如何从数组中删除对象? 我希望从someArray中删除包含名称Kristian的对象。例如:

someArray = [{name:"Kristian", lines:"2,5,10"},
             {name:"John", lines:"1,19,26,96"}];

我想实现:

someArray = [{name:"John", lines:"1,19,26,96"}];

当前回答

我建议使用lodash.js或sugar.js来完成以下常见任务:

// lodash.js
someArray = _.reject(someArray, function(el) { return el.Name === "Kristian"; });

// sugar.js
someArray.remove(function(el) { return el.Name === "Kristian"; });

在大多数项目中,拥有一组由此类库提供的helper方法是非常有用的。

其他回答

性能

今天2021.01.27我在Chrome v88, Safari v13.1.2和Firefox v84上对所选解决方案的MacOs HighSierra 10.13.6进行测试。

结果

对于所有浏览器:

当元素不存在时,最快/最快的解决方案:A和B 快速/最快的大数组解决方案:C 当元素存在时,大数组的快速/最快解决方案:H 对于小数组来说是非常缓慢的解:F和G 对于大数组:D, E和F,非常缓慢的解决方案

细节

我执行4个测试用例:

小数组(10个元素)和元素存在-你可以在这里运行它 小数组(10个元素)和元素不存在-你可以在这里运行它 大数组(百万元素)和元素存在-你可以在这里运行它 大数组(百万元素)和元素不存在-你可以在这里运行它

下面的代码片段展示了解决方案之间的差异 一个 B C D E F G H 我

function A(arr, name) { let idx = arr.findIndex(o => o.name==name); if(idx>=0) arr.splice(idx, 1); return arr; } function B(arr, name) { let idx = arr.findIndex(o => o.name==name); return idx<0 ? arr : arr.slice(0,idx).concat(arr.slice(idx+1,arr.length)); } function C(arr, name) { let idx = arr.findIndex(o => o.name==name); delete arr[idx]; return arr; } function D(arr, name) { return arr.filter(el => el.name != name); } function E(arr, name) { let result = []; arr.forEach(o => o.name==name || result.push(o)); return result; } function F(arr, name) { return _.reject(arr, el => el.name == name); } function G(arr, name) { let o = arr.find(o => o.name==name); return _.without(arr,o); } function H(arr, name) { $.each(arr, function(i){ if(arr[i].name === 'Kristian') { arr.splice(i,1); return false; } }); return arr; } function I(arr, name) { return $.grep(arr,o => o.name!=name); } // Test let test1 = [ {name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}, ]; let test2 = [ {name:"John3", lines:"1,19,26,96"}, {name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}, {name:"Joh2", lines:"1,19,26,96"}, ]; let test3 = [ {name:"John3", lines:"1,19,26,96"}, {name:"John", lines:"1,19,26,96"}, {name:"Joh2", lines:"1,19,26,96"}, ]; console.log(` Test1: original array from question Test2: array with more data Test3: array without element which we want to delete `); [A,B,C,D,E,F,G,H,I].forEach(f=> console.log(` Test1 ${f.name}: ${JSON.stringify(f([...test1],"Kristian"))} Test2 ${f.name}: ${JSON.stringify(f([...test2],"Kristian"))} Test3 ${f.name}: ${JSON.stringify(f([...test3],"Kristian"))} `)); <script src="https://code.jquery.com/jquery-3.5.1.min.js" integrity="sha256-9/aliU8dGd2tb6OSsuzixeV4y/faTqgFtohetphbbj0=" crossorigin="anonymous"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.20/lodash.min.js" integrity="sha512-90vH1Z83AJY9DmlWa8WkjkV79yfS2n2Oxhsi2dZbIv0nC4E6m5AbH8Nh156kkM7JePmqD6tcZsfad1ueoaovww==" crossorigin="anonymous"> </script> This shippet only presents functions used in performance tests - it not perform tests itself!

这里是chrome的示例结果

在你的数组语法中似乎有一个错误,所以假设你的意思是一个数组,而不是一个对象,数组。Splice是你的朋友:

someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];
someArray.splice(1,1)

我做了一个动态函数,以对象数组,键和值,并返回相同的数组后删除所需的对象:

function removeFunction (myObjects,prop,valu)
        {
             return myObjects.filter(function (val) {
              return val[prop] !== valu;
          });

        }

完整示例:DEMO

var obj = {
            "results": [
              {
                  "id": "460",
                  "name": "Widget 1",
                  "loc": "Shed"
              }, {
                  "id": "461",
                  "name": "Widget 2",
                  "loc": "Kitchen"
              }, {
                  "id": "462",
                  "name": "Widget 3",
                  "loc": "bath"
              }
            ]
            };


        function removeFunction (myObjects,prop,valu)
        {
             return myObjects.filter(function (val) {
              return val[prop] !== valu;
          });

        }


console.log(removeFunction(obj.results,"id","460"));

这就是我用的。

Array.prototype.delete = function(pos){
    this[pos] = undefined;
    var len = this.length - 1;
    for(var a = pos;a < this.length - 1;a++){
      this[a] = this[a+1];
    }
    this.pop();
  }

那就像说出来一样简单

var myArray = [1,2,3,4,5,6,7,8,9];
myArray.delete(3);

用任何数字代替3。之后的预期输出应该是:

console.log(myArray); //Expected output 1,2,3,5,6,7,8,9

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;
  }