英文:
Find an item in array of maps and delete it
问题
我有一个包含映射的数组，我想根据其"key"删除其中的一个元素，如果存在的话。
如何做到呢？我希望它的执行速度不要太慢，保持顺序不重要。
myMaps := []map[string]interface{}{
map[string]interface{}{"key": "aaa", "key2": 222, "key3": "aafdsafd"},
map[string]interface{}{"key": "key_to_delete", "key2": 366, "key3": "333aafdsafd"},
map[string]interface{}{"key": "cccc", "key2": 467, "key3": "jhgfjhg"},
}
for i, x := range myMaps {
if x["key"] == "key_to_delete" {
// 如何删除这个元素：a) 作为映射的键 b) 作为数组元素的映射？
myMaps = append(myMaps[:i], myMaps[i+1:]...)
break
}
}
delete(...) 函数：
在遍历数组时，循环体中传递的是它的副本。那么 delete(...) 如何从真正的数组中删除元素呢？
更新：
我需要知道如何删除两种类型的实体，对于我的情况：
- 数组的元素 - 一个映射
- 映射的元素 - 具有特定键的元素
不使用第三方库。
英文:
I have an array of maps, from which I want to delete an element if it exists, which is determined by its "key".
How to do it? I want it not to be slow. Preserving the order isn't important.
myMaps = []map[string]interface{} {
map[string]interface{} {"key": "aaa", "key2": 222, "key3": "aafdsafd"},
map[string]interface{} {"key": "key_to_delete", "key2": 366, "key3": "333aafdsafd"},
map[string]interface{} {"key": "cccc", "key2": 467, "key3": "jhgfjhg"},
}
for _, x := range myMaps {
if x["key"] == "key_to_delete" {
//delete this element as a) key of the map b) the map-element as an element of the array; How?
}
}
The delete(...) function:
when iterating over an array, a copy of it is what gets passed in the body of a loop. No? How would then delete(...) delete an element from the real array?
update:
I need to know of how to delete 2 types of entities, and for my case:
- an element of an array - a map
- an element of a map, with a certain key
Without using a third-party library.
答案1
得分: 0
如果你想从地图中删除键：
for _, x := range myMaps {
if x["key"] == "key_to_delete" {
delete(x, "key")
}
}
如果你想从数组中删除地图，情况就会变得复杂，最好创建一个第二个数组，并在当前地图需要保留时将其插入其中：
myFilteredMaps := make([]map[string]interface{}, 0, len(myMaps))
for _, x := range myMaps {
if x["key"] != "key_to_delete" {
myFilteredMaps = append(myFilteredMaps, x)
}
}
myMaps = myFilteredMaps
这两种方法都很快，只要len(myMaps)不是太大，它们的运行时间与该长度成线性关系。
英文:
If you want to delete the key from the map:
for _, x := range myMaps {
if x["key"] == "key_to_delete" {
delete(x, "key")
}
}
If what you want is to delete the map from the array it gets complicated, you're better off creating a second array and inserting into it if the current map is to be kept:
myFilteredMaps := make([]map[string]interface{}, 0, len(myMaps))
for _, x := range myMaps {
if x["key"] != "key_to_delete" {
myFilteredMaps = append(myFilteredMaps, x)
}
}
myMaps = myFilteredMaps
Both of these are pretty quick so long as len(myMaps) isn't too large, both have linear runtime with respect to that length.
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