如何在数组中获得唯一值的列表?我总是必须使用第二个数组，或者在JavaScript中有类似于java的hashmap的东西吗?
我将只使用JavaScript和jQuery。不能使用其他库。
当前回答
使用jQuery，这是一个数组唯一的函数我做:
Array.prototype.unique = function () {
var arr = this;
return $.grep(arr, function (v, i) {
return $.inArray(v, arr) === i;
});
}
console.log([1,2,3,1,2,3].unique()); // [1,2,3]
其他回答
使用第二阵列的短而甜的解决方案;
var axes2=[1,4,5,2,3,1,2,3,4,5,1,3,4];
var distinct_axes2=[];
for(var i=0;i<axes2.length;i++)
{
var str=axes2[i];
if(distinct_axes2.indexOf(str)==-1)
{
distinct_axes2.push(str);
}
}
console.log("distinct_axes2 : "+distinct_axes2); // distinct_axes2 : 1,4,5,2,3
下面是一个可定制的equals函数的方法，它可以用于原语以及自定义对象:
Array.prototype.pushUnique = function(element, equalsPredicate = (l, r) => l == r) {
let res = !this.find(item => equalsPredicate(item, element))
if(res){
this.push(element)
}
return res
}
用法:
//with custom equals for objects
myArrayWithObjects.pushUnique(myObject, (left, right) => left.id == right.id)
//with default equals for primitives
myArrayWithPrimitives.pushUnique(somePrimitive)
你只需要香草JS找到唯一的数组。some和Array.reduce。在ES2015语法中，它只有62个字符。
a.reduce((c, v) => b.some(w => w === v) ? c : c.concat(v)), b)
数组中。some和Array。IE9+和其他浏览器支持。只需要在不支持ES2015语法的浏览器中修改常规函数的胖箭头函数即可。
var a = [1,2,3];
var b = [4,5,6];
// .reduce can return a subset or superset
var uniques = a.reduce(function(c, v){
// .some stops on the first time the function returns true
return (b.some(function(w){ return w === v; }) ?
// if there's a match, return the array "c"
c :
// if there's no match, then add to the end and return the entire array
c.concat(v)}),
// the second param in .reduce is the starting variable. This is will be "c" the first time it runs.
b);
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
我在想我们能不能用线性搜索来消除重复项
JavaScript:
function getUniqueRadios() {
var x=document.getElementById("QnA");
var ansArray = new Array();
var prev;
for (var i=0;i<x.length;i++)
{
// Check for unique radio button group
if (x.elements[i].type == "radio")
{
// For the first element prev will be null, hence push it into array and set the prev var.
if (prev == null)
{
prev = x.elements[i].name;
ansArray.push(x.elements[i].name);
} else {
// We will only push the next radio element if its not identical to previous.
if (prev != x.elements[i].name)
{
prev = x.elements[i].name;
ansArray.push(x.elements[i].name);
}
}
}
}
alert(ansArray);
}
HTML:
<body>
<form name="QnA" action="" method='post' ">
<input type="radio" name="g1" value="ANSTYPE1"> good </input>
<input type="radio" name="g1" value="ANSTYPE2"> avg </input>
<input type="radio" name="g2" value="ANSTYPE3"> Type1 </input>
<input type="radio" name="g2" value="ANSTYPE2"> Type2 </input>
<input type="submit" value='SUBMIT' onClick="javascript:getUniqueRadios()"></input>
</form>
</body>
我在纯JS中尝试过这个问题。 我遵循了以下步骤1。对给定数组进行排序，2。遍历排序数组，3。用当前值验证前一个值和下一个值
// JS
var inpArr = [1, 5, 5, 4, 3, 3, 2, 2, 2,2, 100, 100, -1];
//sort the given array
inpArr.sort(function(a, b){
return a-b;
});
var finalArr = [];
//loop through the inpArr
for(var i=0; i<inpArr.length; i++){
//check previous and next value
if(inpArr[i-1]!=inpArr[i] && inpArr[i] != inpArr[i+1]){
finalArr.push(inpArr[i]);
}
}
console.log(finalArr);
Demo
