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"}];
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>
干净的解决方案是使用Array.filter:
var filtered = someArray.filter(function(el) { return el.Name != "Kristian"; });
问题是它不能在IE < 9上工作。然而，你可以包括来自Javascript库(例如:underscore.js)的代码，为任何浏览器实现这一点。
最简单的解决方案是创建一个映射，按名称存储每个对象的索引，如下所示:
//adding to array
var newPerson = {name:"Kristian", lines:"2,5,10"}
someMap[ newPerson.name ] = someArray.length;
someArray.push( newPerson );
//deleting from the array
var index = someMap[ 'Kristian' ];
someArray.splice( index, 1 );
你所显示的“数组”是无效的JavaScript语法。花括号{}表示具有属性名/值对的对象，而方括号[]表示数组，就像这样:
someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];
在这种情况下，您可以使用.splice()方法删除一个项。要删除第一项(索引0)，如下所示:
someArray.splice(0,1);
// someArray = [{name:"John", lines:"1,19,26,96"}];
如果你不知道索引，但想通过数组搜索找到名称为“Kristian”的项来删除，你可以这样做:
for (var i =0; i < someArray.length; i++)
if (someArray[i].name === "Kristian") {
someArray.splice(i,1);
break;
}
编辑:我刚刚注意到你的问题被标记为“jQuery”，所以你可以尝试$.grep()方法:
someArray = $.grep(someArray,
function(o,i) { return o.name === "Kristian"; },
true);
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。
虽然这可能不适合这种情况，我发现前几天，如果你不需要改变数组的大小，你也可以使用delete关键字从数组中删除一个项目。
var myArray = [1,2,3];
delete myArray[1];
console.log(myArray[1]); //undefined
console.log(myArray.length); //3 - doesn't actually shrink the array down
使用javascript的splice()函数。
这可能会有帮助:http://www.w3schools.com/jsref/jsref_splice.asp
这个怎么样?
$.each(someArray, function(i){
if(someArray[i].name === 'Kristian') {
someArray.splice(i,1);
return false;
}
});
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]
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"));
someArray = jQuery.grep(someArray , function (value) {
return value.name != 'Kristian';
});
这是一个对我有用的函数:
function removeFromArray(array, value) {
var idx = array.indexOf(value);
if (idx !== -1) {
array.splice(idx, 1);
}
return array;
}
Splice (i, 1)，其中i是数组的增量索引，将删除对象。 但是请记住，splice也会重置数组长度，所以要注意'undefined'。用你的例子，如果你删除了“Kristian”，那么在循环中的下一次执行中，i将是2，但someArray将是1的长度，因此，如果你试图删除“John”，你会得到一个“undefined”错误。解决这个问题的一个方法是使用单独的计数器来跟踪要删除的元素的索引。
你可以使用array.filter()。
如。
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
someArray = someArray.filter(function(returnableObjects){
return returnableObjects.name !== 'Kristian';
});
//someArray will now be = [{name:"John", lines:"1,19,26,96"}];
箭头功能:
someArray = someArray.filter(x => x.name !== 'Kristian')
只从数组中返回属性名不是“Kristian”的对象
var noKristianArray = $.grep(someArray, function (el) { return el.name!= "Kristian"; });
演示:
var someArray = [ {名称:“克里斯蒂安”,行:“2、5、10”}, {名称:“约翰”,行:“1,19日,26日96”}, {名称:“克里斯蒂安”,行:" 58160 "}, {名称:“费利克斯”,行:“96”1,19日,26日} ]; var noKristianArray = $。grep(someArray, function (el){返回el.name!= "克里斯蒂安”;})； console.log (noKristianArray); < script src = " https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js " > < /脚本>
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);
你也可以用一些:
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
someArray.some(item => {
if(item.name === "Kristian") // Case sensitive, will only remove first instance
someArray.splice(someArray.indexOf(item),1)
})
你也可以尝试这样做:
var myArray = [{'name': 'test'}, {'name':'test2'}];
var myObject = {'name': 'test'};
myArray.splice(myArray.indexOf(myObject),1);
这个概念使用剑道网格
var grid = $("#addNewAllergies").data("kendoGrid");
var selectedItem = SelectedCheckBoxList;
for (var i = 0; i < selectedItem.length; i++) {
if(selectedItem[i].boolKendoValue==true)
{
selectedItem.length= 0;
}
}
ES2015
let someArray = [
{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"},
{name:"Kristian", lines:"2,58,160"},
{name:"Felix", lines:"1,19,26,96"}
];
someArray = someArray.filter(person => person.name != 'John');
它会除掉约翰!
具有es6箭头功能
let someArray = [
{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}
];
let arrayToRemove={name:"Kristian", lines:"2,5,10"};
someArray=someArray.filter((e)=>e.name !=arrayToRemove.name && e.lines!= arrayToRemove.lines)
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
如果你想删除给定对象的所有出现(基于某些条件)，那么使用for循环中的javascript splice方法。
因为删除对象会影响数组长度，所以请确保减少计数器一步，以保持长度检查不变。
var objArr=[{Name:"Alex", Age:62},
{Name:"Robert", Age:18},
{Name:"Prince", Age:28},
{Name:"Cesar", Age:38},
{Name:"Sam", Age:42},
{Name:"David", Age:52}
];
for(var i = 0;i < objArr.length; i ++)
{
if(objArr[i].Age > 20)
{
objArr.splice(i, 1);
i--; //re-adjust the counter.
}
}
上面的代码片段删除了年龄大于20的所有对象。
这个答案
for (var i =0; i < someArray.length; i++)
if (someArray[i].name === "Kristian") {
someArray.splice(i,1);
}
对于满足条件的多个记录不工作。如果您有两个这样的连续记录，则只删除第一个记录，跳过另一个记录。你必须使用:
for (var i = someArray.length - 1; i>= 0; i--)
...
代替。
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;
}
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"
}
]
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
如果你在数组中的对象上没有任何你知道的属性(或者可能是唯一的)，但是你有一个你想要删除的对象的引用，你可以执行下面unregisterObject方法中的操作:
let registeredObjects = []; function registerObject(someObject) { registeredObjects.push(someObject); } function unregisterObject(someObject) { registeredObjects = registeredObjects.filter(obj => obj !== someObject); } let myObject1 = {hidden: "someValue1"}; // Let's pretend we don't know the hidden attribute let myObject2 = {hidden: "someValue2"}; registerObject(myObject1); registerObject(myObject2); console.log(`There are ${registeredObjects.length} objects registered. They are: ${JSON.stringify(registeredObjects)}`); unregisterObject(myObject1); console.log(`There are ${registeredObjects.length} objects registered. They are: ${JSON.stringify(registeredObjects)}`);
const someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];
我们得到对象的索引它的name属性值为"Kristian"
const index = someArray.findIndex(key => key.name === "Kristian");
console.log(index); // 0
通过使用拼接函数我们删除了name属性值为“Kristian”的对象
someArray.splice(index,1);
console.log(someArray); // [{name:"John", lines:"1,19,26,96"}]
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)
