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。
您可以使用过滤器,然后为过滤器函数使用过滤数组的约简,当它找到匹配时检查并返回true,然后在返回(!)时反转。filter函数对数组中的每个元素调用一次。在你的文章中,你没有对函数中的任何元素进行比较。
Var a1 = [1,2,3,4],
A2 = [2,3];
Var filter = a1.filter(函数(x) {
返回! a2。Reduce(函数(y, z) {
返回x == y || x == z || y == true;
})
});
document . write(过滤);
/* 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);
如果你需要比较一个对象数组,这在所有情况下都适用:
let arr = [{ id: 1, title: "title1" },{ id: 2, title: "title2" }]
let brr = [{ id: 2, title: "title2" },{ id: 3, title: "title3" }]
const res = arr.filter(f => brr.some(item => item.id === f.id));
console.log(res);