var a = [1, 2, 3];
var b = [3, 2, 1];
var c = new Array(1, 2, 3);
alert(a == b + "|" + b == c);
demo
如何检查这些数组是否相等,并获得一个方法,如果它们相等,则返回true ?
jQuery是否为此提供了任何方法?
var a = [1, 2, 3];
var b = [3, 2, 1];
var c = new Array(1, 2, 3);
alert(a == b + "|" + b == c);
demo
如何检查这些数组是否相等,并获得一个方法,如果它们相等,则返回true ?
jQuery是否为此提供了任何方法?
当前回答
如果您希望检查对象数组的相等性和顺序并不重要,即。
areEqual ([{id:“0”},{id: " 1 "}], [{id: " 1 "}, {id: " 0 "})) / /真实的
首先要对数组进行排序。lodash有你需要的所有工具,通过组合sortBy和isEqual:
// arr1 & arr2: Arrays of objects
// sortProperty: the property of the object with which you want to sort
// Note: ensure every object in both arrays has your chosen sortProperty
// For example, arr1 = [{id: "v-test_id0"}, {id: "v-test_id1"}]
// and arr2 = [{id: "v-test_id1"}, {id: "v-test_id0"}]
// sortProperty should be 'id'
function areEqual (arr1, arr2, sortProperty) {
return _.areEqual(_.sortBy(arr1, sortProperty), _.sortBy(arr2, sortProperty))
}
编辑:由于sortBy返回一个新数组,在排序之前不需要克隆数组。原始数组不会被改变。
注意,对于lodash的isEqual,顺序很重要。如果sortBy没有首先应用于每个数组,上面的例子将返回false。
其他回答
在JavaScript 1.6版中,这很简单:
Array.prototype.equals = function( array ) {
return this.length == array.length &&
this.every( function(this_i,i) { return this_i == array[i] } )
}
例如,[].equals([])给出true,而[1,2,3]。Equals([1,3,2])产生false。
如果您正在使用lodash,并且不想修改任何一个数组,则可以使用_.xor()函数。它将两个数组作为集合进行比较,并返回包含它们之差的集合。如果此差值的长度为零,则两个数组本质上相等:
var a = [1, 2, 3];
var b = [3, 2, 1];
var c = new Array(1, 2, 3);
_.xor(a, b).length === 0
true
_.xor(b, c).length === 0
true
使用map()和reduce():
function arraysEqual (a1, a2) {
return a1 === a2 || (
a1 !== null && a2 !== null &&
a1.length === a2.length &&
a1
.map(function (val, idx) { return val === a2[idx]; })
.reduce(function (prev, cur) { return prev && cur; }, true)
);
}
它处理所有可能的东西,甚至在对象的结构中引用自己。您可以在代码的末尾看到示例。
var deepCompare = (function() {
function internalDeepCompare (obj1, obj2, objects) {
var i, objPair;
if (obj1 === obj2) {
return true;
}
i = objects.length;
while (i--) {
objPair = objects[i];
if ( (objPair.obj1 === obj1 && objPair.obj2 === obj2) ||
(objPair.obj1 === obj2 && objPair.obj2 === obj1) ) {
return true;
}
}
objects.push({obj1: obj1, obj2: obj2});
if (obj1 instanceof Array) {
if (!(obj2 instanceof Array)) {
return false;
}
i = obj1.length;
if (i !== obj2.length) {
return false;
}
while (i--) {
if (!internalDeepCompare(obj1[i], obj2[i], objects)) {
return false;
}
}
}
else {
switch (typeof obj1) {
case "object":
// deal with null
if (!(obj2 && obj1.constructor === obj2.constructor)) {
return false;
}
if (obj1 instanceof RegExp) {
if (!(obj2 instanceof RegExp && obj1.source === obj2.source)) {
return false;
}
}
else if (obj1 instanceof Date) {
if (!(obj2 instanceof Date && obj1.getTime() === obj2.getTime())) {
return false;
}
}
else {
for (i in obj1) {
if (obj1.hasOwnProperty(i)) {
if (!(obj2.hasOwnProperty(i) && internalDeepCompare(obj1[i], obj2[i], objects))) {
return false;
}
}
}
}
break;
case "function":
if (!(typeof obj2 === "function" && obj1+"" === obj2+"")) {
return false;
}
break;
default: //deal with NaN
if (obj1 !== obj2 && obj1 === obj1 && obj2 === obj2) {
return false;
}
}
}
return true;
}
return function (obj1, obj2) {
return internalDeepCompare(obj1, obj2, []);
};
}());
/*
var a = [a, undefined, new Date(10), /.+/, {a:2}, function(){}, Infinity, -Infinity, NaN, 0, -0, 1, [4,5], "1", "-1", "a", null],
b = [b, undefined, new Date(10), /.+/, {a:2}, function(){}, Infinity, -Infinity, NaN, 0, -0, 1, [4,5], "1", "-1", "a", null];
deepCompare(a, b);
*/
这是你应该做的。请不要使用stringify或< >。
function arraysEqual(a, b) {
if (a === b) return true;
if (a == null || b == null) return false;
if (a.length !== b.length) return false;
// If you don't care about the order of the elements inside
// the array, you should sort both arrays here.
// Please note that calling sort on an array will modify that array.
// you might want to clone your array first.
for (var i = 0; i < a.length; ++i) {
if (a[i] !== b[i]) return false;
}
return true;
}