我想在JavaScript代码中比较2个对象数组。这些对象有8个属性，但是每个对象都没有每个对象的值，并且数组永远不会超过8个项目，所以也许是通过遍历每个对象然后查看值的强力方法。 8个属性是我想做的最简单的方法，但在实现之前，我想看看是否有人有更优雅的解决方案。有什么想法吗？
答案 0 :(得分：42)
编辑：你不能在当前常见的基于浏览器的JavaScript解释器实现中重载运算符。
要回答原始问题，一种方法可以做到这一点，并且请注意，这只是一个黑客攻击，只需serialize the two arrays to JSON然后比较两个JSON字符串。这只会告诉你阵列是否不同，显然你可以这样做数组中对象的每个，看看哪些是不同的。
另一种选择是使用一个具有一些很好的设施来比较对象的库 - 我使用并推荐MochiKit。
编辑： The answer kamens gave也值得考虑，因为比较两个给定对象的单个函数比我建议的任何库要小得多（尽管我的建议肯定会有效）好吧）。
这是一个天真的实现，可能只对你有用 - 请注意这个实现存在潜在的问题：
function objectsAreSame(x, y) {
var objectsAreSame = true;
for(var propertyName in x) {
if(x[propertyName] !== y[propertyName]) {
objectsAreSame = false;
break;
}
}
return objectsAreSame;
}
假设两个对象具有完全相同的属性列表。
哦，很明显，无论好坏，我都属于唯一的回归阵营。 ：）
答案 1 :(得分：17)
我知道这是一个老问题，提供的答案工作正常......但这有点短，不需要任何额外的库（即JSON）：
function arraysAreEqual(ary1,ary2){
return (ary1.join('') == ary2.join(''));
}
答案 2 :(得分：14)
老实说，每个对象最多8个对象，最多8个属性，最好的办法就是遍历每个对象并直接进行比较。它会很快，而且很容易。
如果您经常使用这些类型的比较，那么我同意Jason关于JSON序列化...但是否则无需使用新库或JSON序列化代码减慢您的应用程序。
答案 3 :(得分：10)
我在一个简单的算法上做了一些工作来比较两个对象的内容并返回一个可理解的差异列表。以为我会分享。它借用了jQuery的一些想法，即map函数实现以及对象和数组类型检查。
它返回一个“diff objects”列表，它是带有diff信息的数组。这很简单。
这是：
// compare contents of two objects and return a list of differences
// returns an array where each element is also an array in the form:
// [accessor, diffType, leftValue, rightValue ]
//
// diffType is one of the following:
// value: when primitive values at that index are different
// undefined: when values in that index exist in one object but don't in
// another; one of the values is always undefined
// null: when a value in that index is null or undefined; values are
// expressed as boolean values, indicated wheter they were nulls
// type: when values in that index are of different types; values are
// expressed as types
// length: when arrays in that index are of different length; values are
// the lengths of the arrays
//
function DiffObjects(o1, o2) {
// choose a map() impl.
// you may use $.map from jQuery if you wish
var map = Array.prototype.map?
function(a) { return Array.prototype.map.apply(a, Array.prototype.slice.call(arguments, 1)); } :
function(a, f) {
var ret = new Array(a.length), value;
for ( var i = 0, length = a.length; i < length; i++ )
ret[i] = f(a[i], i);
return ret.concat();
};
// shorthand for push impl.
