Question

我遇到了一个问题，我有一个结构数组，例如

package main

import "log"

type Planet struct {
    Name       string  `json:"name"`
    Aphelion   float64 `json:"aphelion"`   // in million km
    Perihelion float64 `json:"perihelion"` // in million km
    Axis       int64   `json:"Axis"`       // in km
    Radius     float64 `json:"radius"`
}

func main() {
    var mars = new(Planet)
    mars.Name = "Mars"
    mars.Aphelion = 249.2
    mars.Perihelion = 206.7
    mars.Axis = 227939100
    mars.Radius = 3389.5

    var earth = new(Planet)
    earth.Name = "Earth"
    earth.Aphelion = 151.930
    earth.Perihelion = 147.095
    earth.Axis = 149598261
    earth.Radius = 6371.0

    var venus = new(Planet)
    venus.Name = "Venus"
    venus.Aphelion = 108.939
    venus.Perihelion = 107.477
    venus.Axis = 108208000
    venus.Radius = 6051.8

    planets := [...]Planet{*mars, *venus, *earth}
    log.Println(planets)
}

假设您想按Axis对其进行排序。你是怎么做到的？

（注意：我看过http://golang.org/pkg/sort/它似乎有用，但我必须添加大约20行才能通过一个非常简单的键进行简单排序。我有一个python背景，它就像{一样简单{1}} - 在Go中有类似的简单吗？）

Answer 1

从Go 1.8开始，您现在可以使用sort.Slice对切片进行排序：

sort.Slice(planets, func(i, j int) bool {
  return planets[i].Axis < planets[j].Axis
})

通常没有理由使用数组而不是切片，但在您的示例中，使用数组，因此您必须使用切片覆盖它（添加[:] ）使它与sort.Slice：

一起使用

sort.Slice(planets[:], func(i, j int) bool {
  return planets[i].Axis < planets[j].Axis
})

排序会更改数组，因此如果您真的想要，可以在排序后继续使用数组而不是切片。

Answer 2

更新：此答案与较早版本的go有关。对于Go 1.8及更新版本，请参阅AndreKR's answer below。

如果您想要比标准库sort包更简洁的东西，您可以使用第三方github.com/bradfitz/slice包。它使用一些技巧来生成排序切片所需的Len和Swap方法，因此您只需提供Less方法。

使用此软件包，您可以使用以下命令执行排序：

slice.Sort(planets[:], func(i, j int) bool {
    return planets[i].Axis < planets[j].Axis
})

生成覆盖阵列的切片需要planets[:]部分。如果您使planets切片而不是数组，则可以跳过该部分。

Answer 3

从Go 1.8开始，@ AndreKR＆＃39; answer是更好的解决方案。

您可以实现实现sort interface。

的集合类型

Here's an example两种类型允许您按轴或名称排序：

package main

import "log"
import "sort"

// AxisSorter sorts planets by axis.
type AxisSorter []Planet

func (a AxisSorter) Len() int           { return len(a) }
func (a AxisSorter) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a AxisSorter) Less(i, j int) bool { return a[i].Axis < a[j].Axis }

// NameSorter sorts planets by name.
type NameSorter []Planet

func (a NameSorter) Len() int           { return len(a) }
func (a NameSorter) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a NameSorter) Less(i, j int) bool { return a[i].Name < a[j].Name }

type Planet struct {
    Name       string  `json:"name"`
    Aphelion   float64 `json:"aphelion"`   // in million km
    Perihelion float64 `json:"perihelion"` // in million km
    Axis       int64   `json:"Axis"`       // in km
    Radius     float64 `json:"radius"`
}

func main() {
    var mars Planet
    mars.Name = "Mars"
    mars.Aphelion = 249.2
    mars.Perihelion = 206.7
    mars.Axis = 227939100
    mars.Radius = 3389.5

