我想答案是非常分支和复杂的。

可以使用下面的路径删除与现代JavaScript术语中给出的对象相匹配的数组对象。

coordinates = [
    { lat: 36.779098444109145, lng: 34.57202827508546 },
    { lat: 36.778754712956506, lng: 34.56898128564454 },
    { lat: 36.777414146732426, lng: 34.57179224069215 }
];

coordinate = { lat: 36.779098444109145, lng: 34.57202827508546 };

removeCoordinate(coordinate: Coordinate): Coordinate {
    const found = this.coordinates.find((coordinate) => coordinate == coordinate);
    if (found) {
      this.coordinates.splice(found, 1);
    }
    return coordinate;
  }

为简单的数组工作投票给UndercoreJS。

_.without()函数帮助删除一个元素:

 _.without([1, 2, 1, 0, 3, 1, 4], 0, 1);
    => [2, 3, 4]

你可以使用以下几种方法从数组中删除项:

//1
someArray.shift(); // first element removed
//2
someArray = someArray.slice(1); // first element removed
//3
someArray.splice(0, 1); // first element removed
//4
someArray.pop(); // last element removed
//5
someArray = someArray.slice(0, someArray.length - 1); // last element removed
//6
someArray.length = someArray.length - 1; // last element removed

如果你想移除x位置的元素，使用:

someArray.splice(x, 1);

Or

someArray = someArray.slice(0, x).concat(someArray.slice(-x));

回复@chill182的评论:您可以使用array从数组中删除一个或多个元素。过滤器或数组。拼接结合数组。findIndex(参见MDN)。

请看这个Stackblitz项目或下面的代码片段:

// non destructive filter > noJohn = John removed, but someArray will not change let someArray = getArray(); let noJohn = someArray.filter( el => el.name !== "John" ); log(`let noJohn = someArray.filter( el => el.name !== "John")`, `non destructive filter [noJohn] =`, format(noJohn)); log(`**someArray.length ${someArray.length}`); // destructive filter/reassign John removed > someArray2 = let someArray2 = getArray(); someArray2 = someArray2.filter( el => el.name !== "John" ); log("", `someArray2 = someArray2.filter( el => el.name !== "John" )`, `destructive filter/reassign John removed [someArray2] =`, format(someArray2)); log(`**someArray2.length after filter ${someArray2.length}`); // destructive splice /w findIndex Brian remains > someArray3 = let someArray3 = getArray(); someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1); someArray3.splice(someArray3.findIndex(v => v.name === "John"), 1); log("", `someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1),`, `destructive splice /w findIndex Brian remains [someArray3] =`, format(someArray3)); log(`**someArray3.length after splice ${someArray3.length}`); // if you're not sure about the contents of your array, // you should check the results of findIndex first let someArray4 = getArray(); const indx = someArray4.findIndex(v => v.name === "Michael"); someArray4.splice(indx, indx >= 0 ? 1 : 0); log("", `someArray4.splice(indx, indx >= 0 ? 1 : 0)`, `check findIndex result first [someArray4] = (nothing is removed)`, format(someArray4)); log(`**someArray4.length (should still be 3) ${someArray4.length}`); // -- helpers -- function format(obj) { return JSON.stringify(obj, null, " "); } function log(...txt) { document.querySelector("pre").textContent += `${txt.join("\n")}\n` } function getArray() { return [ {name: "Kristian", lines: "2,5,10"}, {name: "John", lines: "1,19,26,96"}, {name: "Brian", lines: "3,9,62,36"} ]; } <pre> **Results** </pre>

你也可以使用地图功能。

someArray = [{name:"Kristian", lines:"2,5,10"},{name:"John",lines:"1,19,26,96"}];
newArray=[];
someArray.map(function(obj, index){
    if(obj.name !== "Kristian"){
       newArray.push(obj);
    }
});
someArray = newArray;
console.log(someArray);

这里有一个映射和拼接的例子

const arrayObject = [ {name: "name1"， value: "value1"}， {name: "name2"， value: "value2"}， {name: "name3"， value: "value3"}， ]; let index = arrayObject.map((item) => item.name).indexOf("name1"); If(索引> -1){ arrayObject。拼接(指数(1); console.log(“结果”,arrayObject); ｝

输出

Result [
  {
    "name": "name2",
    "value": "value2"
  },
  {
    "name": "name3",
    "value": "value3"
  }
]

你可以这样过滤:

const someArray = [{ 名称:“克里斯蒂安”, 行:“2、5、10” }， ｛ 名称:“约翰”, :“96”1,19日,26日 ｝ ]; var filtered = someArray.filter((el) => el.name != "Kristian"); console.log(过滤)