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在go中循环切片/数组的有效方法

slice go loops performance arrays

juancsr · 技术社区 · 7 年前

基本上我有:

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:

Eliot Anderson real(true)
Tyrell Wellick real(true)

但我不想在s2内循环s1,然后在s3内循环s1

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,还有什么其他方法可以做到这一点呢?

3 回复 | 直到 7 年前

Ravi R 7 年前

正确的方法是将它们放入散列(在Golang中称为map)。这样,您就可以获得性能,并且只需在id上迭代一个循环。

以下是示例数据:

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 在操场上。希望这有帮助!

p、最后,如果您可以删除某些ID的所有结构条目并添加新的ID,您可以考虑将有效ID添加到映射中 map[int]bool ( mymap[id] = true )并使用 range 而不是 for i.. 同上。

Daniel Tran 7 年前

使用地图按人名索引数据。

type Person struct {
     firstName: string
     lastName: string 
     real: string
}

var data map[int]Person

然后查找并将数据放到地图上。

Eugene Lisitsky 7 年前

事实上,要连接所有第一个元素,然后连接所有第二个元素等等,你不需要在一个循环中做一个循环:

将每个切片准备为地图:

m1 := make(map[int]string)
for i:=0; i < len(s1); i ++ {
    m1[s1[i].id] = s1[i].Name
}

其他两片也一样。最后迭代一次:

for i:=0; i < len(s1); i ++ {
    fmt.Println(m1[i], m2[i], m3[i])
}

该解决方案假设所有切片都有相应的项。如果没有,您应该决定如何处理没有片段的元素:忽略、用一些占位符替换空白、中断整个循环等 firstName 是强制性的(我们将对此进行迭代), secondName 是可选的,应替换为 ? 如果不存在,则整个循环中绝对必须使用实数-如果不存在实数,则中断进一步的工作并返回空切片:

type Person struct { // corrected struct from the neighbour recipe
     firstName: string
     lastName: string 
     real: bool
}

var persons []Person // slice for result

for i, firstName := range m1 { // iterate by keys and values of m1
    secondName, ok := m2[i]
    if !ok {
       secondName = "?"
    }
    real, ok := m3[i]
    if !ok {
       persons = []Person
       break
    }
    person := Person{firstName, secondName, real}
    persons = append(persons, person)
}