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如何判断两个数组是否具有相同的值

javascriptarrayscompare
187

我有两个数组：一个用于存储AJAX请求返回的信息，另一个则记录用户点击的按钮。我使用以下代码（示例数字已填充）：

var array1 = [2, 4];
var array2 = [4, 2]; //It cames from the user button clicks, so it might be disordered.
array1.sort(); //Sorts both Ajax and user info.
array2.sort();
if (array1==array2) {
    doSomething();
}else{
    doAnotherThing();
}

但是它总是返回false，即使两个数组相同，只是名称不同。（我在Chrome的JS控制台中检查了这一点）。那么有没有办法可以知道这两个数组是否相同？为什么会返回false？我如何知道第一个数组中哪些值不在第二个数组中？

1
我非常确定您需要遍历数组的每个元素。 - Thomas Li
你知道为什么它返回false吗？好奇。 - RobW
请查看 @Andrew 的答案 https://dev59.com/om025IYBdhLWcg3wFxqa#6229263 - Carlos Precioso
可能是重复的问题：如何检查 JavaScript 数组是否相等？ - Palec
32个回答
3

另一个一行解决方案：

array1.concat(array2).filter((item, index, currentArr) => currentArr.lastIndexOf(item) == currentArr.indexOf(item)).length == 0;

或者

[...array1, ...array2].filter((item, index, currentArr) => currentArr.lastIndexOf(item) == currentArr.indexOf(item)).length == 0;
3
  1. 检查两个数组是否具有相同的值，但可能不按相同顺序排列

function compareArr(arr1, arr2) {
  return arr1.sort().join(',') == arr2.sort().join(',');
}

console.log(compareArr([1,2,3],[1,2,3])); // returns true
console.log(compareArr([1,2,3],[3,2,1])); // returns true
console.log(compareArr([1,2,3],[3,4,1])); // returns false

检查两个数组是否具有相同的值和相同的顺序（不要排序）。 

function compareArrOrder(arr1, arr2) {
  return arr1.join(',') == arr2.join(',');
}

console.log(compareArrOrder([1,2,3],[1,2,3])); // returns true
console.log(compareArrOrder([1,2,3],[3,2,1])); // returns false
console.log(compareArrOrder([1,2,3],[3,4,1])); // returns false

这段代码适用于我的使用情况，我需要检查两个数组中是否存在相同且仅相同的值，无论顺序如何。这是一段简洁明了的代码。 - Jan
3

使用ES6

我们将使用Ramda的equals函数，但是我们也可以使用Lodash或Underscore的isEqual函数：

const R = require('ramda');

const arraysHaveSameValues = (arr1, arr2) => R.equals( [...arr1].sort(), [...arr2].sort() )

使用扩展运算符，我们避免了对原始数组的改变，并保持了函数的纯洁性。
我不认为这与ES6有任何关系。关键是在排序之前克隆数组，可以使用arr1.slice().sort()来完成。最终是Ramda完成了所有的工作。 - VLAZ
3

我在一个游戏项目中有简单的整数值
每个数组中的值较少，同时需要保留原始数组不受影响
因此，我做了以下操作，效果很好。（代码编辑以在此处粘贴）

var sourceArray = [1, 2, 3];
var targetArray = [3, 2, 1];

if (sourceArray.length !== targetArray.length) {
    // not equal
    // did something
    return false;
}

var newSortedSourceArray = sourceArray.slice().sort();
var newSortedTargetArray = targetArray.slice().sort();

if (newSortedSourceArray.toString() !== newSortedTargetArray.toString()) { // MAIN CHECK
    // not equal
    // did something
    return false;
}
else {
    // equal
    // did something
    // continued further below
}

// did some more work

return true;

希望这能帮到您。
3

你似乎不太关心性能，其他回答者可能也不关心。

以下是sameArrayMembers（即[1,2,2]！=[1,1,2]）和sameArrayMembersAsSet（即[1,2,2] == [1,1,2]）的一些基准测试结果。

我没有检查[1,1,2]是否与[1,2]相同，但鉴于你声称要检查集合，那么应该不需要进行长度检查。

const tests = {
  'Maciej Krawczyk': (_arr1, _arr2) => {

    if (!Array.isArray(_arr1) || ! Array.isArray(_arr2) || _arr1.length !== _arr2.length)
      return false;

    var arr1 = _arr1.concat().sort();
    var arr2 = _arr2.concat().sort();

    for (var i = 0; i < arr1.length; i++) {

        if (arr1[i] !== arr2[i])
            return false;

    }

    return true;
  },
  'canbax': (a1, a2) => {
    const superSet = {};
    for (const i of a1) {
      const e = i + typeof i;
      superSet[e] = 1;
    }

    for (const i of a2) {
      const e = i + typeof i;
      if (!superSet[e]) {
        return false;
      }
      superSet[e] = 2;
    }

    for (let e in superSet) {
      if (superSet[e] === 1) {
        return false;
      }
    }

    return true;
  },
  'kennebec': (array1, array2) => {
    return array1.slice().sort().join(',')=== array2.slice().sort().join(',');
  },
  'Max Heiber': function() {
    const containsAll = (arr1, arr2) => 
        arr2.every(arr2Item => arr1.includes(arr2Item));
    return (arr1, arr2) => 
        containsAll(arr1, arr2) && containsAll(arr2, arr1);
  }(),
  'gman': (a, b) => {
    if (a.length !== b.length) { return false; }
    const counts = new Map();
    for (const v of a) {
      const count = counts.get(v) || 0;
      counts.set(v, count + 1);
    }
    for (const v of b) {
      const count = counts.get(v);
      if (!count) {   // undefined or 0, both mean arrays are not the same
        return false;
      }
      counts.set(v, count - 1);
    }
    return true;
  },
  'Bemmu': (a, b) => {
    if (Array.isArray(a) && Array.isArray(b) && a.length == b.length) {
      a = a.concat().sort()
      b = b.concat().sort()
      return a.reduce((acc,e,i) => acc && e === b[i], true)
    } else {
      return false;
    }
  },
  'Sandeep': (array1, array2) => {
    return JSON.stringify(array1.sort()) === JSON.stringify(array2.sort());
  },
  'camslice': (arr1, arr2) => {
    const arr1test = arr1.slice().sort();
    const arr2test = arr2.slice().sort();
    return !arr1test.some((val, idx) => val !== arr2test[idx]);
  },
  'Dimitrios Stefos': (arr1, arr2) => {
    if (!Array.isArray(arr1) || !Array.isArray(arr2) || arr1.length!=arr2.length)
        return false;

    return arr1.length==arr1.filter(word => arr2.includes(word)).length;
  },
  'SC1000': (arr1, arr2, opts) => {
    function vKey(i, v) {
      return (opts?.enforceOrder ? `${i}-` : '') + `${typeof v}-${v}`
    }

    if (arr1.length !== arr2.length) return false;

    const d1 = {};
    const d2 = {};  
    for (let i = arr1.length - 1; i >= 0; i--) {
      d1[vKey(i, arr1[i])] = true;
      d2[vKey(i, arr2[i])] = true;
    }

    for (let i = arr1.length - 1; i >= 0; i--) {
      const v = vKey(i, arr1[i]);
      if (d1[v] !== d2[v]) return false;
    }

    for (let i = arr2.length - 1; i >= 0; i--) {
      const v = vKey(i, arr2[i]);
      if (d1[v] !== d2[v]) return false;
    }

    return true
  },
  'Magnus Fohlström': (arr1, arr2) => {
        let count = (arr, val) => arr.reduce((count, curr) => (curr === val ? 1:0) + count, 0);
        return arr1.length === arr2.length && arr1.reduce((checks, val) =>
            checks.concat(count(arr1, val) === count(arr2, val)), []).every(check => check);
   },
};

// ----------------------------

function createExposedPromise() {
 const p = {};
 p.promise = new Promise((resolve, reject) => {
   p.resolve = resolve;
   p.reject = reject;
 });
 return p;
}

function assert(cond) {
  if (!cond) {
    log('assert');
    throw new Error();
  }
}

async function main() {
  await testResults(true, 'sameArrayMembers');
  await testResults(false, 'sameArrayMemembersAsSet');
  
  async function testResults(asSet, msg) {
    log(`\n=======[ ${msg} ] ============`);
    const suite = new Benchmark.Suite();
    let test;

    // reject if they fail simple test
    const a = [1,1,2];
    const b = [1,2,2];
    for (const [name, fn] of Object.entries(tests)) {
      if (fn(a, b) === asSet) {
        log(`${name} fails for ${msg}([${a}], [${b}])`);
      } else {    
        suite.add(name, () => test(fn));
      }
    }

    let endPromise;

    suite.on('cycle', event => log(String(event.target)));
    suite.on('complete', function() {
      log('Fastest is ' + this.filter('fastest').map('name'));
      endPromise.resolve();
    });

    async function runWith(num, title) {
      log(`------[ ${title} ] -----------`);

      const a = [];
      for (let i = 0; i < num; ++i) {
        a[i] = Math.random();
      }
      const b = [...a];
      const c = [...a]; c[c.length / 2 | 0]++;

      endPromise = createExposedPromise();

      test = (fn) => {
        assert(fn(a, b))
        assert(!fn(a, c));
      };

      suite.reset();
      suite.run({'async': true});
      await endPromise.promise;
    }

    await runWith(10, 'small (10)');
    await runWith(100, 'medium (100)');
    await runWith(10000, 'large (10000)');
  }
}
main();

function log(...args) {
  const elem = document.createElement('pre');
  elem.textContent = args.join(' ');
  document.body.appendChild(elem);
}
pre { margin: 0; }
<script src="https://unpkg.com/lodash@4.17.20/lodash.js"></script>
<script src="https://unpkg.com/benchmark@2.1.4/benchmark.js"></script>

由于JS引擎会不断更新，因此结果可能会随时间而改变。以下是2020/09/28的一些结果：

Chrome 87

=======[ sameArrayMembers ] ============
canbax fails for sameArrayMembers([1,1,2], [1,2,2])
Max Heiber fails for sameArrayMembers([1,1,2], [1,2,2])
Dimitrios Stefos fails for sameArrayMembers([1,1,2], [1,2,2])
SC1000 fails for sameArrayMembers([1,1,2], [1,2,2])
------[ small (10) ] -----------
Maciej Krawczyk x 246,129 ops/sec ±0.84% (66 runs sampled)
kennebec x 182,234 ops/sec ±0.56% (64 runs sampled)
gman x 377,356 ops/sec ±1.55% (64 runs sampled)
Bemmu x 244,850 ops/sec ±0.76% (64 runs sampled)
Sandeep x 100,529 ops/sec ±1.53% (63 runs sampled)
camslice x 542,577 ops/sec ±0.68% (64 runs sampled)
Fastest is camslice
------[ medium (100) ] -----------
Maciej Krawczyk x 12,121 ops/sec ±1.40% (63 runs sampled)
kennebec x 10,219 ops/sec ±1.60% (63 runs sampled)
gman x 41,225 ops/sec ±1.63% (62 runs sampled)
Bemmu x 12,400 ops/sec ±1.10% (63 runs sampled)
Sandeep x 12,470 ops/sec ±0.50% (64 runs sampled)
camslice x 57,126 ops/sec ±0.54% (64 runs sampled)
Fastest is camslice
------[ large (10000) ] -----------
Maciej Krawczyk x 30.75 ops/sec ±0.86% (42 runs sampled)
kennebec x 27.35 ops/sec ±1.11% (38 runs sampled)
gman x 376 ops/sec ±0.46% (62 runs sampled)
Bemmu x 30.91 ops/sec ±0.77% (42 runs sampled)
Sandeep x 80.33 ops/sec ±0.54% (53 runs sampled)
camslice x 166 ops/sec ±0.44% (61 runs sampled)
Fastest is gman

=======[ sameArrayMemembersAsSet ] ============
Maciej Krawczyk fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
kennebec fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
gman fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
Bemmu fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
Sandeep fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
camslice fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
------[ small (10) ] -----------
canbax x 110,826 ops/sec ±2.07% (63 runs sampled)
Max Heiber x 2,699,807 ops/sec ±1.31% (63 runs sampled)
Dimitrios Stefos x 2,910,096 ops/sec ±0.65% (62 runs sampled)
SC1000 x 59,989 ops/sec ±2.61% (63 runs sampled)
Fastest is Dimitrios Stefos
------[ medium (100) ] -----------
canbax x 9,624 ops/sec ±2.20% (53 runs sampled)
Max Heiber x 88,945 ops/sec ±0.71% (64 runs sampled)
Dimitrios Stefos x 94,425 ops/sec ±0.51% (65 runs sampled)
SC1000 x 5,742 ops/sec ±0.74% (33 runs sampled)
Fastest is Dimitrios Stefos
------[ large (10000) ] -----------
canbax x 59.85 ops/sec ±1.69% (46 runs sampled)
Max Heiber x 13.50 ops/sec ±0.87% (26 runs sampled)
Dimitrios Stefos x 15.40 ops/sec ±0.89% (30 runs sampled)
SC1000 x 37.42 ops/sec ±1.47% (40 runs sampled)
Fastest is canbax

火狐浏览器80版

=======[ sameArrayMembers ] ============
canbax fails for sameArrayMembers([1,1,2], [1,2,2])
Max Heiber fails for sameArrayMembers([1,1,2], [1,2,2])
Dimitrios Stefos fails for sameArrayMembers([1,1,2], [1,2,2])
SC1000 fails for sameArrayMembers([1,1,2], [1,2,2])
------[ small (10) ] -----------
Maciej Krawczyk x 118,391 ops/sec ±0.52% (65 runs sampled)
kennebec x 70,254 ops/sec ±0.40% (67 runs sampled)
gman x 201,659 ops/sec ±3.23% (57 runs sampled)
Bemmu x 118,133 ops/sec ±0.67% (64 runs sampled)
Sandeep x 69,484 ops/sec ±1.40% (65 runs sampled)
camslice x 130,443 ops/sec ±0.55% (65 runs sampled)
Fastest is gman
------[ medium (100) ] -----------
Maciej Krawczyk x 11,418 ops/sec ±2.81% (61 runs sampled)
kennebec x 7,278 ops/sec ±1.37% (41 runs sampled)
gman x 19,748 ops/sec ±6.60% (53 runs sampled)
Bemmu x 11,535 ops/sec ±1.09% (62 runs sampled)
Sandeep x 8,234 ops/sec ±1.46% (45 runs sampled)
camslice x 14,277 ops/sec ±3.08% (60 runs sampled)
Fastest is gman
------[ large (10000) ] -----------
Maciej Krawczyk x 65.25 ops/sec ±2.13% (49 runs sampled)
kennebec x 47.73 ops/sec ±0.82% (51 runs sampled)
gman x 210 ops/sec ±3.54% (54 runs sampled)
Bemmu x 66.90 ops/sec ±0.53% (50 runs sampled)
Sandeep x 63.13 ops/sec ±1.59% (48 runs sampled)
camslice x 115 ops/sec ±1.36% (56 runs sampled)
Fastest is gman

=======[ sameArrayMemembersAsSet ] ============
Maciej Krawczyk fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
kennebec fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
gman fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
Bemmu fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
Sandeep fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
camslice fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
------[ small (10) ] -----------
canbax x 79,433 ops/sec ±1.11% (57 runs sampled)
Max Heiber x 1,822,200 ops/sec ±0.92% (65 runs sampled)
Dimitrios Stefos x 2,258,820 ops/sec ±0.48% (65 runs sampled)
SC1000 x 35,784 ops/sec ±1.42% (63 runs sampled)
Fastest is Dimitrios Stefos
------[ medium (100) ] -----------
canbax x 6,726 ops/sec ±0.60% (38 runs sampled)
Max Heiber x 41,620 ops/sec ±1.08% (65 runs sampled)
Dimitrios Stefos x 53,041 ops/sec ±1.61% (64 runs sampled)
SC1000 x 3,450 ops/sec ±0.56% (64 runs sampled)
Fastest is Dimitrios Stefos
------[ large (10000) ] -----------
canbax x 44.18 ops/sec ±5.87% (42 runs sampled)
Max Heiber x 5.62 ops/sec ±4.79% (19 runs sampled)
Dimitrios Stefos x 6.77 ops/sec ±1.21% (21 runs sampled)
SC1000 x 24.18 ops/sec ±3.50% (36 runs sampled)
Fastest is canbax

Safari 14

=======[ sameArrayMembers ] ============
canbax fails for sameArrayMembers([1,1,2], [1,2,2])
Max Heiber fails for sameArrayMembers([1,1,2], [1,2,2])
Dimitrios Stefos fails for sameArrayMembers([1,1,2], [1,2,2])
SC1000 fails for sameArrayMembers([1,1,2], [1,2,2])
------[ small (10) ] -----------
Maciej Krawczyk x 142,798 ops/sec ±0.50% (65 runs sampled)
kennebec x 118,073 ops/sec ±1.12% (63 runs sampled)
gman x 760,109 ops/sec ±0.46% (66 runs sampled)
Bemmu x 136,265 ops/sec ±0.48% (63 runs sampled)
Sandeep x 69,868 ops/sec ±0.44% (64 runs sampled)
camslice x 155,548 ops/sec ±0.45% (64 runs sampled)
Fastest is gman
------[ medium (100) ] -----------
Maciej Krawczyk x 8,479 ops/sec ±0.52% (46 runs sampled)
kennebec x 5,992 ops/sec ±2.54% (34 runs sampled)
gman x 83,390 ops/sec ±0.37% (64 runs sampled)
Bemmu x 8,615 ops/sec ±0.56% (63 runs sampled)
Sandeep x 5,943 ops/sec ±0.67% (64 runs sampled)
camslice x 8,753 ops/sec ±0.45% (47 runs sampled)
Fastest is gman
------[ large (10000) ] -----------
Maciej Krawczyk x 62.66 ops/sec ±0.87% (51 runs sampled)
kennebec x 46.46 ops/sec ±0.66% (48 runs sampled)
gman x 615 ops/sec ±2.33% (60 runs sampled)
Bemmu x 60.98 ops/sec ±1.28% (52 runs sampled)
Sandeep x 49.11 ops/sec ±2.07% (47 runs sampled)
camslice x 66.33 ops/sec ±4.44% (50 runs sampled)
Fastest is gman

=======[ sameArrayMemembersAsSet ] ============
Maciej Krawczyk fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
kennebec fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
gman fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
Bemmu fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
Sandeep fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
camslice fails for sameArrayMemembersAsSet([1,1,2], [1,2,2])
------[ small (10) ] -----------
canbax x 167,041 ops/sec ±0.63% (64 runs sampled)
Max Heiber x 1,281,054 ops/sec ±0.74% (59 runs sampled)
Dimitrios Stefos x 1,127,639 ops/sec ±0.98% (64 runs sampled)
SC1000 x 112,824 ops/sec ±0.37% (64 runs sampled)
Fastest is Max Heiber
------[ medium (100) ] -----------
canbax x 8,784 ops/sec ±0.53% (47 runs sampled)
Max Heiber x 37,824 ops/sec ±0.52% (65 runs sampled)
Dimitrios Stefos x 41,238 ops/sec ±0.85% (63 runs sampled)
SC1000 x 6,181 ops/sec ±0.61% (35 runs sampled)
Fastest is Dimitrios Stefos
------[ large (10000) ] -----------
canbax x 63.83 ops/sec ±2.46% (49 runs sampled)
Max Heiber x 5.41 ops/sec ±0.52% (18 runs sampled)
Dimitrios Stefos x 6.02 ops/sec ±1.32% (19 runs sampled)
SC1000 x 42.25 ops/sec ±1.45% (45 runs sampled)
Fastest is canbax
3
你可以使用 reduce 替代循环来显得聪明，但这样做的风险是让你的同事觉得你是一个自以为是的人。
function isArrayContentSame(a, b) {
  if (Array.isArray(a) && Array.isArray(b) && a.length == b.length) {
    a = a.concat().sort()
    b = b.concat().sort()
    return a.reduce((acc,e,i) => acc && e === b[i], true)
  } else {
    return false;
  }
}
3

大多数其他解决方案使用排序（O(n*log n)）、使用库或具有O(n^2)复杂度。

这里提供了一个纯Javascript解决方案，具有线性复杂度O(n):

/**
 * Check if two arrays of strings or numbers have the same values
 * @param {string[]|number[]} arr1
 * @param {string[]|number[]} arr2
 * @param {Object} [opts]
 * @param {boolean} [opts.enforceOrder] - By default (false), the order of the values in the arrays doesn't matter.
 * @return {boolean}
 */
function compareArrays(arr1, arr2, opts) {

  function vKey(i, v) {
    return (opts?.enforceOrder ? `${i}-` : '') + `${typeof v}-${v}`
  }

  if (arr1.length !== arr2.length) return false;

  const d1 = {};
  const d2 = {};  
  for (let i = arr1.length - 1; i >= 0; i--) {
    d1[vKey(i, arr1[i])] = true;
    d2[vKey(i, arr2[i])] = true;
  }
  
  for (let i = arr1.length - 1; i >= 0; i--) {
    const v = vKey(i, arr1[i]);
    if (d1[v] !== d2[v]) return false;
  }

  for (let i = arr2.length - 1; i >= 0; i--) {
    const v = vKey(i, arr2[i]);
    if (d1[v] !== d2[v]) return false;
  }

  return true
}

测试：

arr1= [1, 2]
arr2= [1, 2]
compareArrays(arr1, arr2) => true
compareArrays(arr1, arr2, {enforceOrder: true}) => true
-------
arr1= [1, 2]
arr2= [2, 1]
compareArrays(arr1, arr2) => true
compareArrays(arr1, arr2, {enforceOrder: true}) => false
-------
arr1= [2, 1]
arr2= [1, 2]
compareArrays(arr1, arr2) => true
compareArrays(arr1, arr2, {enforceOrder: true}) => false
-------
arr1= [2, 2]
arr2= [1, 2]
compareArrays(arr1, arr2) => false
compareArrays(arr1, arr2, {enforceOrder: true}) => false
-------
arr1= [1, 2]
arr2= [1, 2, 3]
compareArrays(arr1, arr2) => false
compareArrays(arr1, arr2, {enforceOrder: true}) => false
-------
arr1= ["1"]
arr2= [1]
compareArrays(arr1, arr2) => false
compareArrays(arr1, arr2, {enforceOrder: true}) => false
-------
arr1= ["1", 2]
arr2= [2, "1"]
compareArrays(arr1, arr2) => true
compareArrays(arr1, arr2, {enforceOrder: true}) => false
-------
arr1= []
arr2= []
compareArrays(arr1, arr2) => true
compareArrays(arr1, arr2, {enforceOrder: true}) => true
1
嗨，这是一个很棒的条目！它相当复杂，但确实有意义。我有一个问题：我不太熟悉大O符号表示法，但这个算法肯定是O（2n）吧？不过我认为这并没有什么区别。 - Carlos Precioso
1
@CarlosPrecioso 是正确的，O(2n) = O(n)。复杂度在乘以常数因子时不会改变。 - ingdc
这个无法处理像 compareArrays(['1'], [1]) 或者 compareArrays([2, 2], [1, 2]) 这样的情况。 - Chris Kent
@SC1000 我的示例仅包含字符串或数字数组。 - Chris Kent
@SC1000 的确如此。在每个示例中，数组具有不同的值，但函数返回 true。将 ['1'] 与 [1] 进行比较的预期结果应为 false，因为数组具有不同的值。同样，[2, 2] 和 [1, 2] 是具有不同值的数组，因此预期结果为 false。另外，如果您执行 compareArrays([1, 2], [2, 2])，它会返回 false，因此您的函数不是可交换的；我觉得这在这样的函数中是意外的。 - Chris Kent
@ChrisKent 更新后的函数处理了所有情况。现在如果需要，它还有一个可选参数来强制排序。感谢您的反馈。 - ingdc
2
最简单的获取结果的方法如下： 

const array1 = [1,2,3,4]

const array2 = [3,4,5,6]

const result = array1.filter(arr => array2.includes(arr))

console.log(result)

2

代码和解释

function arrayEquals(arr1, arr2) {
  // compare array lengths, if not equal then skip.
  if (arr1.length !== arr2.length) return false;

  // setup a variable of the array length
  let i = arr1.length;

  // iterate through every element in arr1 basically...
  while (i--) {
    // if arr2 doesn't include x element from arr1, return false
    if (!arr2.includes(arr1[i])) return false;

    // passes tests and checks out
  };
  return true;
}

// works with multiple types
let foo = ["foo", "bar", 1, 2, true];
let bar = ["bar", 2, true, "foo", 1];

console.log(foo,bar,"Should return true: " + arrayEquals(foo, bar));

let fob = ["bar", "foo"];
let oar = ["oar", "foo"];

console.log(fob,oar,"Should return false (bar ≠ oar): " + arrayEquals(fob, oar));

免责声明

此功能仅支持简单类型，如字符串、数字、布尔等。它不会深入分析您的对象或数组，除非您编辑代码使其递归，但这似乎并不影响回答问题本身。您提供了数字，因此我的答案将适用于数字。

2

试试这个

function arraysEqual(arr1, arr2){
    if (!Array.isArray(arr1) || !Array.isArray(arr2) || arr1.length!=arr2.length)
        return false;

    return arr1.length==arr1.filter(word => arr2.includes(word)).length;
}
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