如何快速获取二维数组中的第一维

Question

How can get the first dimension in a two-dimensional array in swift,what I mean is like this: Here is a two-dimensional array with the type of string:

[["1","2"],["4","5"],["8","9"]]

what I want is array like this:

["1","4","8"]

Answer 1

You can invoke the first instance property on each sub-array as part of a compactMap(_:) invocation of the outermost array.

let arr = [["1", "2"], ["4", "5"], ["8", "9"]]
let firstElements = arr.compactMap { $0.first } // ["1", "4", "8"]

Note however that first is an Optional property, that is nil for empty collections, and that a nil result in transform of the compactMap(_:) invocation will be removed. Eg:

let arr = [["1", "2"], [], ["8", "9"]]
let firstElements = arr.compactMap { $0.first } // ["1", "8"]

For the general case, accessing the nth index in each sub-array, you can make use of the non-optional subscript(_:) accessor as part of a map(_:) invocation on the outermost array, carefully noting however that an attempt to access a non-existing element ( index out of bounds ) will lead to a run-time exception.

let arr = [["1", "2"], ["4", "5"], ["8", "9"]]
let idx = 1

// proceed only if idx is a valid index for all sub-arrays
if idx >= 0 && (!arr.contains { idx >= $0.count }) {
    let subElements = arr.map { $0[idx] } // ["2", "5", "9"]
    // ...
}
else {
    // this would correspond to an index that is invalid in at
    // at least one of the sub-arrays.
}

Alternatively, you could simply filter out sub-array subscript accesses that would correspond to index out of bounds, eg using compactMap(_:) :

let arr = [["1", "2", "3"], ["4", "5"], ["8", "9", "10"]]
let idx = 2
let subElements = arr
    .compactMap { 0..<$0.count ~= idx ? $0[idx] : nil } // ["3", "10"]