Question

说我有两个阵列：

var arrayA = ["Yes", "Yes2", "Not Answered", "No"]
var arrayB = ["Yes", "NA", "Yes2", "NA"]

我想通过执行以下操作从arrayB中删除“NA”：

var filtered = arrayB.filter({$0 != "NA"})

如何删除在arrayA中删除的相同索引处的项目。我想过使用find（）函数，但只返回字符串出现的第一个索引。您可以通过以下方式从阵列中删除重叠：

let res = arrayA.filter { !contains(arrayB, $0) }

但是如何根据另一个数组的过滤来过滤数组呢？

结果会有：

arrayBFiltered = ["Yes", "Yes2"]
arrayAFiltered = ["Yes", "Not Answered"]

有什么想法吗？

Answer 1

您可能更喜欢使用zip：

var arrayA = ["Yes", "Yes2", "Not Answered", "No"]
var arrayB = ["Yes", "NA", "Yes2", "NA"]

let result = filter(zip(arrayA, arrayB)) { (a, b) in b != "NA" }

for (a, b) in result {
    println("A: \(a) -> B: \(b)")
}

编辑：SWIFT 2.0

在Swift 2.0中，获取一个合并后的广告struct（比如...... Foo）的数组会更加容易，以获得更清晰的后续代码：

struct Foo {
    let a: String
    let b: String
}
// Foo.init will be the function automatically generated by the default initialiser

let result = zip(arrayA, arrayB)
    .filter { (a, b) in b != "NA" }
    .map(Foo.init)
// Order of a and b is important

// result is an Array<Foo> suitable for a clearer subsequent code
for item in result {
    print("A: \(item.a) -> B: \(item.b)")
}

希望这有帮助

Answer 2

来自How can I sort multiple arrays based on the sorted order of another array的想法可以在这里使用，这将适用于两个或更多阵列：

let arrayA = ["Yes", "Yes2", "Not Answered", "No"]
let arrayB = ["Yes", "NA", "Yes2", "NA"]

// Determine array indices that should be kept:
let indices = map(filter(enumerate(arrayB), { $1 != "NA" } ), { $0.0 } )

// Filter arrays based on the indices:
let arrayAFiltered = Array(PermutationGenerator(elements: arrayA, indices: indices))
let arrayBFiltered = Array(PermutationGenerator(elements: arrayB, indices: indices))

println(arrayAFiltered) // [Yes, Not Answered]
println(arrayBFiltered) // [Yes, Yes2]

Answer 3

另一种解决方案是直接在filter闭包内部删除代码：

// both are vars so you can mutate them directly
var arrayA = ["Yes", "Yes2", "Not Answered", "No"]
var arrayB = ["Yes", "NA", "Yes2", "NA"]

arrayA = filter(enumerate(arrayA)){
    arrayB.removeAtIndex($0)
    return $1 != "Na"
}

// use filtered arrayA and arrayB

基于另一个数组Swift的过滤过滤数组

3 个答案: