严格相等运算符将告诉您两个对象类型是否相等。然而,是否有一种方法来判断两个对象是否相等,就像Java中的哈希码值一样?
堆栈溢出问题JavaScript中有hashCode函数吗?类似于这个问题,但需要一个更学术的答案。上面的场景说明了为什么有必要有一个,我想知道是否有等效的解决方案。
严格相等运算符将告诉您两个对象类型是否相等。然而,是否有一种方法来判断两个对象是否相等,就像Java中的哈希码值一样?
堆栈溢出问题JavaScript中有hashCode函数吗?类似于这个问题,但需要一个更学术的答案。上面的场景说明了为什么有必要有一个,我想知道是否有等效的解决方案。
当前回答
这里有很多好的想法!这是我对深度相等的理解。我把它发布在github上,并围绕它写了一些测试。很难涵盖所有可能的情况,有时也没有必要这样做。
我介绍了NaN !== NaN以及循环依赖关系。
https://github.com/ryancat/simple-deep-equal/blob/master/index.js
其他回答
如果两个对象的所有属性都具有相同的值,并且所有嵌套对象和数组都递归地具有相同的值,那么将它们视为相等是很有用的。我也认为以下两个对象是相等的:
var a = {p1: 1};
var b = {p1: 1, p2: undefined};
类似地,数组可以有“缺失”元素和未定义的元素。我也会同样对待它们:
var c = [1, 2];
var d = [1, 2, undefined];
函数:实现等式定义的函数:
function isEqual(a, b) {
if (a === b) {
return true;
}
if (generalType(a) != generalType(b)) {
return false;
}
if (a == b) {
return true;
}
if (typeof a != 'object') {
return false;
}
// null != {}
if (a instanceof Object != b instanceof Object) {
return false;
}
if (a instanceof Date || b instanceof Date) {
if (a instanceof Date != b instanceof Date ||
a.getTime() != b.getTime()) {
return false;
}
}
var allKeys = [].concat(keys(a), keys(b));
uniqueArray(allKeys);
for (var i = 0; i < allKeys.length; i++) {
var prop = allKeys[i];
if (!isEqual(a[prop], b[prop])) {
return false;
}
}
return true;
}
源代码(包括辅助函数,generalType和uniqueArray): 这里是单元测试和测试运行器。
这是我的版本。它正在使用new Object。ES5中引入的keys特性以及+、+和+的想法/测试:
function objectEquals(x, y) { 'use strict'; if (x === null || x === undefined || y === null || y === undefined) { return x === y; } // after this just checking type of one would be enough if (x.constructor !== y.constructor) { return false; } // if they are functions, they should exactly refer to same one (because of closures) if (x instanceof Function) { return x === y; } // if they are regexps, they should exactly refer to same one (it is hard to better equality check on current ES) if (x instanceof RegExp) { return x === y; } if (x === y || x.valueOf() === y.valueOf()) { return true; } if (Array.isArray(x) && x.length !== y.length) { return false; } // if they are dates, they must had equal valueOf if (x instanceof Date) { return false; } // if they are strictly equal, they both need to be object at least if (!(x instanceof Object)) { return false; } if (!(y instanceof Object)) { return false; } // recursive object equality check var p = Object.keys(x); return Object.keys(y).every(function (i) { return p.indexOf(i) !== -1; }) && p.every(function (i) { return objectEquals(x[i], y[i]); }); } /////////////////////////////////////////////////////////////// /// The borrowed tests, run them by clicking "Run code snippet" /////////////////////////////////////////////////////////////// var printResult = function (x) { if (x) { document.write('<div style="color: green;">Passed</div>'); } else { document.write('<div style="color: red;">Failed</div>'); } }; var assert = { isTrue: function (x) { printResult(x); }, isFalse: function (x) { printResult(!x); } } assert.isTrue(objectEquals(null,null)); assert.isFalse(objectEquals(null,undefined)); assert.isFalse(objectEquals(/abc/, /abc/)); assert.isFalse(objectEquals(/abc/, /123/)); var r = /abc/; assert.isTrue(objectEquals(r, r)); assert.isTrue(objectEquals("hi","hi")); assert.isTrue(objectEquals(5,5)); assert.isFalse(objectEquals(5,10)); assert.isTrue(objectEquals([],[])); assert.isTrue(objectEquals([1,2],[1,2])); assert.isFalse(objectEquals([1,2],[2,1])); assert.isFalse(objectEquals([1,2],[1,2,3])); assert.isTrue(objectEquals({},{})); assert.isTrue(objectEquals({a:1,b:2},{a:1,b:2})); assert.isTrue(objectEquals({a:1,b:2},{b:2,a:1})); assert.isFalse(objectEquals({a:1,b:2},{a:1,b:3})); assert.isTrue(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assert.isFalse(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); Object.prototype.equals = function (obj) { return objectEquals(this, obj); }; var assertFalse = assert.isFalse, assertTrue = assert.isTrue; assertFalse({}.equals(null)); assertFalse({}.equals(undefined)); assertTrue("hi".equals("hi")); assertTrue(new Number(5).equals(5)); assertFalse(new