var push = Array.prototype.push;
// check for null/undefined values
if ((o1 == null) || (o2 == null)) {
if (o1 != o2)
return [["", "null", o1!=null, o2!=null]];
return undefined; // both null
}
// compare types
if ((o1.constructor != o2.constructor) ||
(typeof o1 != typeof o2)) {
return [["", "type", Object.prototype.toString.call(o1), Object.prototype.toString.call(o2) ]]; // different type
}
// compare arrays
if (Object.prototype.toString.call(o1) == "[object Array]") {
if (o1.length != o2.length) {
return [["", "length", o1.length, o2.length]]; // different length
}
var diff =[];
for (var i=0; i<o1.length; i++) {
// per element nested diff
var innerDiff = DiffObjects(o1[i], o2[i]);
if (innerDiff) { // o1[i] != o2[i]
// merge diff array into parent's while including parent object name ([i])
push.apply(diff, map(innerDiff, function(o, j) { o[0]="[" + i + "]" + o[0]; return o; }));
}
}
// if any differences were found, return them
if (diff.length)
return diff;
// return nothing if arrays equal
return undefined;
}
// compare object trees
if (Object.prototype.toString.call(o1) == "[object Object]") {
var diff =[];
// check all props in o1
for (var prop in o1) {
// the double check in o1 is because in V8 objects remember keys set to undefined
if ((typeof o2[prop] == "undefined") && (typeof o1[prop] != "undefined")) {
// prop exists in o1 but not in o2
diff.push(["[" + prop + "]", "undefined", o1[prop], undefined]); // prop exists in o1 but not in o2
}
else {
// per element nested diff
var innerDiff = DiffObjects(o1[prop], o2[prop]);
if (innerDiff) { // o1[prop] != o2[prop]
// merge diff array into parent's while including parent object name ([prop])
push.apply(diff, map(innerDiff, function(o, j) { o[0]="[" + prop + "]" + o[0]; return o; }));
}
}
}
for (var prop in o2) {
// the double check in o2 is because in V8 objects remember keys set to undefined
if ((typeof o1[prop] == "undefined") && (typeof o2[prop] != "undefined")) {
// prop exists in o2 but not in o1
diff.push(["[" + prop + "]", "undefined", undefined, o2[prop]]); // prop exists in o2 but not in o1
}
}
// if any differences were found, return them
if (diff.length)
return diff;
// return nothing if objects equal
return undefined;
}
// if same type and not null or objects or arrays
// perform primitive value comparison
if (o1 != o2)
return [["", "value", o1, o2]];
// return nothing if values are equal
return undefined;
}
答案 4 :(得分：6)
我尝试了JSON.stringify()并为我工作。
let array1 = [1,2,{value:'alpha'}] , array2 = [{value:'alpha'},'music',3,4];
JSON.stringify(array1) // "[1,2,{"value":"alpha"}]"
JSON.stringify(array2) // "[{"value":"alpha"},"music",3,4]"
JSON.stringify(array1) === JSON.stringify(array2); // false
答案 5 :(得分：2)
由于序列化通常不起作用（仅当属性顺序匹配：JSON.stringify({a:1,b:2}) !== JSON.stringify({b:2,a:1})时），您必须检查属性计数并同时比较每个属性：
const objectsEqual = (o1, o2) =>
Object.keys(o1).length === Object.keys(o2).length
&& Object.keys(o1).every(p => o1[p] === o2[p]);
const obj1 = { name: 'John', age: 33};
const obj2 = { age: 33, name: 'John' };
const obj3 = { name: 'John', age: 45 };
console.log(objectsEqual(obj1, obj2)); // true
console.log(objectsEqual(obj1, obj3)); // false
如果需要深入比较，可以递归调用该函数：
const obj1 = { name: 'John', age: 33, info: { married: true, hobbies: ['sport', 'art'] } };
const obj2 = { age: 33, name: 'John', info: { hobbies: ['sport', 'art'], married: true } };
const obj3 = { name: 'John', age: 33 };
const objectsEqual = (o1, o2) =>
typeof o1 === 'object' && Object.keys(o1).length > 0
? Object.keys(o1).length === Object.keys(o2).length
&& Object.keys(o1).every(p => objectsEqual(o1[p], o2[p]))
: o1 === o2;
console.log(objectsEqual(obj1, obj2)); // true
console.log(objectsEqual(obj1, obj3)); // false
然后很容易使用此函数比较数组中的对象：
const arr1 = [obj1, obj1];
const arr2 = [obj1, obj2];
const arr3 = [obj1, obj3];
const arraysEqual = (a1, a2) =>
a1.length === a2.length && a1.every((o, idx) => objectsEqual(o, a2[idx]));
console.log(arraysEqual(arr1, arr2)); // true
console.log(arraysEqual(arr1, arr3)); // false
答案 6 :(得分：2)
当函数需要等于空数组（在这种情况下返回false）时，有一个优化的代码
const objectsEqual = (o1, o2) => {
if (o2 === null && o1 !== null) return false;
return o1 !== null && typeof o1 === 'object' && Object.keys(o1).length > 0 ?
