斯威夫特：搜索在数组中包含数字的数组索引的更好方法？

Question

Curious to know if there is a better way to do this. I have an array of arrays that contain integers. All the integers are unique. I want to search for a given number and return the index of the first array. I don't care what the index is of the array containing the numbers.

This does what I want but I'm new to swift and figured there is a better way.

let arr:[[Int]] = [[0], [1], [2, 3], [4], [11, 12, 13, 14, 15, 16], [5], [6], [7], [8], [9], [10]]

var t = -1
for q in arr {
    t += 1
    if let x = q.index(of: 13) { // <-- looking for the index that contains 13
        print ("t: \(t)") // <-- this is what I want. Returns '4'
    }
}

Answer 1

Here's one way:

let arr: [[Int]] = [[0], [1], [2, 3], [4], [11, 12, 13, 14, 15, 16], [5], [6], [7], [8], [9], [10]]

for (i, a) in arr.enumerated() {
    if let _ = a.index(of: 13) {
        print(i)
    }
}

It makes use of the enumerated() method of collections. It returns a tuple containing the index of the array i and the actual element of the array a .

Using functional programming

Here's the functional programming version if that's more your style:

arr.enumerated().map { $0 }.map { i, a in
    if let _ = a.index(of: 13) { 
        print(i) 
    }
}

Answer 2

A solution without loop using index(where:

let arr: [[Int]] = [[0], [1], [2, 3], [4], [11, 12, 13, 14, 15, 16], [5], [6], [7], [8], [9], [10]]

if let index = arr.index(where: { $0.contains(13) }) {
    print(index)
} else {
    print("not found")
}