ios - 希望在 swift 中使用过滤器来过滤结构的内部数组而不接触外部数组

Question

这是我创建的一个 Playground ，用于尝试创建正确的过滤器:我想要返回的是所有外部结构和任何内部数组值，这些值满足将从搜索栏入站的“文本” block 。

//: Playground - noun: a place where people can play

import UIKit
import Foundation

struct events{
    var title:String
    var unique:Int
    var swimmers_info:[swim_info]

}

struct swim_info{
    var name:String
    var team:String
}

var arrayEV:[events]
var si1:[swim_info]
var si2:[swim_info]

si1 = [
    swim_info(name:"Smith, John",team:"PHNX"),
        swim_info(name:"Tenay, Teresa",team:"CRIM")
]

si2 = [
    swim_info(name:"Test, Two",team:"PHNX"),
    swim_info(name:"Female, Swimmer",team:"SSA")
]


arrayEV = [
    events(title: "First Event",unique: 5,swimmers_info:si1),
    events(title: "Second Event",unique: 2,swimmers_info:si2)
]

//print(arrayEV)

var newarrayEV1 = arrayEV.filter({(value) -> Bool in
    return value.swimmers_info.contains(where: {$0.team == "SSA"})
})

var newarrayEV2 = arrayEV.map({$0.swimmers_info.filter({(value) -> Bool in
    return value.team == "SSA"
})})


print(newarrayEV1)

YEILDS [__lldb_expr_42.events(title: "Second Event", unique: 2, swimmers_info: [__lldb_expr_42.swim_info(name: "Test, Two", team: "PHNX"), __lldb_expr_42.swim_info(name: "Female") , 游泳者", 团队: "SSA")])]

**不正确，因为它只产生第二个主结构和两个子结构

print("  ")
print(newarrayEV2   )

[[], [__lldb_expr_42.swim_info(name: "Female, Swimmer", team: "SSA")]]

不正确，因为删除了所有主要结构信息

print("11")

有点正确

我可以通过以下方式获得所需的结果:

for (index,loop) in arrayEV.enumerated(){
    //print(String(describing: index))
    //print(loop.swimmers_info)
    var new_si = loop.swimmers_info.filter({(value) -> Bool in
        return value.team == "SSA"
    })
    //print(new_si)
    arrayEV[index].swimmers_info = new_si

}

但是我不确定这是否是解决此问题的最佳方式。

Answer 1

根据您建议的“解决方案”，我认为这就是您所追求的:

arrayEV = arrayEV.map {
    var ev = $0
    ev.swimmers_info = ev.swimmers_info.filter {$0.team == "SSA"}
    return ev
}

之所以使用看起来相当笨拙的三行代码，是因为我们不能改变结构的属性，除非对结构本身的引用是 var 引用——但是 map为我们提供了一个 let 引用 ($0)。