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
我写这个方法只是为了确保数组和对象都能被清晰地比较。
这应该也能做到!:)
public class Objects {
/**
* Checks whether a value is of type Object
* @param value the value
*/
public static isObject = (value: any): boolean => {
return value === Object(value) && Object.prototype.toString.call(value) !== '[object Array]'
}
/**
* Checks whether a value is of type Array
* @param value the value
*/
public static isArray = (value: any): boolean => {
return Object.prototype.toString.call(value) === '[object Array]' && !Objects.isObject(value)
}
/**
* Check whether two values are equal
*/
public static isEqual = (objectA: any, objectB: any) => {
// Objects
if (Objects.isObject(objectA) && !Objects.isObject(objectB)) {
return false
}
else if (!Objects.isObject(objectA) && Objects.isObject(objectB)) {
return false
}
// Arrays
else if (Objects.isArray(objectA) && !Objects.isArray(objectB)) {
return false
}
else if (!Objects.isArray(objectA) && Objects.isArray(objectB)) {
return false
}
// Primitives
else if (!Objects.isArray(objectA) && !Objects.isObject(objectA)) {
return objectA === objectB
}
// Object or array
else {
const compareObject = (objectA: any, objectB: any): boolean => {
if (Object.keys(objectA).length !== Object.keys(objectB).length) return false
for (const propertyName of Object.keys(objectA)) {
const valueA = objectA[propertyName]
const valueB = objectB[propertyName]
if (!Objects.isEqual(valueA, valueB)) {
return false
}
}
return true
}
const compareArray = (arrayA: any[], arrayB: any[]): boolean => {
if (arrayA.length !== arrayB.length) return false
for (const index in arrayA) {
const valueA = arrayA[index]
const valueB = arrayB[index]
if (!Objects.isEqual(valueA, valueB)) {
return false
}
}
return true
}
if (Objects.isObject(objectA)) {
return compareObject(objectA, objectB)
} else {
return compareArray(objectA, objectB)
}
}
}
}
其他回答
这是一个非常干净的CoffeeScript版本,你可以这样做:
Object::equals = (other) ->
typeOf = Object::toString
return false if typeOf.call(this) isnt typeOf.call(other)
return `this == other` unless typeOf.call(other) is '[object Object]' or
typeOf.call(other) is '[object Array]'
(return false unless this[key].equals other[key]) for key, value of this
(return false if typeof this[key] is 'undefined') for key of other
true
下面是测试:
describe "equals", ->
it "should consider two numbers to be equal", ->
assert 5.equals(5)
it "should consider two empty objects to be equal", ->
assert {}.equals({})
it "should consider two objects with one key to be equal", ->
assert {a: "banana"}.equals {a: "banana"}
it "should consider two objects with keys in different orders to be equal", ->
assert {a: "banana", kendall: "garrus"}.equals {kendall: "garrus", a: "banana"}
it "should consider two objects with nested objects to be equal", ->
assert {a: {fruit: "banana"}}.equals {a: {fruit: "banana"}}
it "should consider two objects with nested objects that are jumbled to be equal", ->
assert {a: {a: "banana", kendall: "garrus"}}.equals {a: {kendall: "garrus", a: "banana"}}
it "should consider two objects with arrays as values to be equal", ->
assert {a: ["apple", "banana"]}.equals {a: ["apple", "banana"]}
it "should not consider an object to be equal to null", ->
assert !({a: "banana"}.equals null)
it "should not consider two objects with different keys to be equal", ->
assert !({a: "banana"}.equals {})
it "should not consider two objects with different values to be equal", ->
assert !({a: "banana"}.equals {a: "grapefruit"})
排序对象(字典) 比较JSON字符串 函数areTwoDictsEqual(dictA, dictB) { 函数sortDict(dict) { var keys = Object.keys(dict); keys.sort (); var newDict = {}; For (var i=0;我< keys.length;我+ +){ Var key = keys[i]; Var值= dict[key]; newDict[key] = value; } 返回newDict; } 返回JSON.stringify(sortDict(dictA)) == JSON.stringify(sortDict(dictB)); }
这是对以上所有内容的补充,而不是替代。如果需要快速浅比较对象,而不需要检查额外的递归情况。这是一个镜头。
这比较:1)自己的属性数量相等,2)键名相等,3)如果bCompareValues == true,对应的属性值及其类型相等(三重相等)
var shallowCompareObjects = function(o1, o2, bCompareValues) {
var s,
n1 = 0,
n2 = 0,
b = true;
for (s in o1) { n1 ++; }
for (s in o2) {
if (!o1.hasOwnProperty(s)) {
b = false;
break;
}
if (bCompareValues && o1[s] !== o2[s]) {
b = false;
break;
}
n2 ++;
}
return b && n1 == n2;
}
为了比较简单的键/值对对象实例的键,我使用:
function compareKeys(r1, r2) {
var nloops = 0, score = 0;
for(k1 in r1) {
for(k2 in r2) {
nloops++;
if(k1 == k2)
score++;
}
}
return nloops == (score * score);
};
一旦比较了键,一个简单的for. in循环就足够了。
复杂度是O(N*N), N是键的个数。
我希望/猜测我定义的对象不会拥有超过1000个属性…
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。