    var earth Planet
    earth.Name = "Earth"
    earth.Aphelion = 151.930
    earth.Perihelion = 147.095
    earth.Axis = 149598261
    earth.Radius = 6371.0

    var venus Planet
    venus.Name = "Venus"
    venus.Aphelion = 108.939
    venus.Perihelion = 107.477
    venus.Axis = 108208000
    venus.Radius = 6051.8

    planets := []Planet{mars, venus, earth}
    log.Println("unsorted:", planets)

    sort.Sort(AxisSorter(planets))
    log.Println("by axis:", planets)

    sort.Sort(NameSorter(planets))
    log.Println("by name:", planets)
}

Answer 4

您可以，而不是在Sort interface上实现[]Planet，而不是在包含集合的类型和将进行比较的闭包上实现。您必须为每个属性提供比较闭包的实现。

我觉得这个方法比为struct的每个属性实现Sort类型要好。

这个答案几乎是从sort docs直接撕掉的，所以我不能为此付出太多的荣誉

package main

import (
    "log"
    "sort"
)

type Planet struct {
    Name       string  `json:"name"`
    Aphelion   float64 `json:"aphelion"`   // in million km
    Perihelion float64 `json:"perihelion"` // in million km
    Axis       int64   `json:"Axis"`       // in km
    Radius     float64 `json:"radius"`
}

type By func(p1, p2 *Planet) bool

func (by By) Sort(planets []Planet) {
    ps := &planetSorter{
        planets: planets,
        by:      by, 
    }
    sort.Sort(ps)
}

type planetSorter struct {
    planets []Planet
    by      func(p1, p2 *Planet) bool 
}

func (s *planetSorter) Len() int {
    return len(s.planets)
}

func (s *planetSorter) Swap(i, j int) {
    s.planets[i], s.planets[j] = s.planets[j], s.planets[i]
}

func (s *planetSorter) Less(i, j int) bool {
    return s.by(&s.planets[i], &s.planets[j])
}

如何调用它。

func main() {
    /* Same code as in the question */

    planets := []Planet{*mars, *venus, *earth}

    By(func(p1, p2 *Planet) bool {
        return p1.Name < p2.Name
    }).Sort(planets)

    log.Println(planets)

    By(func(p1, p2 *Planet) bool {
        return p1.Axis < p2.Axis
    }).Sort(planets)

    log.Println(planets)
}

Here is a Demo

Answer 5

这是另一种减少锅炉板的方法。免责声明，它使用反射和损失类型安全。

Here is a Demo

所有的魔法都发生在Prop函数中。它需要对struct属性进行排序以及对其进行排序（升序，降序）的顺序，并返回将执行比较的函数。

package main

import (
    "log"
    "reflect"
    "sort"
)

func test(planets []Planet) {
    log.Println("Sort Name")
    By(Prop("Name", true)).Sort(planets)
    log.Println(planets)

    log.Println("Sort Aphelion")
    By(Prop("Aphelion", true)).Sort(planets)
    log.Println(planets)

    log.Println("Sort Perihelion")
    By(Prop("Perihelion", true)).Sort(planets)
    log.Println(planets)

    log.Println("Sort Axis")
    By(Prop("Axis", true)).Sort(planets)
    log.Println(planets)

    log.Println("Sort Radius")
    By(Prop("Radius", true)).Sort(planets)
    log.Println(planets)
}

func Prop(field string, asc bool) func(p1, p2 *Planet) bool {
    return func(p1, p2 *Planet) bool {

        v1 := reflect.Indirect(reflect.ValueOf(p1)).FieldByName(field)
        v2 := reflect.Indirect(reflect.ValueOf(p2)).FieldByName(field)

        ret := false

        switch v1.Kind() {
        case reflect.Int64:
            ret = int64(v1.Int()) < int64(v2.Int())
        case reflect.Float64:
            ret = float64(v1.Float()) < float64(v2.Float())
        case reflect.String:
            ret = string(v1.String()) < string(v2.String())
        }

