javascript - 从对象数组javascript中删除元素

Question

我有以下对象数组。

[{"rId":24,"gId":40,"sId":20,"disabled":false},
 {"rId":24,"gId":40,"sId":19,"disabled":false},
 {"rId":24,"gId":40,"sId":50,"disabled":false},
 {"rId":24,"gId":40,"sId":20,"disabled":true},
 {"rId":24,"gId":40,"sId":19,"disabled":true},
 {"rId":24,"gId":40,"sId":50,"disabled":true},
 {"rId":24,"gId":39,"sId":18,"disabled":false}]

其中一些记录是对立的。第一个元素和第四个元素具有相同的 rId、gId 和 sId 但禁用标志相反。 我想消除所有此类记录。

我期望的数组是 {"rId":24,"gId":39,"sId":18,"disabled":false}(消除所有对偶记录)

我尝试了以下代码，但它给了我错误的输出。

arrOfObj=[{"rId":24,"gId":40,"sId":20,"disabled":false},
 {"rId":24,"gId":40,"sId":19,"disabled":false},
 {"rId":24,"gId":40,"sId":50,"disabled":false},
 {"rId":24,"gId":40,"sId":20,"disabled":true},
 {"rId":24,"gId":40,"sId":19,"disabled":true},
 {"rId":24,"gId":40,"sId":50,"disabled":true},
 {"rId":24,"gId":39,"sId":18,"disabled":false}]


$.each(arrOfObj,function (index1,firstObj) {
    $.each(arrOfObj,function (index2,secondObj) {
        if(index1>= index2){
            return true;
        }
        var areObjAntithesis=firstObj.rId===secondObj.rId && firstObj.gId===secondObj.gId
           && firstObj.sId===secondObj.sId && firstObj.disabled!==secondObj.disabled;

        if(areObjAntithesis){
            arrOfObj.splice(index1,1);
            arrOfObj.splice(index2,1)
            return false;
        }
    })
})

有什么优雅的方法可以达到预期的输出吗？

Answer 1

您可以使用 map() 和 filter()

var data = [{"rId":24,"gId":40,"sId":20,"disabled":false},
 {"rId":24,"gId":40,"sId":19,"disabled":false},
 {"rId":24,"gId":40,"sId":50,"disabled":false},
 {"rId":24,"gId":40,"sId":20,"disabled":true},
 {"rId":24,"gId":40,"sId":19,"disabled":true},
 {"rId":24,"gId":40,"sId":50,"disabled":true},
 {"rId":24,"gId":39,"sId":18,"disabled":false}]
 
var ar = data.map(function(e) {
  return e.rId + '|' + e.gId + '|' + e.sId;
});
 
var result = data.filter(function(e) {
  var key = e.rId + '|' + e.gId + '|' + e.sId;
  return ar.indexOf(key) == ar.lastIndexOf(key);
});

console.log(result)