Number(5).equals(10)); assertFalse(new Number(1).equals("1")); assertTrue([].equals([])); assertTrue([1,2].equals([1,2])); assertFalse([1,2].equals([2,1])); assertFalse([1,2].equals([1,2,3])); assertTrue(new Date("2011-03-31").equals(new Date("2011-03-31"))); assertFalse(new Date("2011-03-31").equals(new Date("1970-01-01"))); assertTrue({}.equals({})); assertTrue({a:1,b:2}.equals({a:1,b:2})); assertTrue({a:1,b:2}.equals({b:2,a:1})); assertFalse({a:1,b:2}.equals({a:1,b:3})); assertTrue({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assertFalse({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); var a = {a: 'text', b:[0,1]}; var b = {a: 'text', b:[0,1]}; var c = {a: 'text', b: 0}; var d = {a: 'text', b: false}; var e = {a: 'text', b:[1,0]}; var i = { a: 'text', c: { b: [1, 0] } }; var j = { a: 'text', c: { b: [1, 0] } }; var k = {a: 'text', b: null}; var l = {a: 'text', b: undefined}; assertTrue(a.equals(b)); assertFalse(a.equals(c)); assertFalse(c.equals(d)); assertFalse(a.equals(e)); assertTrue(i.equals(j)); assertFalse(d.equals(k)); assertFalse(k.equals(l)); // from comments on stackoverflow post assert.isFalse(objectEquals([1, 2, undefined], [1, 2])); assert.isFalse(objectEquals([1, 2, 3], { 0: 1, 1: 2, 2: 3 })); assert.isFalse(objectEquals(new Date(1234), 1234)); // no two different function is equal really, they capture their context variables // so even if they have same toString(), they won't have same functionality var func = function (x) { return true; }; var func2 = function (x) { return true; }; assert.isTrue(objectEquals(func, func)); assert.isFalse(objectEquals(func, func2)); assert.isTrue(objectEquals({ a: { b: func } }, { a: { b: func } })); assert.isFalse(objectEquals({ a: { b: func } }, { a: { b: func2 } }));
只是想利用一些es6的特性来贡献我的对象比较版本。它不考虑订单。在将所有if/else转换为三元后,我带来了以下内容:
function areEqual(obj1, obj2) {
return Object.keys(obj1).every(key => {
return obj2.hasOwnProperty(key) ?
typeof obj1[key] === 'object' ?
areEqual(obj1[key], obj2[key]) :
obj1[key] === obj2[key] :
false;
}
)
}
我对这个函数做如下假设:
你控制你比较的对象,你只有基本的值(例如。而不是嵌套的对象、函数等)。 您的浏览器支持Object.keys。
这应该被视为一个简单策略的示范。
/**
* Checks the equality of two objects that contain primitive values. (ie. no nested objects, functions, etc.)
* @param {Object} object1
* @param {Object} object2
* @param {Boolean} [order_matters] Affects the return value of unordered objects. (ex. {a:1, b:2} and {b:2, a:1}).
* @returns {Boolean}
*/
function isEqual( object1, object2, order_matters ) {
var keys1 = Object.keys(object1),
keys2 = Object.keys(object2),
i, key;
// Test 1: Same number of elements
if( keys1.length != keys2.length ) {
return false;
}
// If order doesn't matter isEqual({a:2, b:1}, {b:1, a:2}) should return true.
// keys1 = Object.keys({a:2, b:1}) = ["a","b"];
// keys2 = Object.keys({b:1, a:2}) = ["b","a"];
// This is why we are sorting keys1 and keys2.
if( !order_matters ) {
keys1.sort();
keys2.sort();
}
// Test 2: Same keys
for( i = 0; i < keys1.length; i++ ) {
if( keys1[i] != keys2[i] ) {
return false;
}
}
// Test 3: Values
for( i = 0; i < keys1.length; i++ ) {
key = keys1[i];
if( object1[key] != object2[key] ) {
return false;
}
}
return true;
}
这取决于你对平等的定义。因此,作为类的开发人员,要由您来定义它们的相等性。
有时会使用一种情况,如果两个实例指向内存中的相同位置,则认为它们是“相等的”,但这并不总是您想要的。例如,如果我有一个Person类,如果两个Person对象具有相同的Last Name、First Name和Social Security Number(即使它们指向内存中的不同位置),我可能会认为它们是“相等的”。
另一方面,我们不能简单地说两个对象是相等的,如果它们的每个成员的值都相同,因为,有时,你并不想这样。换句话说,对于每个类,由类开发人员定义组成对象“标识”的成员并开发适当的相等操作符(通过重载==操作符或Equals方法)。
Saying that two objects are equal if they have the same hash is one way out. However you then have to wonder how the hash is calculated for each instance. Going back to the Person example above, we could use this system if the hash was calculated by looking at the values of the First Name, Last Name, and Social Security Number fields. On top of that, we are then relying on the quality of the hashing method (that's a huge topic on its own, but suffice it to say that not all hashes are created equal, and bad hashing methods can lead to more collisions, which in this case would return false matches).