3 1 2 0 1 0 1 0 1 3 0 3 0 2
*/

Answer 3

Fill an array with the colors and shuffle it with a Fisher-Yates shuffle. Then use the element at an end, remove it, and insert it at a random position at least n positions from the end.
For example, say my array has 10 elements. I shuffle it and take the last. I want at least 2 values to be chosen before I see it again so I generate a random position in the range 0...8 and insert it there.
var colorArray = [
(UIColor.redColor() , "red" ),
(UIColor.greenColor() , "green" ),
(UIColor.blueColor() , "blue" ),
(UIColor.yellowColor() , "yellow"),
(UIColor.orangeColor() , "orange"),
(UIColor.lightGrayColor(), "grey" )].shuffle() // shuffle() is from my link above
let spacing = 2 // Pick at least 2 colors before we see it again
if let randomColor = colorArray.popLast() {
colorArray.insert(randomColor,
atIndex: Int(arc4random_uniform(UInt32(colorArray.count - spacing))))
}

Answer 4

One case, described here: https://github.com/dimpiax/GenericSequenceType
Another is functional:
func getRandomItem<T>(arr: [T]) -> (unique: Bool) -> T {
var indexes: [Int]!
return { value in
let uniqIndex: Int
if value {
if indexes?.isEmpty != false {
indexes = [Int](0.stride(to: arr.count, by: 1))
}
uniqIndex = indexes.removeAtIndex(Int(arc4random_uniform(UInt32(indexes.count))))
}
else {
uniqIndex = Int(arc4random_uniform(UInt32(arr.count)))
}
return arr[uniqIndex]
}
}
let generate = getRandomItem(colorArray)
generate(unique: true).0 // greenColor
generate(unique: true).0 // redColor
generate(unique: true).0 // lightGrayColor

Answer 5

Try it it's work for me and 100% tested
let arrString = ["1","2","3","4","5","6"]
var selectedIndix = -1
#IBAction func btnClick(_ sender: Any) {
let randomElementIndex = randomElementString()
}
Call this function when button clicked
func randomElementString() -> Int{
let randomm = Int(arc4random_uniform(UInt32(arrString.count)))
if selectedIndix == randomm{
return randomElementString()
}else{
selectedIndix = randomm
return randomm
}
}
OUTPUT:-
5121242316513126

Answer 6

How about running a while loop with the condition:
while(self.source.backgroundColor == sourceColor) {
// get a new random sourceColor
}
This will keep looping until a new random color has been selected.
edit
Additional Note: The point was the while loop. There are ways to safeguard from an infinite loop, it's up to the coder to find the right solution. I don't think SO is a place to write other's code but instead to offer suggestions .. mine is a start.
But since my answer was given such a negative rating, i'll push instead of nudge in the right direction.
The other answers are unnecessarily bloated. And? The one I offered above offers a less than desirable time complexity. So, here's my new answer (in meta code):
// array of all background colors
var arrayOfColors = [..]
// get a random index
var randomIndex = arc4random(size of arrayOfColors)
// select new background color
var newBGColor = arrayOfColors[randomIndex]
// old background color
var oldBGColor = self.source.backgroundColor
// remove new color from array (so that it's excluded from choices next time)
arrayOfColors.removeAtIndex(randomIndex)
// set the new color to the background
self.source.backgroundColor = newBGColor
// add current color back into the pool of potential colors
arrayOfColors.addObject(oldBGColor)