我需要检查一个JavaScript数组，看看是否有任何重复的值。最简单的方法是什么?我只需要找到复制的值是什么——实际上我不需要它们的索引，也不需要它们复制了多少次。
我知道我可以循环遍历数组并检查所有其他值是否匹配，但似乎应该有更简单的方法。
类似的问题:
获取JavaScript数组中所有唯一的值(删除重复值)
当前回答
http://jsfiddle.net/vol7ron/gfJ28/
var arr = ['hello','goodbye','foo','hello','foo','bar',1,2,3,4,5,6,7,8,9,0,1,2,3];
var hash = [];
// build hash
for (var n=arr.length; n--; ){
if (typeof hash[arr[n]] === 'undefined') hash[arr[n]] = [];
hash[arr[n]].push(n);
}
// work with compiled hash (not necessary)
var duplicates = [];
for (var key in hash){
if (hash.hasOwnProperty(key) && hash[key].length > 1){
duplicates.push(key);
}
}
alert(duplicates);
The result will be the hash array, which will contain both a unique set of values and the position of those values. So if there are 2 or more positions, we can determine that the value has a duplicate. Thus, every place hash[<value>].length > 1, signifies a duplicate. hash['hello'] will return [0,3] because 'hello' was found in node 0 and 3 in arr[]. Note: the length of [0,3] is what's used to determine if it was a duplicate. Using for(var key in hash){ if (hash.hasOwnProperty(key)){ alert(key); } } will alert each unique value.
其他回答
简单的一行方式
Var arr = [9,1,2,4,3,4,9] console.log(arr.filter((ele,indx)=>indx!==arr.indexOf(ele))) //获取副本 console.log(arr.filter((ele,indx)=>indx===arr.indexOf(ele))) //删除重复项
你可以使用下面的代码来获取给定数组中的重复元素:
let name = ['satya', 'amit', 'aditya', 'abhay', 'satya', 'amit'];
let dup = [];
let uniq = [];
name.forEach((item, index) => {
if(!uniq.includes(item)) {
uniq[index] = item;
}
if (name.indexOf(item, index + 1) != -1) {
dup[index] = item;
}
})
打印重复值
var arr = [1,2,3,4,13,2,3,4,3,4]; // non_unique Printing function nonUnique(arr){ var result = []; for(var i =0;i<arr.length;i++){ if(arr.indexOf(arr[i],i+1) > -1){ result.push(arr[i]); } } console.log(result); }nonUnique(arr); // unique Printing function uniqueDuplicateVal(arr){ var result = []; for(var i =0;i<arr.length;i++){ if(arr.indexOf(arr[i],i+1) > -1){ if(result.indexOf(arr[i]) === -1]){ result.push(arr[i]); } } } } uniqueDuplicateVal(arr)
排名较高的答案有一些固有的问题，包括使用遗留的javascript，不正确的排序或只支持2个重复的项目。
这里有一个解决这些问题的现代解决方案:
const arrayNonUniq = array => {
if (!Array.isArray(array)) {
throw new TypeError("An array must be provided!")
}
return array.filter((value, index) => array.indexOf(value) === index && array.lastIndexOf(value) !== index)
}
arrayNonUniq([1, 1, 2, 3, 3])
//=> [1, 3]
arrayNonUniq(["foo", "foo", "bar", "foo"])
//=> ['foo']
你也可以使用npm包array-non-uniq。
这是我的建议(ES6):
let a = [1, 2, 3, 4, 2, 2, 4, 1, 5, 6]
let b = [...new Set(a.sort().filter((o, i) => o !== undefined && a[i + 1] !== undefined && o === a[i + 1]))]
// b is now [1, 2, 4]
