如何知道两个数组是否具有相同的值

Question

我有这两个数组：一个填充来自 ajax 请求的信息，另一个存储用户单击的按钮。我使用此代码（我填写了示例编号）：

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();
}

但它总是给出false，即使两个数组相同，但名称不同。 （我在 Chrome 的 JS 控制台中检查了这个）。那么，有什么方法可以知道这两个数组是否包含相同的内容？为什么给false？如何知道第一个数组中的哪些值不在第二个数组中？

Answer 1

Array.prototype.compare = function(testArr) {
    if (this.length != testArr.length) return false;
    for (var i = 0; i < testArr.length; i++) {
        if (this[i].compare) { //To test values in nested arrays
            if (!this[i].compare(testArr[i])) return false;
        }
        else if (this[i] !== testArr[i]) return false;
    }
    return true;
}

var array1 = [2, 4];
var array2 = [4, 2];
if(array1.sort().compare(array2.sort())) {
    doSomething();
} else {
    doAnotherThing();
}

也许？

Answer 2

当您比较这两个数组时，您是在比较表示数组的对象，而不是内容。

您必须使用函数来比较两者。您可以编写自己的代码，简单地循环一个并在检查长度是否相同后将其与另一个进行比较。

Answer 3

如果您的数组项不是对象-例如，如果它们是数字或字符串，您可以比较它们的连接字符串以查看它们是否以任何顺序具有相同的成员-

var array1= [10, 6, 19, 16, 14, 15, 2, 9, 5, 3, 4, 13, 8, 7, 1, 12, 18, 11, 20, 17];
var array2= [12, 18, 20, 11, 19, 14, 6, 7, 8, 16, 9, 3, 1, 13, 5, 4, 15, 10, 2, 17];

if(array1.sort().join(',')=== array2.sort().join(',')){
    alert('same members');
}
else alert('not a match');

Answer 4

如果你使用Prototype Framework，你可以使用数组的intersect方法找出它们是否相同（不分先后）：

var array1 = [1,2];
var array2 = [2,1];

if(array1.intersect(array2).length === array1.length) {
    alert("arrays are the same!");
}

Answer 5

对象相等检查：JSON.stringify(array1.sort()) === JSON.stringify(array2.sort())

上述测试也适用于对象数组，在这种情况下使用http://www.w3schools.com/jsref/jsref_sort.asp中记录的排序函数

对于具有扁平 JSON 模式的小型数组可能就足够了。

Answer 6

如果您只想检查两个数组是否具有相同的值（不管每个值的出现次数和顺序），您可以使用lodash：

_.isEmpty(_.xor(array1, array2))

简短、简单、漂亮！

Answer 7

我在游戏项目中有简单的整数值
每个数组中的值数量较少，也需要原始数组保持不变
所以，我做了下面的，它工作得很好。 （代码编辑粘贴到这里）

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;

希望对您有所帮助。

Answer 8

为什么您的代码不起作用

JavaScript 有 primitive data types 和非原始数据类型。

对于原始数据类型，== 和=== 检查条形两侧的东西是否具有相同的值。这就是为什么1 === 1 是正确的。

对于数组等非原始数据类型，== 和 === 检查引用相等性。也就是说，他们检查arr1 和arr2 是否是同一个对象。在您的示例中，两个数组以相同的顺序具有相同的对象，但不等价。

解决方案

两个数组 arr1 和 arr2 具有相同的成员当且仅当：

• arr2 中的所有内容都在 arr1 中

与

• arr1 中的所有内容都在arr2 中

所以这可以解决问题（ES2016）：

const containsAll = (arr1, arr2) => 
                arr2.every(arr2Item => arr1.includes(arr2Item))
                
const sameMembers = (arr1, arr2) => 
                        containsAll(arr1, arr2) && containsAll(arr2, arr1);

sameMembers(arr1, arr2); // `true`

使用Underscore 的第二个解决方案更接近您想要做的：

arr1.sort();
arr2.sort();

_.isEqual(arr1, arr2); // `true`

之所以有效，是因为 isEqual 检查“深度相等”，这意味着它不仅仅查看引用相等并比较值。

第三个问题的解答

您还问如何找出arr1 中的哪些内容不包含在arr2 中。

这样就可以了（ES2015）：

const arr1 = [1, 2, 3, 4];
const arr2 = [3, 2, 1];

arr1.filter(arr1Item => !arr2.includes(arr1Item)); // `[4]`

你也可以使用Underscore的difference:方法：

_.difference(arr1, arr2); // `[4]`

更新

查看@Redu 的评论——我的解决方案是sameMembers，但您可能想到的是sameMembersInOrderalso-known-as deepEquals。

更新 2

如果你不关心数组成员的顺序，ES2015+ 的Set 可能是比Array 更好的数据结构。请参阅MDN notes on how to implement isSuperset and difference 使用危险的猴子补丁。

