You can use enumerate to convert a sequence (Array, String, etc.) to a sequence of tuples with an integer counter and and element paired together. That is:

let numbers = [7, 8, 9, 10]
let indexAndNum: [String] = numbers.enumerate().map { (index, element) in
    return "\(index): \(element)"
}
print(indexAndNum)
// ["0: 7", "1: 8", "2: 9", "3: 10"]

Link to enumerate definition

Note that this isn't the same as getting the index of the collection—enumerate gives you back an integer counter. This is the same as the index for an array, but on a string or dictionary won't be very useful. To get the actual index along with each element, you can use zip:

let actualIndexAndNum: [String] = zip(numbers.indices, numbers).map { "\($0): \($1)" }
print(actualIndexAndNum)
// ["0: 7", "1: 8", "2: 9", "3: 10"]

When using an enumerated sequence with reduce, you won't be able to separate the index and element in a tuple, since you already have the accumulating/current tuple in the method signature. Instead, you'll need to use .0 and .1 on the second parameter to your reduce closure:

let summedProducts = numbers.enumerate().reduce(0) { (accumulate, current) in
    return accumulate + current.0 * current.1
    //                          ^           ^
    //                        index      element
}
print(summedProducts)   // 56

Swift 3.0 and above

Since Swift 3.0 syntax is quite different.
Also, you can use short-syntax/inline to map array on dictionary:

let numbers = [7, 8, 9, 10]
let array: [(Int, Int)] = numbers.enumerated().map { (1) }
//                                                     ^   ^
//                                                   index element

That produces:

[(0, 7), (1, 8), (2, 9), (3, 10)]

// must be sorted by 'day'!!!
let arrA0 = [2.45, 2.75, 2.9, 3.1, 3.2, 3.3]
// associated values
let arrA1 = [145.0, 150.0, 160.0, 245.0, 250.0, 260.0]

let arr = Array(zip(arrA0, arrA1))

// now i have an array of tuples, where tuple.0 is key and tuple.1 etc. is associated value
// you can expand the tuple for as much associated values, as you want
print(arr)
// [(2.45, 145.0), (2.75, 150.0), (2.9, 160.0), (3.1, 245.0), (3.2, 250.0), (3.3, 260.0)]




// now i can perform my 'calculations' the most effective way
var res:[Int:(Double,Double)] = [:]
// sorted set of Int 'day' values
let set = Set(arr.map {Int($0.0)}).sort()
// for two int values the sort is redundant, but
// don't be depend on that!

print(set)
// [2, 3]
var g = 0
var j = 0
set.forEach { (i) -> () in
    var sum1 = 0.0
    var sum2 = 0.0
    var t = true
    while t && g < arr.count {
        let v1 = arr[g].0
        let v2 = arr[g].1
        t = i == Int(v1)
        if t {
            g++
            j++
        } else {
            break
        }
        sum1 += v1
        sum2 += v2
    }
    res[i] = (sum1 / Double(j), sum2 / Double(j))
    j = 0
}
print(res)
// [2: (2.7, 151.666666666667), 3: (3.2, 251.666666666667)]

see, that every element of your data is process only once in the 'calculation', independent from size of 'key' set size

use Swift's Double instead of CGFloat! this increase the speed too :-)

and finally what you are looking for

if let (day, value) = res[2] {
    print(day, value) // 2.7 151.666666666667
}

Let's start with a single array, like:

Copylet raceResult = ["one", "two", "four"]

If we want to combine each element with an offset counting from 0, we can use Array.enumerated(), along with map.

Copylet numberedRaceResult = raceResult
    .enumerated()
    .map { offset, element in "\(offset). \(element)" }

for numberedResult in numberedRaceResult {
    print(numberedResult)
}

// Prints:
// 0. one
// 1. two
// 2. four

You can see that I didn't call print inside the closure passed to map. You can do this, but it kind of defeats the purpose of map (which is to create an equal-sized output array from the transformed elements of the input array), because the result would be unused. In that case, it makes more sense to just use a for loop or a call to forEach, like @Sh_Khan showed.

To handle a nested array, it's much the same. We can use the same logic as for one array, but apply it to each sub-array.

Copylet raceResults = [
    ["one", "two", "four"],
    ["two", "one", "five", "six"],
    ["two", "one", "four", "ten"],
    ["one", "two", "four"],
]

let numberedRaceResults = raceResults
    .enumerated()
    .flatMap { outterOffset, raceResult in
        raceResult
            .enumerated()
            .map { innerOffset, element in "\(outterOffset).\(innerOffset). \(element)" }
    }

for numberedResult in numberedRaceResults {
    print(numberedResult)
}

// Prints:
// 0.0. one
// 0.1. two
// 0.2. four
// 1.0. two
// 1.1. one
// 1.2. five
// 1.3. six
// 2.0. two
// 2.1. one
// 2.2. four
// 2.3. ten
// 3.0. one
// 3.1. two
// 3.2. four

You'll notice that I used flatMap on the outter array, instead of a simple map. You can change it back and forth and compare the result. In short, flatMap gives you a single flat array of string results, rather than an array of sub-arrays of strings.

You can filter the indices of the array directly, it avoids the extra mapping.

let items = ["A", "B", "A", "C", "A", "D"]
let filteredIndices = items.indices.filter {items[$0] == "A"}

or as Array extension:

extension Array where Element: Equatable {

    func whatFunction(_ value :  Element) -> [Int] {
        return self.indices.filter {self[$0] == value}
    }

}

items.whatFunction("A") // -> [0, 2, 4]
items.whatFunction("B") // -> [1]

or still more generic

extension Collection where Element: Equatable {

    func whatFunction(_ value :  Element) -> [Index] {
        return self.indices.filter {self[$0] == value}
    }

}

Here's one way of doing this:

print(
    productsList
        .enumerated()
        .filter { $0.offset % 2 == 1 }
        .map { $0.element.name }
)

The enumerated method turns each product into a 2-tuple that contains both the product and its index in the array (offset). You then filter to leave only the products that have an odd (or even) index. After that, you need to map the 2-tuples to the name of the product.

As swift is in some regards more functional than object-oriented (and Arrays are structs, not objects), use the function "find" to operate on the array, which returns an optional value, so be prepared to handle a nil value:

let arr:Array = ["a","b","c"]
find(arr, "c")!              // 2
find(arr, "d")               // nil

Use firstIndex and lastIndex - depending on whether you are looking for the first or last index of the item:

let arr = ["a","b","c","a"]

let indexOfA = arr.firstIndex(of: "a") // 0
let indexOfB = arr.lastIndex(of: "a") // 3