英文:
Reverse a map in <value, key> format in Golang
问题
我有一个程序的映射表,看起来像这样:
fruit_map := map[string]string {
"apple": "likey",
"orange": "no likey",
}
我想要将它反转,使其变成以下形式:
{
"likey": "apple",
"no likey": "orange",
}
值中没有重复项。此外,我的映射表很小,大约有200个键。我没有找到任何内置的方法来反转这样的映射表。有没有办法快速地做到这一点?我对空间复杂度不太关心,但解决方案需要快速。
谢谢。
英文:
I have a map for my program that looks like following:
fruit_map := map[string]string {
"apple": "likey",
"orange": "no likey",
}
I would like to reverse it so that it reads the following:
{
"likey": "apple",
"no likey": "orange",
}
There are no duplicates in values. Also, my map is small - about 200 keys. I did not find any built-in method to reverse a map like this. Is there any way to do this fast? I am not much bothered about space complexity, but the solution needs to be fast.
Thanks.
答案1
得分: 12
你可以编写一个for
循环来遍历原始映射的键值对,并将它们放入一个新的映射中(参见reverseMap
函数)。
代码@https://play.golang.org/p/5y1gABTpdb8
package main
import (
"fmt"
)
func main() {
fruit_map := map[string]string{
"apple": "likey",
"orange": "no likey",
}
reversedMap := reverseMap(fruit_map)
fmt.Println(reversedMap)
}
func reverseMap(m map[string]string) map[string]string {
n := make(map[string]string, len(m))
for k, v := range m {
n[v] = k
}
return n
}
输出:
map[likey:apple no likey:orange]
顺便说一下,用fruit_map
这样的命名方式并不符合Go语言的惯例,你真的应该使用驼峰命名法,比如fruitMap
。
英文:
You may write a for
loop to iterate over the key-value pair of original map, and put them in a new map (see function reverseMap
)
Code@https://play.golang.org/p/5y1gABTpdb8
package main
import (
"fmt"
)
func main() {
fruit_map := map[string]string{
"apple": "likey",
"orange": "no likey",
}
reversedMap := reverseMap(fruit_map)
fmt.Println(reversedMap)
}
func reverseMap(m map[string]string) map[string]string {
n := make(map[string]string, len(m))
for k, v := range m {
n[v] = k
}
return n
}
Output:
map[likey:apple no likey:orange]
BTW, it is not idiomatic to name go variable like fruit_map
, you really should use camel-case, like fruitMap
.
答案2
得分: 4
其他答案提供了基于直接处理映射的简单解决方案。
另一种解决方案是将双向映射封装为一个独立的实用程序,这样你可以为其编写全面的单元测试,并能够通过简单的 API 来依赖它的正确操作。
这是我的示例实现(不完整,尚未具备必要的单元测试):
package main
import (
"fmt"
)
func main() {
biMap := NewBiMap()
biMap.Put("apple", "likey")
biMap.Put("orange", "no likey")
v, _ := biMap.GetByValue("no likey")
fmt.Println(v)
}
type BiMap struct {
ab map[string]string
ba map[string]string
}
func NewBiMap() *BiMap {
return &BiMap{make(map[string]string), make(map[string]string)}
}
func (m *BiMap) Put(key, value string) *BiMap {
m.ab[key] = value
m.ba[value] = key
return m
}
func (m *BiMap) GetByKey(key string) (value string, exists bool) {
value, exists = m.ab[key]
return
}
func (m *BiMap) GetByValue(value string) (key string, exists bool) {
key, exists = m.ba[value]
return
}
func (m *BiMap) Len() int {
return len(m.ab)
}
func (m *BiMap) DeleteKey(key string) *BiMap {
value, exists := m.ab[key]
if exists {
delete(m.ab, key)
delete(m.ba, value)
}
return m
}
func (m *BiMap) DeleteValue(value string) *BiMap {
key, exists := m.ba[value]
if exists {
delete(m.ab, key)
delete(m.ba, value)
}
return m
}
英文:
The other answers offer the simple solution based on handling the map directly.
An alternative solution is to encapsulate a bidirectional map as a self-contained utility, which has the advantage that you could write thorough unit tests for it and then be able to rely on it to operate correctly via a simple API.