更新 3

如果顺序无关紧要，最好还是使用集合。但是如果你必须使用数组，这个解决方案的时间复杂度比我之前给出的要好：

function sameMembers(arr1, arr2) {
    const set1 = new Set(arr1);
    const set2 = new Set(arr2);
    return arr1.every(item => set2.has(item)) &&
        arr2.every(item => set1.has(item))
}

Answer 9

对数组进行排序并一一比较它们的值。

function arrayCompare(_arr1, _arr2) {
    if (
      !Array.isArray(_arr1)
      || !Array.isArray(_arr2)
      || _arr1.length !== _arr2.length
      ) {
        return false;
      }
    
    // .concat() to not mutate arguments
    const arr1 = _arr1.concat().sort();
    const arr2 = _arr2.concat().sort();
    
    for (let i = 0; i < arr1.length; i++) {
        if (arr1[i] !== arr2[i]) {
            return false;
         }
    }
    
    return true;
}

Answer 10

请检查此答案

var arr1= [12,18];
var arr2= [12, 18, 20, 11, 19, 14, 6, 7, 8, 16, 9, 3, 1, 13, 5, 4, 15, 10, 2, 17];
for(i=0;i<arr1.length;i++)
{
var array1=arr1[i];
for(j=0;j<arr2.length;j++)
{
    var array2=arr2[j];
    if(array1==array2)
    {
return true;
    }
}
}

Answer 11

使用 ES6

我们将使用 Ramda 的 equals 函数，但我们可以使用 Lodash 或 Underscore 的 isEqual：

const R = require('ramda');

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

使用扩展运算符，我们避免改变原始数组，并保持函数纯净。

Answer 12

我们的目标基本上是检查两个数组是否相等。 set 是数学定义的集合。最快的渐近排序需要 O(nlog(n)) 时间。因此，如果对数组进行排序，则至少需要 O(nlog(n)) 时间。但是您可以更快地完成这项任务，使用字典数据结构渐近需要 O(n) 时间（平均情况不是最坏情况）。在 JS 中，字典只是一个带有键和值的对象。

/** assumes array elements are primitive types
* check whether 2 arrays are equal sets.
* @param  {} a1 is an array
* @param  {} a2 is an array
*/
function areArraysEqualSets(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;
}

请注意，此函数适用于原始类型的数组，并假定 a1 和 a2 是数组。

Answer 13

很久以后才回答，但希望这对寻求简单解决方案的人和现代新手有所帮助。

现在我们可以使用多个库来实现这一点，例如lodash、underscore 等（由于简单、多种功能和高使用率，这些现在已成为项目的一部分）

您可以使用 lodash 库中的交集。

_.intersection(['2-1', '1'], ['2-2', '3-1', '2-1']); 
// => ['2-1']

这适用于任何数据类型..

Answer 14

如果您想比较两个数组并检查两个数组中是否有相同的对象，它将起作用。 示例：

Array1 = [a,b,c,d]
Array2 = [d,e,f,g]

这里，'d' 在两个数组中都很常见，所以这个函数将返回真值。

  cehckArray(array1, array2) {
    for (let i = 0; i < array1.length; i++) {
      for (let j = 0; j < array2.length; j++) {
        if (array1[i] === array2[j]) {
          return true;
        }
      }
    }
    // Return if no common element exist 
    return false;
  }

Answer 15

这个呢？ ES 2017 我想：

const array1 = [1, 3, 5];
const array2 = [1, 5, 3];

const isEqual = (array1.length === array2.length) && (array1.every(val => array2.includes(val)));
console.log(isEqual);

第一个条件检查两个数组是否具有相同的长度，第二个条件检查第一个数组是否是第二个数组的子集。结合这 2 个条件应该会比较这 2 个数组的所有项，而不管元素的顺序如何。

上面的代码只有在两个数组都有不重复项的情况下才有效。

Answer 16

您可以使用reduce 代替循环来显得聪明，但有可能让您的开发人员认为您是个聪明人。

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;
  }
}

Answer 17

使用 ES6 实现浅层相等的简单解决方案：

const arr1test = arr1.slice().sort()
const arr2test = arr2.slice().sort()
const equal = !arr1test.some((val, idx) => val !== arr2test[idx])

创建每个数组的浅表副本并对其进行排序。然后使用some() 循环遍历arr1test 值，检查每个值与arr2test 中具有相同索引的值。如果所有值都相等，则some() 返回false，然后equal 计算结果为true。

也可以使用every()，但它必须循环遍历数组中的每个元素以满足true 结果，而some() 将在找到不相等的值时立即退出：

const equal = arr1test.every((val, idx) => val === arr2test[idx])

Answer 18

比较两个数组的简单解决方案：

var array1 = [2, 4];
var array2 = [4, 2];

array1.sort();
array2.sort();

if (array1[0] == array2[0]) {
    console.log("Success");
}else{
    console.log("Wrong");
}

Answer 19

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;
}

Answer 20

根据接受的答案，我有另一种方法。

function compareArrays(array1, array2) {

    if (
        !Array.isArray(array1)
        || !Array.isArray(array2)
        || array1.length !== array2.length
    ) return false;

    var first = array1.sort().map(value => (String(value))).join();
    var second = array2.sort().map(value => (String(value))).join();

    return first == second ? true : false;
}

Answer 21

比较两个数组的函数，以检查两者是否具有相同的元素。就算是乱七八糟的……

这对于简单的数组很有用。 [String,Number,Boolean,null,NaN].

