如何从另一个数组的所有元素中过滤数组

Koo*_*op4 102 javascript arrays filter


我想了解从另一个元素的所有元素中过滤数组的最佳方法.我试过过滤器功能,但它没有告诉我如何给它我想要删除的值.
就像是:

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 <,< 
}
Run Code Online (Sandbox Code Playgroud)

如果过滤功能没用,你会如何实现?
编辑:我检查了可能的重复问题,它可能对那些容易理解JavaScript的人有用.检查得很好的答案让事情变得简单.

Sim*_* Hi 122

您可以使用函数的this参数filter()来避免将过滤器数组存储在全局变量中.

var filtered = [1, 2, 3, 4].filter(
    function(e) {
      return this.indexOf(e) < 0;
    },
    [2, 4]
);
console.log(filtered);
Run Code Online (Sandbox Code Playgroud)


Red*_*edu 107

我会这样做;

var arr = [1,2,3,4],
    brr = [2,4],
    res = arr.filter(f => !brr.includes(f));
console.log(res);
Run Code Online (Sandbox Code Playgroud)

  • 摇滚明星.谢谢,这对于解决稍微不同的问题非常有帮助.根据react组件中的值数组过滤掉一个对象数组:```const filteredResults = this.state.cards.filter(result =>!this.state.filterOut.includes(result.category))`` `其中this.state.cards在一个对象数组中,this.state.filterOut是一个值数组,对应于我想要删除的对象中的"category"键. (5认同)
  • 箭头函数 FTW。比老式的回调好得多! (2认同)

Ash*_*ger 33

var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(myCallBack);

function myCallBack(el){
  return anotherOne.indexOf(el) < 0;
}
Run Code Online (Sandbox Code Playgroud)

在回调中,检查每个值array是否在anotherOne

https://jsfiddle.net/0tsyc1sx/

如果您正在使用lodash.js,请使用_.difference

filteredArray = _.difference(array, anotherOne);
Run Code Online (Sandbox Code Playgroud)

演示

如果您有一个对象数组:

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
});
Run Code Online (Sandbox Code Playgroud)

演示对象数组

用lodash演示对象的diff对象数组


小智 19

下面的代码是相对于另一个数组过滤数组的最简单方法。两个数组都可以在其中包含对象而不是值。

let array1 = [1, 3, 47, 1, 6, 7];
let array2 = [3, 6];
let filteredArray1 = array1.filter(el => array2.includes(el));
console.log(filteredArray1); 
Run Code Online (Sandbox Code Playgroud)

输出: [3, 6]


Dav*_*dit 18

        /* 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);
Run Code Online (Sandbox Code Playgroud)


Jac*_*fin 12

上述所有解决方案都“有效”,但都不是最佳性能,并且都以相同的方式解决问题,即使用Array.prototype.indexOfArray.prototype.includes在每个点线性搜索所有条目。一个更快的解决方案(在大多数情况下甚至比二分搜索快得多)是对数组进行排序并向前跳过,如下所示。但是,一个缺点是这要求数组中的所有条目都是数字或字符串。然而,在极少数情况下,二分搜索可能比渐进线性搜索更快。这些情况源于这样一个事实,即我的渐进线性搜索的复杂性为O(2n 1 +n 2 )(仅O(n 1+n 2 )在更快的 C/C++ 版本中)(其中n 1是搜索的数组,n 2是过滤器数组),而二分搜索的复杂度为O(n 1 ceil(log 2 n 2 )) ( ceil = 向上舍入 - 到上限),最后,indexOf 搜索在O(n 1 )O(n 1 n 2 )之间具有高度可变的复杂性,平均为O(n 1 ceil(n 2 ÷2))。因此,平均而言,indexOf 只会是最快的(n 1 ,n 2 )等于{1,2}{1,3}{x,1|x?N}。然而,这仍然不是现代硬件的完美代表。IndexOf 已在大多数现代浏览器中进行了最大程度的本地优化,使其非常受分支预测定律的约束。因此,如果我们对 indexOf 做出与渐进线性和二分搜索相同的假设——数组是预先排序的——那么,根据链接中列出的统计数据,我们可以预期 IndexOf 的速度大约提高 6 倍,将其复杂度转换为O(n 1 ÷6)O(n 1 n 2 ),平均为O(n 1 ceil(n 2 7÷12))。最后,请注意以下解决方案永远不会与对象一起使用,因为 JavaScript 中的对象无法通过 JavaScript 中的指针进行比较。

function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
    var out=[], value=0;
    for (var i=0,  len=array.length|0; i < len; i=i+1|0)
        if (handle(value = array[i])) 
            out.push( value );
    return out;
}

const Math_clz32 = Math.clz32 || (function(log, LN2){
  return function(x) {
    return 31 - log(x >>> 0) / LN2 | 0; // the "| 0" acts like math.floor
  };
})(Math.log, Math.LN2);

/* USAGE:
  filterArrayByAnotherArray(
      [1,3,5],
      [2,3,4]
  ) yields [1, 5], and it can work with strings too
*/
function filterArrayByAnotherArray(searchArray, filterArray) {
    if (
        // NOTE: This does not check the whole array. But, if you know
        //        that there are only strings or numbers (not a mix of
        //        both) in the array, then this is a safe assumption.
        // Always use `==` with `typeof` because browsers can optimize
        //  the `==` into `===` (ONLY IN THIS CIRCUMSTANCE)
        typeof searchArray[0] == "number" &&
        typeof filterArray[0] == "number" &&
        (searchArray[0]|0) === searchArray[0] &&
        (filterArray[0]|0) === filterArray[0]
    ) {filterArray
        // if all entries in both arrays are integers
        searchArray.sort(sortIntArray);
        filterArray.sort(sortIntArray);
    } else {
        searchArray.sort(sortAnyArray);
        filterArray.sort(sortAnyArray);
    }
    var searchArrayLen = searchArray.length, filterArrayLen = filterArray.length;
    var progressiveLinearComplexity = ((searchArrayLen<<1) + filterArrayLen)>>>0
    var binarySearchComplexity= (searchArrayLen * (32-Math_clz32(filterArrayLen-1)))>>>0;
    // After computing the complexity, we can predict which algorithm will be the fastest
    var i = 0;
    if (progressiveLinearComplexity < binarySearchComplexity) {
        // Progressive Linear Search
        return fastFilter(searchArray, function(currentValue){
            while (filterArray[i] < currentValue) i=i+1|0;
            // +undefined = NaN, which is always false for <, avoiding an infinite loop
            return filterArray[i] !== currentValue;
        });
    } else {
        // Binary Search
        return fastFilter(
            searchArray,
            fastestBinarySearch(filterArray)
        );
    }
}

// see https://stackoverflow.com/a/44981570/5601591 for implementation
//  details about this binary search algorithm

function fastestBinarySearch(array){
  var initLen = (array.length|0) - 1 |0;
  
  const compGoto = Math_clz32(initLen) & 31;
  return function(sValue) {
    var len = initLen |0;
    switch (compGoto) {
      case 0:
        if (len & 0x80000000) {
          const nCB = len & 0x80000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 1:
        if (len & 0x40000000) {
          const nCB = len & 0xc0000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 2:
        if (len & 0x20000000) {
          const nCB = len & 0xe0000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 3:
        if (len & 0x10000000) {
          const nCB = len & 0xf0000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 4:
        if (len & 0x8000000) {
          const nCB = len & 0xf8000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 5:
        if (len & 0x4000000) {
          const nCB = len & 0xfc000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 6:
        if (len & 0x2000000) {
          const nCB = len & 0xfe000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 7:
        if (len & 0x1000000) {
          const nCB = len & 0xff000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 8:
        if (len & 0x800000) {
          const nCB = len & 0xff800000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 9:
        if (len & 0x400000) {
          const nCB = len & 0xffc00000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 10:
        if (len & 0x200000) {
          const nCB = len & 0xffe00000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 11:
        if (len & 0x100000) {
          const nCB = len & 0xfff00000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 12:
        if (len & 0x80000) {
          const nCB = len & 0xfff80000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 13:
        if (len & 0x40000) {
          const nCB = len & 0xfffc0000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 14:
        if (len & 0x20000) {
          const nCB = len & 0xfffe0000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 15:
        if (len & 0x10000) {
          const nCB = len & 0xffff0000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 16:
        if (len & 0x8000) {
          const nCB = len & 0xffff8000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 17:
        if (len & 0x4000) {
          const nCB = len & 0xffffc000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 18:
        if (len & 0x2000) {
          const nCB = len & 0xffffe000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 19:
        if (len & 0x1000) {
          const nCB = len & 0xfffff000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 20:
        if (len & 0x800) {
          const nCB = len & 0xfffff800;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 21:
        if (len & 0x400) {
          const nCB = len & 0xfffffc00;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 22:
        if (len & 0x200) {
          const nCB = len & 0xfffffe00;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 23:
        if (len & 0x100) {
          const nCB = len & 0xffffff00;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 24:
        if (len & 0x80) {
          const nCB = len & 0xffffff80;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 25:
        if (len & 0x40) {
          const nCB = len & 0xffffffc0;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 26:
        if (len & 0x20) {
          const nCB = len & 0xffffffe0;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 27:
        if (len & 0x10) {
          const nCB = len & 0xfffffff0;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 28:
        if (len & 0x8) {
          const nCB = len & 0xfffffff8;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 29:
        if (len & 0x4) {
          const nCB = len & 0xfffffffc;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 30:
        if (len & 0x2) {
          const nCB = len & 0xfffffffe;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 31:
        if (len & 0x1) {
          const nCB = len & 0xffffffff;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
    }
    // MODIFICATION: Instead of returning the index, this binary search
    //                instead returns whether something was found or not.
    if (array[len|0] !== sValue) {
       return true; // preserve the value at this index
    } else {
       return false; // eliminate the value at this index
    }
  };
}
Run Code Online (Sandbox Code Playgroud)

请参阅我的其他职位在这里对二进制搜索算法使用的细节,

如果您对文件大小(我尊重)感到拘谨,那么您可以牺牲一点性能以大大减小文件大小并提高可维护性。

function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
    var out=[], value=0;
    for (var i=0,  len=array.length|0; i < len; i=i+1|0)
        if (handle(value = array[i])) 
            out.push( value );
    return out;
}

/* USAGE:
  filterArrayByAnotherArray(
      [1,3,5],
      [2,3,4]
  ) yields [1, 5], and it can work with strings too
*/
function filterArrayByAnotherArray(searchArray, filterArray) {
    if (
        // NOTE: This does not check the whole array. But, if you know
        //        that there are only strings or numbers (not a mix of
        //        both) in the array, then this is a safe assumption.
        typeof searchArray[0] == "number" &&
        typeof filterArray[0] == "number" &&
        (searchArray[0]|0) === searchArray[0] &&
        (filterArray[0]|0) === filterArray[0]
    ) {
        // if all entries in both arrays are integers
        searchArray.sort(sortIntArray);
        filterArray.sort(sortIntArray);
    } else {
        searchArray.sort(sortAnyArray);
        filterArray.sort(sortAnyArray);
    }
    // Progressive Linear Search
    var i = 0;
    return fastFilter(searchArray, function(currentValue){
        while (filterArray[i] < currentValue) i=i+1|0;
        // +undefined = NaN, which is always false for <, avoiding an infinite loop
        return filterArray[i] !== currentValue;
    });
}
Run Code Online (Sandbox Code Playgroud)

为了证明速度上的差异,让我们检查一些 JSPerfs。对于过滤 16 个元素的数组,二分查找比 indexOf 快大约 17%,而 filterArrayByAnotherArray 比 indexOf 快大约 93%。对于过滤 256 个元素的数组,二分查找比 indexOf 快大约 291%,而 filterArrayByAnotherArray 比 indexOf 快大约 353%。对于过滤 4096 个元素的数组,二分查找比 indexOf 快大约 2655%,而 filterArrayByAnotherArray 比 indexOf 快大约 4627%。

反向过滤(如与门)

上一节提供了获取数组 A 和数组 B 的代码,并删除 A 中存在于 B 中的所有元素:

filterArrayByAnotherArray(
    [1,3,5],
    [2,3,4]
);
// yields [1, 5]
Run Code Online (Sandbox Code Playgroud)

下一节将提供反向过滤的代码,我们从 A 中删除 B 中不存在的所有元素。 这个过程在功能上等同于只保留 A 和 B 共有的元素,如与门:

reverseFilterArrayByAnotherArray(
    [1,3,5],
    [2,3,4]
);
// yields [3]
Run Code Online (Sandbox Code Playgroud)

这是反向过滤的代码:

function sortAnyArray(a,b) { return a>b ? 1 : (a===b ? 0 : -1); }
function sortIntArray(a,b) { return (a|0) - (b|0) |0; }
function fastFilter(array, handle) {
    var out=[], value=0;
    for (var i=0,  len=array.length|0; i < len; i=i+1|0)
        if (handle(value = array[i])) 
            out.push( value );
    return out;
}

const Math_clz32 = Math.clz32 || (function(log, LN2){
  return function(x) {
    return 31 - log(x >>> 0) / LN2 | 0; // the "| 0" acts like math.floor
  };
})(Math.log, Math.LN2);

/* USAGE:
  reverseFilterArrayByAnotherArray(
      [1,3,5],
      [2,3,4]
  ) yields [3], and it can work with strings too
*/
function reverseFilterArrayByAnotherArray(searchArray, filterArray) {
    if (
        // NOTE: This does not check the whole array. But, if you know
        //        that there are only strings or numbers (not a mix of
        //        both) in the array, then this is a safe assumption.
        // Always use `==` with `typeof` because browsers can optimize
        //  the `==` into `===` (ONLY IN THIS CIRCUMSTANCE)
        typeof searchArray[0] == "number" &&
        typeof filterArray[0] == "number" &&
        (searchArray[0]|0) === searchArray[0] &&
        (filterArray[0]|0) === filterArray[0]
    ) {
        // if all entries in both arrays are integers
        searchArray.sort(sortIntArray);
        filterArray.sort(sortIntArray);
    } else {
        searchArray.sort(sortAnyArray);
        filterArray.sort(sortAnyArray);
    }
    var searchArrayLen = searchArray.length, filterArrayLen = filterArray.length;
    var progressiveLinearComplexity = ((searchArrayLen<<1) + filterArrayLen)>>>0
    var binarySearchComplexity= (searchArrayLen * (32-Math_clz32(filterArrayLen-1)))>>>0;
    // After computing the complexity, we can predict which algorithm will be the fastest
    var i = 0;
    if (progressiveLinearComplexity < binarySearchComplexity) {
        // Progressive Linear Search
        return fastFilter(searchArray, function(currentValue){
            while (filterArray[i] < currentValue) i=i+1|0;
            // +undefined = NaN, which is always false for <, avoiding an infinite loop
            // For reverse filterning, I changed !== to ===
            return filterArray[i] === currentValue;
        });
    } else {
        // Binary Search
        return fastFilter(
            searchArray,
            inverseFastestBinarySearch(filterArray)
        );
    }
}

// see /sf/answers/3148709931/ for implementation
//  details about this binary search algorithim

function inverseFastestBinarySearch(array){
  var initLen = (array.length|0) - 1 |0;
  
  const compGoto = Math_clz32(initLen) & 31;
  return function(sValue) {
    var len = initLen |0;
    switch (compGoto) {
      case 0:
        if (len & 0x80000000) {
          const nCB = len & 0x80000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 1:
        if (len & 0x40000000) {
          const nCB = len & 0xc0000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 2:
        if (len & 0x20000000) {
          const nCB = len & 0xe0000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 3:
        if (len & 0x10000000) {
          const nCB = len & 0xf0000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 4:
        if (len & 0x8000000) {
          const nCB = len & 0xf8000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 5:
        if (len & 0x4000000) {
          const nCB = len & 0xfc000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 6:
        if (len & 0x2000000) {
          const nCB = len & 0xfe000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 7:
        if (len & 0x1000000) {
          const nCB = len & 0xff000000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 8:
        if (len & 0x800000) {
          const nCB = len & 0xff800000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 9:
        if (len & 0x400000) {
          const nCB = len & 0xffc00000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 10:
        if (len & 0x200000) {
          const nCB = len & 0xffe00000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 11:
        if (len & 0x100000) {
          const nCB = len & 0xfff00000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 12:
        if (len & 0x80000) {
          const nCB = len & 0xfff80000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 13:
        if (len & 0x40000) {
          const nCB = len & 0xfffc0000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 14:
        if (len & 0x20000) {
          const nCB = len & 0xfffe0000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 15:
        if (len & 0x10000) {
          const nCB = len & 0xffff0000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 16:
        if (len & 0x8000) {
          const nCB = len & 0xffff8000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 17:
        if (len & 0x4000) {
          const nCB = len & 0xffffc000;
          len ^= (len ^ (nCB-1)) & ((array[nCB] <= sValue |0) - 1 >>>0);
        }
      case 18:
        if (len & 0x2000) {
          const nCB = len & 

  • 感谢您的回答,我相信迟早会对某人有所帮助,即使它应该在边缘情况下使用(仅当出现性能问题时才使用 IE)。在所有其他情况下,我建议使用可维护/可读的版本。 (6认同)

小智 11

如果您需要比较对象数组,这适用于所有情况:

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);
Run Code Online (Sandbox Code Playgroud)


小智 7

您的问题有很多答案,但是我看不到有人使用lambda表达式:

var array = [1,2,3,4];
var anotherOne = [2,4];
var filteredArray = array.filter(x => anotherOne.indexOf(x) < 0);
Run Code Online (Sandbox Code Playgroud)