        if asc {
            return ret
        }
        return !ret
    }
}

type Planet struct {
    Name       string  `json:"name"`
    Aphelion   float64 `json:"aphelion"`   // in million km
    Perihelion float64 `json:"perihelion"` // in million km
    Axis       int64   `json:"Axis"`       // in km
    Radius     float64 `json:"radius"`
}

type By func(p1, p2 *Planet) bool

func (by By) Sort(planets []Planet) {
    ps := &planetSorter{
        planets: planets,
        by:      by, // The Sort method's receiver is the function (closure) that defines the sort order.
    }
    sort.Sort(ps)
}

type planetSorter struct {
    planets []Planet
    by      func(p1, p2 *Planet) bool // Closure used in the Less method.
}

// Len is part of sort.Interface.
func (s *planetSorter) Len() int { return len(s.planets) }

// Swap is part of sort.Interface.
func (s *planetSorter) Swap(i, j int) {
    s.planets[i], s.planets[j] = s.planets[j], s.planets[i]
}

// Less is part of sort.Interface. It is implemented by calling the "by" closure in the sorter.
func (s *planetSorter) Less(i, j int) bool {
    return s.by(&s.planets[i], &s.planets[j])
}

func main() {
    test(dataSet())
}

func dataSet() []Planet {

    var mars = new(Planet)
    mars.Name = "Mars"
    mars.Aphelion = 249.2
    mars.Perihelion = 206.7
    mars.Axis = 227939100
    mars.Radius = 3389.5

    var earth = new(Planet)
    earth.Name = "Earth"
    earth.Aphelion = 151.930
    earth.Perihelion = 147.095
    earth.Axis = 149598261
    earth.Radius = 6371.0

    var venus = new(Planet)
    venus.Name = "Venus"
    venus.Aphelion = 108.939
    venus.Perihelion = 107.477
    venus.Axis = 108208000
    venus.Radius = 6051.8

    return []Planet{*mars, *venus, *earth}
}

Answer 6

您还可以使用快速排序来实现，并且可以在分区函数内选择要排序的字段，例如，我选择“名称”。

package main

import (
    "fmt"
)

type Planet struct {
    Name       string  `json:"name"`
    Aphelion   float64 `json:"aphelion"`   // in million km
    Perihelion float64 `json:"perihelion"` // in million km
    Axis       int64   `json:"Axis"`       // in km
    Radius     float64 `json:"radius"`
}

func main() {
    var mars Planet
    mars.Name = "Mars"
    mars.Aphelion = 249.2
    mars.Perihelion = 206.7
    mars.Axis = 227939100
    mars.Radius = 3389.5

    var earth Planet
    earth.Name = "Earth"
    earth.Aphelion = 151.930
    earth.Perihelion = 147.095
    earth.Axis = 149598261
    earth.Radius = 6371.0

    var venus Planet
    venus.Name = "Venus"
    venus.Aphelion = 108.939
    venus.Perihelion = 107.477
    venus.Axis = 108208000
    venus.Radius = 6051.8

    planets := []Planet{mars, venus, earth}
    fmt.Println(quickSort(&planets,0,len(planets)-1))

}

func quickSort(arr *[]Planet, start, end int)[]Planet{
    if start < end{
        partitionIndex := partition(*arr,start,end)
        quickSort(arr,start,partitionIndex-1)
        quickSort(arr,partitionIndex+1, end)
    }
    return *arr
}

func partition(arr []Planet, start, end int) int{
    pivot := arr[end].Name
    pIndex := start
    for i:= start; i<end; i++{
        if arr[i].Name <= pivot{
            //  swap
            arr[i],arr[pIndex] = arr[pIndex],arr[i]
            pIndex++
        }
    }
    arr[pIndex],arr[end] = arr[end],arr[pIndex]
    return pIndex
}

通过（任意）字段名称简单地对结构数组进行排序的最短方法是什么？

6 个答案: