英文:
Check if key exists in multiple maps in one condition
问题
我需要检查两个映射中是否存在相同的键：
if v1, ok1 := map1["aaa"]; ok1 {...}if v2, ok2 := map2["aaa"]; ok2 {...}
是否可能将这两个条件合并为一个？我设法做到了这样：
v1, ok1 := map1["aaa"]v2, ok2 := map2["aaa"]if ok1 && ok2 {...}
但我想知道是否可以在一个 if 条件中完成赋值和检查。
英文:
I need to check if the same key exists in two maps:
if v1, ok1 := map1["aaa"]; ok1 {...}if v2, ok2 := map2["aaa"]; ok2 {...}
Is it possible to join these two conditions into one? I managed to do something like this:
v1, ok1 := map1["aaa"]v2, ok2 := map2["aaa"]if ok1 && ok2 {...}
but I'm curious whether it (assigning and checking) can be done in one if condition.
答案1
得分: 8
不，这是不可能的。规范：索引表达式：
在特殊形式的赋值或初始化中，对类型为map[K]V的映射a进行索引表达式操作，例如：
v, ok = a[x]v, ok := a[x]var v, ok = a[x]
会产生一个额外的无类型布尔值。如果键x存在于映射中，ok的值为true，否则为false。
因此，只有在没有其他赋值操作时，才能使用特殊的v, ok := m[k]形式。
然而，如果你不使用映射值类型的零值，你可以使用简单的元组赋值来进行检查；通过不使用特殊形式，而是使用两个简单的索引表达式。
例如，如果你的值类型是某个接口类型（例如interface{}），并且你知道你不使用nil值，你可以这样做：
if v1, v2 := m1["aaa"], m2["aaa"]; v1 != nil && v2 != nil {fmt.Printf("Both map contains key 'aaa': %v, %v\n", v1, v2)}
当然，通过一个辅助函数，你可以一步完成：
func idx(m1, m2 map[string]interface{}, k string) (v1, v2 interface{}, ok1, ok2 bool) {v1, ok1 = m1[k]v2, ok2 = m2[k]return}
使用它：
if v1, v2, ok1, ok2 := idx(m1, m2, "aaa"); ok1 && ok2 {fmt.Printf("Both map contains key 'aaa': %v, %v\n", v1, v2)}
在Go Playground上尝试这些示例。
英文:
No, it can't be done. Spec: Index expressions:
> An index expression on a map a of type map[K]V used in an assignment or initialization of the special form
>
> v, ok = a[x]
> v, ok := a[x]
> var v, ok = a[x]
>
> yields an additional untyped boolean value. The value of ok is true if the key x is present in the map, and false otherwise.
So you can use the special v, ok := m[k] form only if nothing else gets assigned.
However, if you don't use the zero value of the value type of the map, you can do the check using a simple tuple-assignment; by not using the special form but 2 simple index expressions.
For example if your value type is some interface type (e.g. interface{}), and you know you don't use the nil value, you may do the following:
if v1, v2 := m1["aaa"], m2["aaa"]; v1 != nil && v2 != nil {fmt.Printf("Both map contains key '%s': %v, %v\n", "aaa", v1, v2)}
Of course with a helper function, you can do it in one step:
func idx(m1, m2 map[string]interface{}, k string) (v1, v2 interface{}, ok1, ok2 bool) {v1, ok1 = m1[k]v2, ok2 = m2[k]return}
Using it:
if v1, v2, ok1, ok2 := idx(m1, m2, "aaa"); ok1 && ok2 {fmt.Printf("Both map contains key '%s': %v, %v\n", "aaa", v1, v2)}
Try the examples on the Go Playground.
答案2
得分: 0
你也可以使用可变参数（三个点）来检查多个键：
// 检查 map1 和 map2 是否为空func checkMap(m1, m2 map[string]interface{}, keys ...string) []bool {var isExist []boolfor _, key := range keys {// 从 map1 检查值_, ok := m1[key]if ok {isExist = append(isExist, true) // 第一次追加以避免 panic} else {isExist = append(isExist, false) // 第一次追加以避免 panic}// 从 map2 检查值_, ok = m2[key]if ok {isExist[len(isExist)-1] = true} else {isExist[len(isExist)-1] = false}}return isExist}
然后你可以按顺序检查你的键，像这样：
result := checkMap(myMap, myMap2, "a", "b", "c", "d", "e", "f", "g")fmt.Printf("result = %+v\n", result) // 打印结果if result[0] {fmt.Println("key a 在两个 map 中存在")}if result[1] {fmt.Println("key b 在两个 map 中存在")}
英文:
you could also use variadic parameter(three dots) to check multiple keys :
// check map 1 and map2 is null or notfunc checkMap(m1, m2 map[string]interface{}, keys ...string) []bool {var isExist []boolfor key := range keys {//checking value from map 1_, ok := m1[keys[key]]if ok {isExist = append(isExist, true) // append for the first time to avoid panic} else {isExist = append(isExist, false) // append for the first time to avoid panic}// checking value from map2_, ok = m2[keys[key]]if ok {isExist[key] = true} else {isExist[key] = false}}return isExist}
And then you can check your keys in order like this :
result := checkMap(myMap, myMap2, "a", "b", "c", "d", "e", "f", "g")fmt.Printf("result = %+v\n", result) // print the resultif result[0] {fmt.Println("key a exist in both map")}if result[1] {fmt.Println("key b exist in both map")}
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