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
下面是一个例子
让firstArray =[1、2、3、4、5); 让secondArray =(2、3); let filteredArray = firstArray.filter((a) => secondArray.indexOf(a)<0); console.log (filteredArray);//上面的行给出[1,4,5]
其他回答
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)
函数的arr (arr1 arr2) { 函数filt(价值){ 返回arr2.indexOf(value) == -1; ｝ 返回arr1.filter (filt) ｝ . getelementbyid (p)。innerHTML = arr([1,2,3,4]，[2,4]) < p id = p > < / p >
/* Here's an example that uses (some) ES6 Javascript semantics to filter an object array by another object array. */ // x = full dataset // y = filter dataset let x = [ {"val": 1, "text": "a"}, {"val": 2, "text": "b"}, {"val": 3, "text": "c"}, {"val": 4, "text": "d"}, {"val": 5, "text": "e"} ], y = [ {"val": 1, "text": "a"}, {"val": 4, "text": "d"} ]; // Use map to get a simple array of "val" values. Ex: [1,4] let yFilter = y.map(itemY => { return itemY.val; }); // Use filter and "not" includes to filter the full dataset by the filter dataset's val. let filteredX = x.filter(itemX => !yFilter.includes(itemX.val)); // Print the result. console.log(filteredX);
如果你想过滤一个具有一些匹配属性的不同结构的数组，你应该这样做。
let filteredArray = [];
array1.map(array1Item => {
array2.map(array2Item => {
if (array1.property1 === array2.property2) {
filteredArray.push(array1Item);
}
});
这会让你的生活变得轻松!
var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(myCallBack);
function myCallBack(el){
return anotherOne.indexOf(el) < 0;
}
在回调中，检查数组的每个值是否在另一个数组中
https://jsfiddle.net/0tsyc1sx/
如果使用lodash.js，请使用_.difference
filteredArray = _.difference(array, anotherOne);
Demo
如果你有一个对象数组:
var array = [{id :1, name :"test1"},{id :2, name :"test2"},{id :3, name :"test3"},{id :4, name :"test4"}];
var anotherOne = [{id :2, name :"test2"}, {id :4, name :"test4"}];
var filteredArray = array.filter(function(array_el){
return anotherOne.filter(function(anotherOne_el){
return anotherOne_el.id == array_el.id;
}).length == 0
});
对象的演示数组
用lodash演示不同的对象数组