我不使用 .sort()，它会修改原始数组。有人说它不好...

小心。这个函数是有限的，它不能比较 Objects"[],{}" 或者这些数组中的函数，数组本身就是 Objects。

   let arraysHasSameElements = (arr1, arr2) => {
        let count =
            // returns counting of occurrences.
            (arr, val) => arr.reduce((count, curr) => (curr === val ? 1 : 0) + count, 0);

        /* this will return true if lengths of the arrays is equal.
           then compare them.*/
        return arr1.length === arr2.length

            // compare arr1 against arr2.
            && arr1.reduce((checks, val) =>

                /*  creating array of checking if a value has equal amount of occurrences
                    in both arrays, then adds true 'check'. */
                checks.concat(count(arr1, val) === count(arr2, val)), [])

                // checking if each check is equal to true, then .every() returns true.
                .every(check => check);
    }

    let arr1 = ['',-99,true,NaN,21,null,false,'help',-99,'help',NaN], 
        arr2 = [null,-99,'',NaN,NaN,false,true,-99,'help',21,'help'];
    arraysHasSameElements(arr1, arr2); //true

    let arr3 = [false,false,false,false,false,false], 
        arr4 = [false,false,false,false,false,false]
    arraysHasSameElements(arr3, arr4); //true


    // here we have uncommented version.
    let arraysHasSameElements = (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);
    }

Answer 22

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。

这是一个纯 Javascript 的解决方案，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

Answer 23

您似乎并不关心性能，不确定其他答案是否也如此。

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

没有检查 [1,1,2] 与 [1,2] 相同，如果您声称检查集合而不是长度检查，我可能应该给出。

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();

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<script src="https://unpkg.com/lodash@4.17.20/lodash.js"></script>
<script src="https://unpkg.com/benchmark@2.1.4/benchmark.js"></script>

随着 JS 引擎的更新，结果可能会随着时间而改变。以下是 2020/09/28 的一些结果

铬 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80

=======[ 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

Answer 24

如果你只想测试原始值，你可以试试：

if(JSON.stringify(arr1.sort()) !== JSON.stringify(arr2.sort())) {
  console.log('invalid');
}

Answer 25

另一种解决方案：

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

或

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

Answer 26

这是我没有排序的解决方案：

function isDifferent(t1,t2)
{
    return ((t1.length !== t2.length) || t1.some(tag1 => !t2.some(t=>t===tag1)));
}

let a1=['t','','b','xx','i','p','i'];
let a2=['','o','xx','b','t','p','i'];

console.log("DIFFERENT=",isDifferent(a1,a2)); //false

a1=['a','c','d'];
a2=['a','c'];

console.log("DIFFERENT=",isDifferent(a1,a2)); //true

a1=['a','c','d'];
a2=['a','c','c'];

console.log("DIFFERENT=",isDifferent(a1,a2)); //true

a1=['a','b','c'];
a2=['c','b','a'];

console.log("DIFFERENT=",isDifferent(a1,a2)); //false

Answer 27

1. 检查两个数组是否具有相同的值但顺序可能不同

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

1. 检查两个数组是否具有相同的值和相同的顺序（只是不排序。）

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

Answer 28

代码及说明

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

这仅支持简单类型，如字符串、数字、布尔值等。除非您编辑代码以使其递归，否则它不会真正深入您的对象或数组，但这似乎不是一个问题就回答问题本身而言。您提供了数字，所以我的答案将适用于数字。

Answer 29

这对我来说很好，但我不明白为什么有些人在数组上使用 concat 方法。

/**
 * 
 * @param {Array} firstArray 
 * @param {Array} secondArray 
 * @returns {boolean}
 */
function arrayIsEqual(firstArray, secondArray)
{
    if (
        !Array.isArray(firstArray) 
        || !Array.isArray(secondArray) 
        || firstArray.length !== secondArray.length
    ) {
        console.log('Bad params!\n');
        return false;
    }
    $bool = firstArray === secondArray || firstArray.every((x,i)=> Array.isArray(x) ? arrayIsEqual(x, secondArray[i]) : x === secondArray[i]);
    console.log($bool);
    return $bool;
}

let a = [1, 2];
let b = [3, 4];
let c = "Me and you";
let d = [1, [2, 3, 5], 4];
let e = [1, [2, 3, 5], 4];
let f = [1, [2, 3, 5], [4]];
let g = [1, [2, 3, 5], [4]];

arrayIsEqual(a, b);
arrayIsEqual(b, c);
arrayIsEqual(c, a);
arrayIsEqual(a, a);
arrayIsEqual(d, e);
arrayIsEqual(e, f);
arrayIsEqual(e, e);
arrayIsEqual(g, f);

输出

false
Bad params!

Bad params!

true
true
true
true
false
true
true
true
true