Question

假设我有两个map[string]([]string)

 MAP1 := map[string]([]string) {
    "User" : []string{"11", "33"},
    "Type" : []string{"A"},
    }


MAP2 := map[string]([]string) {
    "User" : []string{"11", "17"},
    "Type" : []string{"B"},
    }

此处，MAP1部分匹配MAP2。

User = 11 is in both map

如何以简单的方式检查？

Answer 1

For example:

package main

import "fmt"

func Max(x, y int) int {
    if x > y {
        return x
    }
    return y
}

func Intersect(as, bs []string) []string {
    i := make([]string, 0, Max(len(as), len(bs)))
    for _, a := range as {
        for _, b := range bs {
            if a == b {
                i = append(i, a)
            }
        }
    }
    return i
}

func main() {

    MAP1 := map[string][]string{
        "User": []string{"11", "33"},
        "Type": []string{"A"},
    }

    MAP2 := map[string][]string{
        "User": []string{"11", "17"},
        "Type": []string{"B"},
    }

    MAP3 := make(map[string][]string)

    for k, _ := range MAP1 {
        MAP3[k] = Intersect(MAP1[k], MAP2[k])
    }

    fmt.Println(MAP3) // MAP3 contains commonalities between MAP1 and MAP2
}

Note that this solution does not exploit any potential performance optimizations (like assuming that the string arrays will be sorted in some way, or else) and hence has a runtime performance of O(m • n²), where:

m is the number of keys in the map (assumed to be the same for both maps)
n is the number of elements in each string slice (assumed to be the same for both corresponding map entries)

Which is okay but not great.

Answer 2

检查这两个地图之间交叉点的基数是否大于0。

请参阅set.go Intersect中的deckarep/golang-set。

// Returns a new set containing only the elements
// that exist only in both sets.
//
// Note that the argument to Intersect
// must be of the same type as the receiver
// of the method. Otherwise, Intersect will
// panic.
Intersect(other Set) Set

如何在golang中检查地图是否可以部分匹配

2 个答案: