我想了解从另一个数组中过滤掉所有元素的最佳方法。我尝试使用筛选函数，但是我不知道如何给它要移除的值。类似于：
var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(myCallback);
// filteredArray should now be [1,3]
function myCallBack(){
return element ! filteredArray;
//which clearly can't work since we don't have the reference <,<
}
如果过滤函数无法发挥作用，您将如何实现此操作？
编辑：我检查了可能的重复问题，并且对于那些容易理解JavaScript的人可能会有用。被选为好答案的答案使事情变得容易。
const filtered = [1, 2, 3, 4].filter(e => {
return this.indexOf(e) < 0;
},[2, 4]);
以下是一个例子
let firstArray=[1,2,3,4,5];
let secondArray=[2,3];
let filteredArray = firstArray.filter((a) => secondArray.indexOf(a)<0);
console.log(filteredArray); //above line gives [1,4,5]这完全取决于数组的类型。
对于简单的数组，例如字符串数组。您可以像@Redu和@Hugolpz指出的那样使用以下代码：
var arr1 = [1,2,3,4],
arr2 = [2,4],
res = arr1.filter(item => !arr2.includes(item));
console.log(res);
对于更复杂的数组过滤器，例如从另一个对象数组中过滤对象数组，您可以使用以下代码
function filterArray(arr1, arr2) {
return arr1.filter(item1 => !arr2.some(item2 => item1.id === item2.id));
}
或者
function filterArray(arr1, arr2) {
const set = new Set(arr2.map(item => item.id));
return arr1.reduce((filtered, item) => {
if (!set.has(item.id)) {
filtered.push(item);
}
return filtered;
}, []);
}
function filterArray(arr1, arr2) {
const map = new Map(arr2.map(item => [item.id, item]));
return arr1.filter(item => !map.has(item.id));
}
哈希集合
function filterArray(arr1, arr2) {
const set = new Set(arr2.map(item => item.id));
return arr1.filter(item => !set.has(item.id));
}
var filtered = [1,2,3,4,].filter(byIndex(element => element, [2,4]))
byIndex函数需要元素函数和一个数组，它的样子像这样：
byIndex = (getter: (e:number) => number, arr: number[]) => (x: number) => {
var i = getter(x);
return arr.indexOf(i);
}
filtered = [1,3]
new Set()创建一个仅包含唯一元素的过滤数组：const a = [1, 2, 3, 4];
const b = [3, 4, 5];
const c = Array.from(new Set(a.concat(b)));
对象数组：
const a = [{id:1}, {id: 2}, {id: 3}, {id: 4}];
const b = [{id: 3}, {id: 4}, {id: 5}];
const stringifyObject = o => JSON.stringify(o);
const parseString = s => JSON.parse(s);
const c = Array.from(new Set(a.concat(b).map(stringifyObject)), parseString);
var arr = [1, 2, 3 ,4 ,5, 6, 7];
var filter = [4, 5, 6];
var filtered = arr.filter(
function(val) {
for (var i = 0; i < filter.length; i++) {
if (val == filter[i]) {
return false;
}
}
return true;
}
);
function filterFn(array, diffArray, prop, propDiff) {
diffArray = !propDiff ? diffArray : diffArray.map(d => d[propDiff])
this.fn = f => diffArray.indexOf(f) === -1
if (prop) {
return array.map(r => r[prop]).filter(this.fn)
} else {
return array.filter(this.fn)
}
}
//You can use it like this;
var arr = [];
for (var i = 0; i < 10; i++) {
var obj = {}
obj.index = i
obj.value = Math.pow(2, i)
arr.push(obj)
}
var arr2 = [1, 2, 3, 4, 5]
var sec = [{t:2}, {t:99}, {t:256}, {t:4096}]
var log = console.log.bind(console)
var filtered = filterFn(arr, sec, 'value', 't')
var filtered2 = filterFn(arr2, sec, null, 't')
log(filtered, filtered2)var a1 = [1, 2, 3, 4],
a2 = [2, 3];
var filtered = a1.filter(function(x) {
return !a2.reduce(function(y, z) {
return x == y || x == z || y == true;
})
});
document.write(filtered);var arr1= [1,2,3,4];
var arr2=[2,4]
function fil(value){
return value !=arr2[0] && value != arr2[1]
}
document.getElementById("p").innerHTML= arr1.filter(fil)<!DOCTYPE html>
<html>
<head>
</head>
<body>
<p id="p"></p>function arr(arr1,arr2){
function filt(value){
return arr2.indexOf(value) === -1;
}
return arr1.filter(filt)
}
document.getElementById("p").innerHTML = arr([1,2,3,4],[2,4])<p id="p"></p>
return arrTwo.indexOf(e) === -1;。 代码：var filteredArr = firstArr.filter(el => secondArr.indexOf(el) === -1);（将第一个数组中不在第二个数组中的元素筛选出来的代码，可以使用上述语句实现） - Tushar