Here's my example implementation (which is incomplete and doesn't yet have the necessary unit tests):
package main
import (
"fmt"
)
func main() {
biMap := NewBiMap()
biMap.Put("apple", "likey")
biMap.Put("orange", "no likey")
v, _ := biMap.GetByValue("no likey")
fmt.Println(v)
}
type BiMap struct {
ab map[string]string
ba map[string]string
}
func NewBiMap() *BiMap {
return &BiMap{make(map[string]string), make(map[string]string)}
}
func (m *BiMap) Put(key, value string) *BiMap {
m.ab[key] = value
m.ba[value] = key
return m
}
func (m *BiMap) GetByKey(key string) (value string, exists bool) {
value, exists = m.ab[key]
return
}
func (m *BiMap) GetByValue(value string) (key string, exists bool) {
key, exists = m.ba[value]
return
}
func (m *BiMap) Len() int {
return len(m.ab)
}
func (m *BiMap) DeleteKey(key string) *BiMap {
value, exists := m.ab[key]
if exists {
delete(m.ab, key)
delete(m.ba, value)
}
return m
}
func (m *BiMap) DeleteValue(value string) *BiMap {
key, exists := m.ba[value]
if exists {
delete(m.ab, key)
delete(m.ba, value)
}
return m
}
答案3
得分: 0
你是对的,没有内置的方法可以实现这个,但是非常简单:
package main
import "fmt"
func main() {
fruit_map := map[string]string{
"apple": "likey",
"orange": "no likey",
}
//创建一个新的空映射,用于保存反转后的内容。
reversed_fruit_map := make(map[string]string)
for k, v := range fruit_map {
reversed_fruit_map[v] = k
}
fmt.Println(reversed_fruit_map)
}
这将输出以下内容:
map[likey:apple no likey:orange]
在playground上查看。如果这是常见的操作,你可以将其提取到自己的函数中。
英文:
You are right, there is nothing built-in to accomplish this but it is really straightforward:
package main
import "fmt"
func main() {
fruit_map := map[string]string{
"apple": "likey",
"orange": "no likey",
}
//create your new empty map that will hold your reversed contents.
reversed_fruit_map := make(map[string]string)
for k, v := range fruit_map{
reversed_fruit_map[v] = k
}
fmt.Println(reversed_fruit_map)
}
This outputs the following:
map[likey:apple no likey:orange]
Check it out on the playground. If this is common you can always extract out into your own function.
答案4
得分: 0
没有内置的函数可以做到这一点,但使用for
循环非常简单。
fruit_map := map[string]string {
"apple": "likey",
"orange": "no likey",
}
reversed_map := make(map[string]string)
for key,value := range fruit_map {
reversed_map[value] = key
}
参考:http://play.golang.org/p/BQjqUsf9aU
英文:
There isn't a built in function to do that, but it's simple enough with a for
loop.
fruit_map := map[string]string {
"apple": "likey",
"orange": "no likey",
}
reversed_map := make(map[string]string)
for key,value := range fruit_map {
reversed_map[value] = key
}
答案5
得分: 0
使用实现了Reverse()
方法的自定义类型是另一种选择。
package main
import "fmt"
type (
MapA map[string]string
MapB map[int]string
)
// Reverse返回MapA对象的反向映射。
func (m MapA) Reverse() map[string]string {
n := make(map[string]string, len(m))
for k, v := range m {
n[v] = k
}
return n
}
// Reverse返回MapB对象的反向映射。
func (m MapB) Reverse() map[string]int {
n := make(map[string]int, len(m))
for k, v := range m {
n[v] = k
}
return n
}
func main() {
myMapA := &MapA{
"apple": "likey",
"orange": "no likey",
}
myMapB := &MapB{
10: "likey",
5: "not sure",
0: "no likey",
}
fmt.Println(myMapA.Reverse())
fmt.Println(myMapB.Reverse())
}
英文:
Using a custom type with a Reverse()
method implemented is another option.
package main
import "fmt"
type (
MapA map[string]string
MapB map[int]string
)
// Reverse returns a reverse map of the MapA object.
func (m MapA) Reverse() map[string]string {
n := make(map[string]string, len(m))
for k, v := range m {
n[v] = k
}
return n
}
// Reverse returns a reverse map of the MapB object.
func (m MapB) Reverse() map[string]int {
n := make(map[string]int, len(m))
for k, v := range m {
n[v] = k
}
return n
}
func main() {
myMapA := &MapA{
"apple": "likey",
"orange": "no likey",
}
myMapB := &MapB{
10: "likey",
5: "not sure",
0: "no likey",
}
fmt.Println(myMapA.Reverse())
fmt.Println(myMapB.Reverse())
}
答案6
得分: 0
扩展@Dave C的答案,使用泛型
package main
import (
"fmt"
)
func main() {
myMap := map[int]string{
1: "one",
2: "two",
3: "three",
}
result := reverseMap(myMap)
fmt.Println(result)
}
func reverseMap[M ~map[K]V, K comparable, V comparable](m M) map[V]K {
reversedMap := make(map[V]K)
for key, value := range m {
reversedMap[value] = key
}
return reversedMap
}
英文:
Extending @Dave C's answer , with generic
package main
import (
"fmt"
)
func main() {
myMap := map[int]string{
1: "one",
2: "two",
3: "three",
}
result := reverseMap(myMap)
fmt.Println(result)
}
func reverseMap[M ~map[K]V, K comparable, V comparable](m M) map[V]K {
reversedMap := make(map[V]K)
for key, value := range m {
reversedMap[value] = key
}
return reversedMap
}
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