arrays - 在 go 中循环 slice/数组的有效方法

Question

基本上我有这个:

package main

import "fmt"

type Struct1 struct {
    id   int
    name string
}

type Struct2 struct {
    id       int
    lastname string
}

type Struct3 struct {
    id   int
    real bool
}

func main() {
    var (
     s1 []Struct1
     s2 []Struct2
     s3 []Struct3
    )
    s1 = append(s1, Struct1{id: 1, name: "Eliot"}, Struct1{id: 2, name: "Tyrell"}, Struct1{id: 3, name: "Mr Robot"})
    s2 = append(s2, Struct2{id: 1, lastname: "Anderson"}, Struct2{id: 2, lastname: "Wellick"})
    s3 = append(s3, Struct3{id: 1, real: true}, Struct3{id: 2, real: true}, Struct3{id: 3, real: false})
}

我想展示这样的东西:

• 艾略特安德森真实(真实)
• Tyrell Wellick 真实(真实)

但我不想在 s2 中循环 s1，然后在 s3 中循环

例子:

for i := 0; i < len(s1); i++ {
        for j := 0; j < len(s2); j++ {
            if s1[i].id == s2[j].id {
                for k := 0; k < len(s3); k++ {
                    if s2[j].id == s3[k].id {
                        // some code ...
                    }
                }
            }
        }
    }

那么，还有哪些其他方法可以做到这一点？

Answer 1

正确的方法是将它们放在一个散列中(在 Golang 中称为 map)。这样您可以获得性能，并且只需一个循环遍历 id 即可。

这是您的示例数据的示例:

package main

import (
    "fmt"
)

type Struct1 struct {
    id   int
    name string
}

type Struct2 struct {
    id       int
    lastname string
}

type Struct3 struct {
    id   int
    real bool
}

func main() {
    //var (
    //s1 []Struct1
    //      s2 []Struct2
    //  s3 []Struct3
    //  )
    s1Hash := make(map[int]Struct1)
    s2Hash := make(map[int]Struct2)
    s3Hash := make(map[int]Struct3)

    s11 := Struct1{id: 1, name: "Eliot"}
    s12 := Struct1{id: 2, name: "Tyrell"}
    s13 := Struct1{id: 3, name: "Mr Robot"}
    s1Hash[s11.id] = s11
    s1Hash[s12.id] = s12
    s1Hash[s13.id] = s13

    s21 := Struct2{id: 1, lastname: "Anderson"}
    s22 := Struct2{id: 2, lastname: "Wellick"}
    s2Hash[s21.id] = s21
    s2Hash[s22.id] = s22

    s31 := Struct3{id: 1, real: true}
    s32 := Struct3{id: 2, real: true}
    s33 := Struct3{id: 3, real: false}
    s3Hash[s31.id] = s31
    s3Hash[s32.id] = s32
    s3Hash[s33.id] = s33

    //s1 = append(s1, Struct1{id: 1, name: "Eliot"}, Struct1{id: 2, name: "Tyrell"}, Struct1{id: 3, name: "Mr Robot"})
    //s2 = append(s2, Struct2{id: 1, lastname: "Anderson"}, Struct2{id: 2, lastname: "Wellick"})
    //s3 = append(s3, Struct3{id: 1, real: true}, Struct3{id: 2, real: true}, Struct3{id: 3, real: false})

    //i to loop over possible id range
    for i := 1; i <= len(s1Hash); i++ {
        fmt.Println("i is ", i)
        if _, ok := s1Hash[i]; ok {
            fmt.Printf("Name: %s ", s1Hash[i].name)
        }

        if _, ok := s2Hash[i]; ok {
            fmt.Printf(" Lastname: %s ", s2Hash[i].lastname)
        }

        if _, ok := s3Hash[i]; ok {
            fmt.Printf(" Real: %t\n", s3Hash[i].real)
        }
        //fmt.Printf("%s %s real:%t\n", s1Hash[i].name, s2[i].lastname, s3[i].real)
    }

}

输出:

i is  1
Name: Eliot  Lastname: Anderson  Real: true
i is  2
Name: Tyrell  Lastname: Wellick  Real: true
i is  3
Name: Mr Robot  Real: false

检查 this在 Playground 上。希望这对您有所帮助!

附注:最终，如果您可能删除某些 ID 的所有结构条目并添加更新的 ID - 您可以考虑将有效 ID 添加到映射 map[int]bool (mymap[id] = true) 并使用 range 代替上面的 for i.. 遍历 map 。