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
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)
其他回答
Jack Giffin的解决方案很好，但不适用于大于2^32的数组。下面是基于Jack的解决方案来过滤数组的重构快速版本，但它适用于64位数组。
const Math_clz32 = Math.clz32 || ((log, LN2) => x => 31 - log(x >>> 0) / LN2 | 0)(Math.log, Math.LN2);
const filterArrayByAnotherArray = (searchArray, filterArray) => {
searchArray.sort((a,b) => a > b);
filterArray.sort((a,b) => a > b);
let searchArrayLen = searchArray.length, filterArrayLen = filterArray.length;
let progressiveLinearComplexity = ((searchArrayLen<<1) + filterArrayLen)>>>0
let binarySearchComplexity = (searchArrayLen * (32-Math_clz32(filterArrayLen-1)))>>>0;
let i = 0;
if (progressiveLinearComplexity < binarySearchComplexity) {
return searchArray.filter(currentValue => {
while (filterArray[i] < currentValue) i=i+1|0;
return filterArray[i] !== currentValue;
});
}
else return searchArray.filter(e => binarySearch(filterArray, e) === null);
}
const binarySearch = (sortedArray, elToFind) => {
let lowIndex = 0;
let highIndex = sortedArray.length - 1;
while (lowIndex <= highIndex) {
let midIndex = Math.floor((lowIndex + highIndex) / 2);
if (sortedArray[midIndex] == elToFind) return midIndex;
else if (sortedArray[midIndex] < elToFind) lowIndex = midIndex + 1;
else highIndex = midIndex - 1;
} return null;
}
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)
/* 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);
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);
使用对象筛选结果
[{id:1},{id:2},{id:3},{id:4}].filter(v=>!([{id:2},{id:4}].some(e=>e.id === v.id)))
