使用iOS swift合并两个结构数组

Question

如何根据结构（名称）中的键合并两个（第一个和第二个）结构数组。合并时，我需要用第二个数组元素替换第一个数组的现有元素，如果该元素中有任何值发生变化。

struct Example: Codable {
    var name: String
    var dob: String
    var address: String
}

var first: [Example] = []
var second: [Example] = []

first.append(Example(name: "Arun", dob: "01-01-1994", address: "Tirupati"))
first.append(Example(name: "Balaji", dob: "01-01-1994", address: "Tirupati"))
first.append(Example(name: "Prasanth", dob: "01-01-1994", address: "Tirupati"))
    first.append(Example(name: "Satish", dob: "01-01-1994", address: "Tirupati"))


second.append(Example(name: "Arun", dob: "01-01-1994", address: "Kadapa"))
second.append(Example(name: "Balaji", dob: "01-01-1994", address: "Tirupati"))
second.append(Example(name: "Prasanth", dob: "01-01-1994", address: "Tirupati"))
second.append(Example(name: "Harsha", dob: "01-01-1994", address: "Tirupati"))

/*    let merged: [Example] = merge(first, second, with: name)
'merged' must contain four elements Arun, Balaji, Prasanth, Satish, Harsha, 
but Arun details are from 'second', because 'address' changed in second. So, I need element from 'second'*/

请告诉我简单的方法，谢谢。

Answer 1

import Foundation

struct Example: Codable {
    var name: String
    var dob: String
    var address: String
}

var first: [Example] = []
var second: [Example] = []

first.append(Example(name: "Arun", dob: "01-01-1994", address: "Tirupati"))
first.append(Example(name: "Balaji", dob: "01-01-1994", address: "Tirupati"))
first.append(Example(name: "Prasanth", dob: "01-01-1994", address: "Tirupati"))

second.append(Example(name: "Arun", dob: "01-01-1994", address: "Kadapa"))
second.append(Example(name: "Balaji", dob: "01-01-1994", address: "Tirupati"))
second.append(Example(name: "Prasanth", dob: "01-01-1994", address: "Tirupati"))
second.append(Example(name: "Harsha", dob: "01-01-1994", address: "Tirupati"))

first = second + first.filter { element in
    return !second.contains { $0.name == element.name }
}

[{name "Arun", dob "01-01-1994", address "Kadapa"},
{name "Balaji", dob "01-01-1994", 地址 "Tirupati"},
{name "Prasanth", dob "01-01-1994", 地址 "Tirupati"},
{name "Harsha", dob "01-01-1994", 地址“蒂鲁帕蒂”}]

Answer 2

如果合并数组中元素的顺序无关紧要，那么您可以使用 将每个名称映射到具有该名称的（最近的）元素的字典 （类似于user2760845 suggested）。

var dict: [String: Example] = [:]
for elem in first { dict[elem.name] = elem }
for elem in second { dict[elem.name] = elem }
let merged = Array(dict.values)

遍历第二个数组会覆盖同名的条目 从第一个数组。

或者，作为（混淆的？）单行：

let merged = Array(Dictionary([first, second].joined().map { ($0.name, $0)}, uniquingKeysWith: { $1 }).values)

Answer 3

下面是一个简单但非常高效的实现。如果您需要保持元素的相对顺序，则需要对解决方案进行一些调整。

var secondDict = [String: Example]()
var merge = [Example]()
for eg2 in second {
    secondDict[eg2.name] = eg2
}
for eg1 in first {
    if let eg2 = secondDict.removeValue(forKey: eg1.name) {
        merge.append(eg2)
    } else {
        merge.append(eg1)
    }
}
for eg2 in secondDict.values {
    merge.append(eg2)
}
// 'merge' is the merged array

Answer 4

您可以通过使用以下函数使用第二个数组中的信息更新重复名称来合并两个 Example 数组。如果second 不包含first 的所有元素，此功能也有效。

func merge(first: [Example], second: [Example]) -> [Example] {
    var secondCopy = second
    let updatedFirst = first.map({ person -> Example in
        let updatedIndex = secondCopy.index(where: {$0.name == person.name})
        if let updatedIndex = updatedIndex {
            let updated = secondCopy[updatedIndex]
            secondCopy.remove(at: updatedIndex)
            return updated
        } else {
            return person
        }
    })
    return updatedFirst + secondCopy
}

let merged = merge(first: first, second: second)

[{name "Arun", dob "01-01-1994", address "Kadapa"},

{name "Balaji", dob "01-01-1994", address "Tirupati"},

{name "Prasanth", dob "01-01-1994", address "Tirupati"},

{姓名“Harsha”，出生日期“01-01-1994”，地址“Tirupati”}]

Answer 5

你可以试试这个，不知道是不是你需要的。

struct Example: Codable, Hashable {

    var hashValue: Int {
        return self.name.hashValue
    }

    var name: String
    var dob: String
    var address: String
}

func ==(lhs: Example, rhs: Example) -> Bool {
    return lhs.name == rhs.name
}

func merge(first: [Example], second: [Example]) -> [Example] {
    return Array(Set(second).union(first))
}

let merged = merge(first: first, second: second) // is this what you need?