英文:
Iterating over all the keys of a map
问题
有没有办法获取Go语言地图中所有键的列表？元素的数量可以通过len()函数获得，但如果我有一个地图，像这样：
m := map[string]string{ "key1":"val1", "key2":"val2" };
我如何遍历所有的键？
英文:
Is there a way to get a list of all the keys in a Go language map? The number of elements is given by len(), but if I have a map like:
m := map[string]string{ "key1":"val1", "key2":"val2" };
How do I iterate over all the keys?
答案1
得分: 734
https://play.golang.org/p/JGZ7mN0-U-
for k, v := range m {
fmt.Printf("key[%s] value[%s]\n", k, v)
}
或者
for k := range m {
fmt.Printf("key[%s] value[%s]\n", k, m[k])
}
Go语言for语句的规范指定第一个值是键，第二个变量是值，但不一定要存在。
英文:
https://play.golang.org/p/JGZ7mN0-U-
for k, v := range m {
fmt.Printf("key[%s] value[%s]\n", k, v)
}
or
for k := range m {
fmt.Printf("key[%s] value[%s]\n", k, m[k])
}
Go language specs for for statements specifies that the first value is the key, the second variable is the value, but doesn't have to be present.
答案2
得分: 21
这是一种获取地图键的简单方法。
// 返回给定地图的键
func Keys(m map[string]interface{}) (keys []string) {
for k := range m {
keys = append(keys, k)
}
return keys
}
// 使用 `Keys` 函数
func main() {
m := map[string]interface{}{
"foo": 1,
"bar": true,
"baz": "baz",
}
fmt.Println(Keys(m)) // [foo bar baz]
}
英文:
Here's some easy way to get slice of the map-keys.
// Return keys of the given map
func Keys(m map[string]interface{}) (keys []string) {
for k := range m {
keys = append(keys, k)
}
return keys
}
// use `Keys` func
func main() {
m := map[string]interface{}{
"foo": 1,
"bar": true,
"baz": "baz",
}
fmt.Println(Keys(m)) // [foo bar baz]
}
答案3
得分: 19
有没有办法获取Go语言中map的所有键的列表？
ks := reflect.ValueOf(m).MapKeys()
如何遍历所有的键？
使用接受的答案：
for _, k := range m { ... }
英文:
> Is there a way to get a list of all the keys in a Go language map?
ks := reflect.ValueOf(m).MapKeys()
> how do I iterate over all the keys?
Use the accepted answer:
for _, k := range m { ... }
答案4
得分: 6
一个类型不可知的解决方案：
for _, key := range reflect.ValueOf(yourMap).MapKeys() {
value := yourMap.MapIndex(key).Interface()
fmt.Println("键：", key, "值：", value)
}
英文:
A Type agnostic solution:
for _, key := range reflect.ValueOf(yourMap).MapKeys() {
value := yourMap.MapIndex(key).Interface()
fmt.Println("Key:", key, "Value:", value)
}
答案5
得分: 4
使用泛型：
func Keys[K comparable, V any](m map[K]V) []K {
keys := make([]K, 0, len(m))
for k := range m {
keys = append(keys, k)
}
return keys
}
英文:
Using Generics:
func Keys[K comparable, V any](m map[K]V) []K {
keys := make([]K, 0, len(m))
for k := range m {
keys = append(keys, k)
}
return keys
}
答案6
得分: 0
对于map[string]string的排序键。
package main
import (
"fmt"
"sort"
)
func main() {
m := map[string]string{"key1": "val1", "key2": "val2"}
sortStringMap(m)
}
// sortStringMap按键排序打印[string]string
func sortStringMap(m map[string]string) {
var keys []string
for key := range m {
keys = append(keys, key)
}
sort.Strings(keys) // 对键进行排序
for _, key := range keys {
fmt.Printf("%s\t:%s\n", key, m[key])
}
}
输出：
key1 :val1
key2 :val2
英文:
For sorted keys of map[string]string.
package main
import (
"fmt"
"sort"
)
func main() {
m := map[string]string{"key1": "val1", "key2": "val2"}
sortStringMap(m)
}
// sortStringMap prints the [string]string as keys sorted
func sortStringMap(m map[string]string) {
var keys []string
for key := range m {
keys = append(keys, key)
}
sort.Strings(keys) // sort the keys
for _, key := range keys {
fmt.Printf("%s\t:%s\n", key, m[key])
}
}
output:
key1 :val1
key2 :val2
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