Object.keys(o1).length === Object.keys(o2).length &&
Object.keys(o1).every(p => objectsEqual(o1[p], o2[p]))
: (o1 !== null && Array.isArray(o1) && Array.isArray(o2) && !o1.length &&
!o2.length) ? true : o1 === o2;
}
答案 7 :(得分：1)
@ JasonBunting的回答中提到的objectsAreSame函数对我来说很好。但是，有一点问题：如果x[propertyName]和y[propertyName]是对象（typeof x[propertyName] == 'object'），则需要递归调用该函数以进行比较。
答案 8 :(得分：1)
请尝试这个：
function used_to_compare_two_arrays(a, b)
{
// This block will make the array of indexed that array b contains a elements
var c = a.filter(function(value, index, obj) {
return b.indexOf(value) > -1;
});
// This is used for making comparison that both have same length if no condition go wrong
if (c.length !== a.length) {
return 0;
} else{
return 1;
}
}
答案 9 :(得分：1)
以下是我的尝试，使用Node's assert module + npm package object-hash。
我想你想检查两个数组是否包含相同的对象，即使这些对象在两个数组之间的排序方式不同。
var assert = require('assert');
var hash = require('object-hash');
var obj1 = {a: 1, b: 2, c: 333},
obj2 = {b: 2, a: 1, c: 444},
obj3 = {b: "AAA", c: 555},
obj4 = {c: 555, b: "AAA"};
var array1 = [obj1, obj2, obj3, obj4];
var array2 = [obj3, obj2, obj4, obj1]; // [obj3, obj3, obj2, obj1] should work as well
// calling assert.deepEquals(array1, array2) at this point FAILS (throws an AssertionError)
// even if array1 and array2 contain the same objects in different order,
// because array1[0].c !== array2[0].c
// sort objects in arrays by their hashes, so that if the arrays are identical,
// their objects can be compared in the same order, one by one
var array1 = sortArrayOnHash(array1);
var array2 = sortArrayOnHash(array2);
// then, this should output "PASS"
try {
assert.deepEqual(array1, array2);
console.log("PASS");
} catch (e) {
console.log("FAIL");
console.log(e);
}
// You could define as well something like Array.prototype.sortOnHash()...
function sortArrayOnHash(array) {
return array.sort(function(a, b) {
return hash(a) > hash(b);
});
}
答案 10 :(得分：1)
使用lodash中的_.some：https://lodash.com/docs/4.17.11#some
const array1AndArray2NotEqual =
_.some(array1, (a1, idx) => a1.key1 !== array2[idx].key1
|| a1.key2 !== array2[idx].key2
|| a1.key3 !== array2[idx].key3);
答案 11 :(得分：1)
我的实践实施、测试和工作。
const obj1 = { name: 'John', age: 33};
const obj2 = { age: 33, name: 'John' };
const obj3 = { name: 'John', age: 45 };
const equalObjs = ( obj1, obj2 ) => {
let keyExist = false;
for ( const [key, value] of Object.entries(obj1) ) {
// Search each key in reference object and attach a callback function to
// compare the two object keys
if( Object.keys(obj2).some( ( e ) => e == key ) ) {
keyExist = true;
}
}
return keyExist;
}
console.info( equalObjs( obj1, obj2 ) );
比较你的数组
const equalArrays = ( arr1, arr2 ) => {
// If the arrays are different length we an eliminate immediately
if( arr1.length !== arr2.length ) {
return false;
} else if ( arr1.every(( obj, index ) => equalObjs( obj, arr2[index] ) ) ) {
return true;
}
}
console.info( equalArrays( arr1, arr2 ) );
答案 12 :(得分：0)
与 json 相比非常糟糕。试试这个包来比较嵌套数组并得到差异。
<块引用>答案 13 :(得分：0)
这是我的解决方案。它将比较也有对象和数组的数组。元素可以停留在任何位置。 示例：
const array1 = [{a: 1}, {b: 2}, { c: 0, d: { e: 1, f: 2, } }, [1,2,3,54]];
const array2 = [{a: 1}, {b: 2}, { c: 0, d: { e: 1, f: 2, } }, [1,2,3,54]];
const arraysCompare = (a1, a2) => {
if (a1.length !== a2.length) return false;
const objectIteration = (object) => {
const result = [];
const objectReduce = (obj) => {
for (let i in obj) {
if (typeof obj[i] !== 'object') {
result.push(`${i}${obj[i]}`);
} else {
objectReduce(obj[i]);
}
}
};
objectReduce(object);
return result;
};
const reduceArray1 = a1.map(item => {
if (typeof item !== 'object') return item;
return objectIteration(item).join('');
});
const reduceArray2 = a2.map(item => {
if (typeof item !== 'object') return item;
return objectIteration(item).join('');
});
const compare = reduceArray1.map(item => reduceArray2.includes(item));
return compare.reduce((acc, item) => acc + Number(item)) === a1.length;
};
console.log(arraysCompare(array1, array2));