Putting the count of a loop in an array index

Question

Hello am new to SO and i need some help. Is there a way i can put the value of count of a loop as an index for this:

var groups = [[String]]()

groupA.append(arr[randoms[0]])
groupA.append(arr[randoms[1]])

groupB.append(arr[randoms[2]])
groupB.append(arr[randoms[3]])

groupC.append(arr[randoms[4]])
groupC.append(arr[randoms[5]])

groupD.append(arr[randoms[6]])
groupD.append(arr[randoms[7]])

groupE.append(arr[randoms[8]])
groupE.append(arr[randoms[9]])

groups.append(groupA)
groups.append(groupB)
groups.append(groupC)
groups.append(groupD)
groups.append(groupE)

i want the index 0-10 of the random array to be from a loop or be dynamic in some way is there a way i can achieve this with swift?

Why do you need randoms at all? Just generate the random numbers. — matt
– matt, Commented Mar 6, 2017 at 19:53
I generated random number that won't repeat the same number and stored it in an array — Hafeez Sagay
– Hafeez Sagay, Commented Mar 6, 2017 at 19:59
In that case, why not just shuffle arr? Now the first 10 items of the shuffled array are your items to be dealt out into groups. — matt
– matt, Commented Mar 6, 2017 at 20:01
After shuffling all the items how can i group it then? because am using a multidimensional array here — Hafeez Sagay
– Hafeez Sagay, Commented Mar 6, 2017 at 20:05

harrisg · Accepted Answer · 2017-03-06 20:09:41Z

1

I think you are looking for the enumerated() function...

import Foundation
let randoms = [arc4random(),arc4random(),arc4random(),arc4random(),]
randoms.enumerated().forEach {
    print($0,$1)
}

Try that out in an Xcode Playground.

edited Mar 6, 2017 at 20:09

answered Mar 6, 2017 at 20:01

harrisg

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3 Comments

Hafeez Sagay Over a year ago

This did not work and how am i supposed to use it with my code?

BallpointBen Over a year ago

Don't forget to import Foundation

harrisg Over a year ago

Hafeez Sagay, I'm just trying to show you how to incorporate the index of an item in an array into a loop.

Community · Accepted Answer · 2017-05-23 12:00:21Z

You might be looking for this sort of thing. Let's get rid of random and just shuffle arr to generate a list of actual values in random order. (There's lots of array-shuffling code on Stack Overflow so I won't bother to tell you how to do that.) Then we can also get rid of your intermediate arrays and just deal directly into the inner arrays of the array-of-arrays groups:

var groups : [[String]] = Array(repeating: [], count: 5)
var shuffled = ["hey", "ho", "nonny", "no", "sis", "boom", "bah", "yo", "hi", "lo"]

var next = 0
for ix in stride(from:0, to:9, by:2) {
    for ixx in 0..<2 {
        groups[next].append(shuffled[ix+ixx])
    }
    next += 1
}

groups // [["hey", "ho"], ["nonny", "no"], ["sis", "boom"], ["bah", "yo"], ["hi", "lo"]]

We can actually express that more concisely using map (as explained Swift: what is the right way to split up a [String] resulting in a [[String]] with a given subarray size?):

let shuffled = ["hey", "ho", "nonny", "no", "sis", "boom", "bah", "yo", "hi", "lo"]
let chunkSize = 2
let groups = stride(from: 0, to: shuffled.count, by: chunkSize).map {
    Array(shuffled[$0..<min($0 + chunkSize, shuffled.count)])
}

groups // [["hey", "ho"], ["nonny", "no"], ["sis", "boom"], ["bah", "yo"], ["hi", "lo"]]

Even better, we can make this an extension on Array:

extension Array {
    func chunked(by chunkSize:Int) -> [[Element]] {
        let groups = stride(from: 0, to: self.count, by: chunkSize).map {
            Array(self[$0..<[$0 + chunkSize, self.count].min()!])
        }
        return groups
    }
}


let shuffled = ["hey", "ho", "nonny", "no", "sis", "boom", "bah", "yo", "hi", "lo"]
let groups = shuffled.chunked(by:2)

groups // [["hey", "ho"], ["nonny", "no"], ["sis", "boom"], ["bah", "yo"], ["hi", "lo"]]

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