我正在寻找一种有效的方法,从javascript数组中删除所有元素,如果它们存在于另一个数组中。

// If I have this array:
var myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];

// and this one:
var toRemove = ['b', 'c', 'g'];

我想对myArray进行操作,使其处于这种状态:['a', 'd', 'e', 'f']

与jQuery,我使用grep()和inArray(),这工作得很好:

myArray = $.grep(myArray, function(value) {
    return $.inArray(value, toRemove) < 0;
});

有没有一个纯javascript的方法来做到这一点没有循环和剪接?


当前回答

var myArray = [
  {name: 'deepak', place: 'bangalore'}, 
  {name: 'chirag', place: 'bangalore'}, 
  {name: 'alok', place: 'berhampur'}, 
  {name: 'chandan', place: 'mumbai'}
];
var toRemove = [
  {name: 'deepak', place: 'bangalore'},
  {name: 'alok', place: 'berhampur'}
];



myArray = myArray.filter(ar => !toRemove.find(rm => (rm.name === ar.name && ar.place === rm.place) ))

其他回答

如果你正在使用Typescript并且想要匹配单个属性值,这应该基于上面Craciun Ciprian的答案。

您还可以通过允许非对象匹配和/或多属性值匹配使其更通用。

/**
 *
 * @param arr1 The initial array
 * @param arr2 The array to remove
 * @param propertyName the key of the object to match on
 */
function differenceByPropVal<T>(arr1: T[], arr2: T[], propertyName: string): T[] {
  return arr1.filter(
    (a: T): boolean =>
      !arr2.find((b: T): boolean => b[propertyName] === a[propertyName])
  );
}

我只是实现为:

Array.prototype.exclude = function(list){
        return this.filter(function(el){return list.indexOf(el)<0;})
}

使用:

myArray.exclude(toRemove);

高性能和不可变的解决方案

Javascript

const excludeFromArr = (arr, exclude) => {
  const excludeMap = exclude.reduce((all, item) => ({ ...all, [item]: true }), {});
  return arr.filter((item) => !excludeMap?.[item]);
};

打字稿:

const excludeFromArr = (arr: string[], exclude: string[]): string[] => {
  const excludeMap = exclude.reduce<Record<string, boolean>>((all, item) => ({ ...all, [item]: true }), {});
  return arr.filter((item) => !excludeMap?.[item]);
};

你可以使用_。by和lodash的区别

const myArray = [
  {name: 'deepak', place: 'bangalore'}, 
  {name: 'chirag', place: 'bangalore'}, 
  {name: 'alok', place: 'berhampur'}, 
  {name: 'chandan', place: 'mumbai'}
];
const toRemove = [
  {name: 'deepak', place: 'bangalore'},
  {name: 'alok', place: 'berhampur'}
];
const sorted = _.differenceBy(myArray, toRemove, 'name');

示例代码:CodePen

我没有使用任何内置方法构建逻辑,请让我知道任何优化或修改。 我在JS编辑器中测试了,它工作得很好。

var myArray = [
            {name: 'deepak', place: 'bangalore'},
            {name: 'alok', place: 'berhampur'},
            {name: 'chirag', place: 'bangalore'},
            {name: 'chandan', place: 'mumbai'},

        ];
        var toRemove = [

            {name: 'chirag', place: 'bangalore'},
            {name: 'deepak', place: 'bangalore'},
            /*{name: 'chandan', place: 'mumbai'},*/
            /*{name: 'alok', place: 'berhampur'},*/


        ];
        var tempArr = [];
        for( var i=0 ; i < myArray.length; i++){
            for( var j=0; j<toRemove.length; j++){
                var toRemoveObj = toRemove[j];
                if(myArray[i] && (myArray[i].name === toRemove[j].name)) {
                    break;
                }else if(myArray[i] && (myArray[i].name !== toRemove[j].name)){
                        var fnd = isExists(tempArr,myArray[i]);
                        if(!fnd){
                            var idx = getIdex(toRemove,myArray[i])
                            if (idx === -1){
                                tempArr.push(myArray[i]);
                            }

                        }

                    }

                }
        }
        function isExists(source,item){
            var isFound = false;
            for( var i=0 ; i < source.length; i++){
                var obj = source[i];
                if(item && obj && obj.name === item.name){
                    isFound = true;
                    break;
                }
            }
            return isFound;
        }
        function getIdex(toRemove,item){
            var idex = -1;
            for( var i=0 ; i < toRemove.length; i++){
                var rObj =toRemove[i];
                if(rObj && item && rObj.name === item.name){
                    idex=i;
                    break;
                }
            }
            return idex;
        }