我想了解从另一个数组的所有元素中过滤一个数组的最佳方法。我尝试了过滤功能，但它不来我如何给它的值，我想删除。喜欢的东西:
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 <,<
}
如果过滤器函数没有用处，您将如何实现它? 编辑:我检查了可能的重复问题，这可能对那些容易理解javascript的人有用。如果答案勾选“好”，事情就简单多了。
当前回答
下面的例子使用new Set()创建一个只有唯一元素的过滤数组:
数组的基本数据类型:字符串，数字，布尔，空，未定义，符号:
const a = [1, 2, 3, 4];
const b = [3, 4, 5];
const c = Array.from(new Set(a.concat(b)));
以对象为项的数组:
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);
其他回答
对过滤功能最好的描述是https://developer.mozilla.org/pl/docs/Web/JavaScript/Referencje/Obiekty/Array/filter
你应该简单地条件函数:
function conditionFun(element, index, array) {
return element >= 10;
}
filtered = [12, 5, 8, 130, 44].filter(conditionFun);
在变量值被赋值之前，您不能访问它
下面的例子使用new Set()创建一个只有唯一元素的过滤数组:
数组的基本数据类型:字符串，数字，布尔，空，未定义，符号:
const a = [1, 2, 3, 4];
const b = [3, 4, 5];
const c = Array.from(new Set(a.concat(b)));
以对象为项的数组:
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 arr1= [1,2,3,4]; var arr2 =(2、4) 函数费尔(价值){ 返回value !=arr2[0] && value !=arr2[1] ｝ . getelementbyid (p)。innerHTML = arr1.filter (fil) <！DOCTYPE html > < html > < >头 > < /头 身体< > < p id = p > < / p >
来自另一个包含对象属性的数组的更灵活的过滤数组
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)
我会这样做;
Var arr1 = [1,2,3,4]， Arr2 = [2,4]， Res = arr1。Filter (item => !arr2.includes(item)); console.log (res);
