Question

我有2个数组。在第一个数组中，我有一个名单，我想从第二个数组中删除。

第一个数组只是字符串数组：

var arrayNames = ["apple", "apricot", "cucumber"]

第二个数组是自定义structs：

的数组

struct fruitsData {
var name: String?
}

var secondArray = [fruitsData(name: "apple"),fruitsData(name: "apricot"), fruitsData(name: "mango"), fruitsData(name: "grapes"), fruitsData(name: "tomato"), fruitsData(name: "lichi"), fruitsData(name: "cucumber"), fruitsData(name: "brinjal")]

那么我怎样才能得到只包含这些数据的数组：

var finalData = [fruitsData(name: "mango"), fruitsData(name: "grapes"), fruitsData(name: "tomato"), fruitsData(name: "lichi"), fruitsData(name: "brinjal")]

，其中不包含arrayNames的任何名称？

Answer 1

有几种方法可以做到这一点：

最好的方法是使用filter方法：

var arrayNames = ["apple", "apricot", "cucumber"]
var secondArray = [fruitsData(name: "apple"),fruitsData(name: "apricot"), fruitsData(name: "mango"), fruitsData(name: "grapes"), fruitsData(name: "tomato"), fruitsData(name: "lichi"), fruitsData(name: "cucumber"), fruitsData(name: "brinjal")]

secondArray = secondArray.filter({$0.name != nil && !arrayNames.contains($0.name!)})

或者，如果您想为了可读性而牺牲效率，可以在辅助数组旁边使用for-in循环：

var arrayNames = ["apple", "apricot", "cucumber"]
var secondArray = [fruitsData(name: "apple"),fruitsData(name: "apricot"), fruitsData(name: "mango"), fruitsData(name: "grapes"), fruitsData(name: "tomato"), fruitsData(name: "lichi"), fruitsData(name: "cucumber"), fruitsData(name: "brinjal")]
var helperArray = [fruitsData]()

for fruit in secondArray {

    if fruit.name != nil && !arrayNames.contains(fruit.name!){
        helperArray.append(fruit)
    }

}

secondArray = helperArray

以上内容将清除secondArray中包含arrayNames名称的所有元素。您应该熟悉 Map, Filter and Reduce 。

Answer 2

finalData secondArray arrayNames除var finalData = secondArray.filter { !arrayNames.contains($0.name!)}之外的kotlinc-jvm可以在此处完成。。

kotlinc

Answer 3

我正在打电话，所以我无法检查，但这应该可以解决问题

secondArray.filter { !firstArray.contains($0.name) }

Answer 4

当想要从给定数组和某个谓词构造子集数组时，自然方法是使用filter。由于name结构的FruitsData是Optional，因此在将其与Optional中的非arrayNames名称进行比较之前，您需要（尝试）展开它。 {1}}列表，例如使用map(_:)的{{1}}方法以及可选链接来处理Optional - 值nil个案例。

E.g：

name

由于// example setup struct FruitsData { var name: String? } let arrayNames = ["apple", "apricot", "cucumber"] let secondArray = [ FruitsData(name: "apple"), FruitsData(name: "apricot"), FruitsData(name: "mango"), FruitsData(name: "grapes"), FruitsData(name: "tomato"), FruitsData(name: "lichi"), FruitsData(name: "cucumber"), FruitsData(name: "brinjal") ] // filter 'secondArray' based on the non-existance of an // associated fruit name in the 'arrayNames' list let finalData = secondArray.filter { $0.name.map { fruitName in !arrayNames.contains(fruitName) } ?? true }符合String，您可能还需要考虑将要删除的水果名称列表（Hashable）设为arrayNames而不是Set {1}}，因为前者在将Array应用于O(1)时会允许contains { ... }查找。 E.g：

let removeNames = Set(arrayNames)
let finalData = secondArray.filter {
    $0.name.map { fruitName in !removeNames.contains(fruitName) } ?? true
}                                        /* ^^^^^^^^- O(1) lookup */

如果您还希望过滤掉FruitsData中secondArray个nil个name个属性的... ?? true个实例，您可以使用... ?? false谓词部分上方为filter：FruitsData操作将过滤掉name属性为nil属性arrayNames或// ... filter out also FruitsData instances with 'nil' valued 'name' let removeNames = Set(arrayNames) let finalData = secondArray.filter { $0.name.map { fruitName in !removeNames.contains(fruitName) } ?? false }中存在的所有{{1}}个实例。

{{1}}

Answer 5

如果名称不是空字符串，则可以使用

let arrayNames = ["apple"]

struct fruitsData {
    var name: String?
}
var secondArray = [fruitsData(name: "apple"),fruitsData(name: "apricot"), fruitsData()]

// [__lldb_expr_98.fruitsData(name: Optional("apple"))]
let flt1 = secondArray.filter { arrayNames.contains($0.name ?? "") }

// [__lldb_expr_87.fruitsData(name: Optional("apricot")), __lldb_expr_87.fruitsData(name: nil)]
let flt2 = secondArray.filter { !arrayNames.contains($0.name ?? "") }

如何根据名称数组过滤数组？

5 个答案: