比较两个 Javascript 数组并删除重复项
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Compare two Javascript Arrays and remove Duplicates
提问by sunleo
It is working well is there any other better way to remove duplicates from one array if it has elements of another array ?.
如果它具有另一个数组的元素，是否还有其他更好的方法可以从一个数组中删除重复项？
<script>
var array1 = new Array("a","b","c","d","e","f");
var array2 = new Array("c","e");
for (var i = 0; i<array2.length; i++) {
var arrlen = array1.length;
for (var j = 0; j<arrlen; j++) {
if (array2[i] == array1[j]) {
array1 = array1.slice(0, j).concat(array1.slice(j+1, arrlen));
}
}
}
alert(array1);
</script>
回答by Aesthete
array1 = array1.filter(function(val) {
return array2.indexOf(val) == -1;
});
Or, with the availability of ES6:
或者，随着 ES6 的推出：
array1 = array1.filter(val => !array2.includes(val));
回答by Benisburgers
The trick, for reasons that are beyond me, is to loop the outer loop downwards (i--) and the inner loop upwards (j++).
出于我无法理解的原因，诀窍是将外循环向下 (i--) 和内循环向上 (j++)。
See the example bellow:
请参阅下面的示例：
function test() {
var array1 = new Array("a","b","c","d","e","f");
var array2 = new Array("c","e");
for (var i = array1.length - 1; i >= 0; i--) {
for (var j = 0; j < array2.length; j++) {
if (array1[i] === array2[j]) {
array1.splice(i, 1);
}
}
}
console.log(array1)
}
How do I know this? See the below:
我怎么知道这个？请参阅以下内容：
for( var i =myArray.length - 1; i>=0; i--){
for( var j=0; j<toRemove.length; j++){
if(myArray[i] === toRemove[j]){
myArray.splice(i, 1);
}
}
}
or
或者
var myArray = [
{name: 'deepak', place: 'bangalore'},
{name: 'chirag', place: 'bangalore'},
{name: 'alok', place: 'berhampur'},
{name: 'chandan', place: 'mumbai'}
];
var toRemove = [
{name: 'deepak', place: 'bangalore'},
{name: 'alok', place: 'berhampur'}
];
for( var i=myArray.length - 1; i>=0; i--){
for( var j=0; j<toRemove.length; j++){
if(myArray[i] && (myArray[i].name === toRemove[j].name)){
myArray.splice(i, 1);
}
}
}
alert(JSON.stringify(myArray));
On that note, would anyone be able to explain why the outer loop needs to be looped downwards (--)?
关于这一点，有人能解释为什么外循环需要向下循环 (--)？
Good luck!
祝你好运！
回答by Flavio
Using the Set.prototype Constructor: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set
使用 Set.prototype 构造函数：https: //developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set
let array1 = Array('a', 'b', 'c', 'd', 'e', 'f')
let array2 = Array('c', 'e', 'g')
let concat = array1.concat(array2) // join arrays => [ 'a', 'b', 'c', 'd', 'e', 'f', 'c', 'e', 'g' ]
// Set will filter out duplicates automatically
let set = new Set(concat) // => Set { 'a', 'b', 'c', 'd', 'e', 'f', 'g' }
// Use spread operator to extend Set to an Array
let result = [...set]
console.log(result) // => [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]
回答by Jijo Paulose
use Array.splice()
用 Array.splice()
var array1 = ['1', '2', '3', '4', '5'];
var array2 = ['4', '5'];
var index;
for (var i=0; i<array2.length; i++) {
index = array1.indexOf(array2[i]);
if (index > -1) {
array1.splice(index, 1);
}
}
回答by Daniel Ortegón
This my solution
这是我的解决方案
array1 = array1.filter(function(val) {
return array2.indexOf(val.toString()) == -1;
});
array1 = array1.filter(function(val) {
return array2.indexOf(val.toString()) == -1;
});
回答by Sachin Jagtap
This is my solution to remove duplicate in ES6.
这是我在 ES6 中删除重复项的解决方案。
let foundDuplicate = false;
existingOptions.some(existingItem => {
result = result.filter(item => {
if (existingItem.value !== item.value) {
return item;
} else {
foundDuplicate = true;
}
});
return foundDuplicate;
});
I used this approach because in my case I was having array of objects and indexOf was having problem with it.
我使用这种方法是因为在我的情况下，我有对象数组，而 indexOf 有问题。
回答by External Purpose
window.onload = function () {
var array1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm'];
var array2 = ['c', 'h', 'k'];
var array3 = [];
var SecondarrayIndexcount = 0;
for (var i = 0; i < array1.length; i++) {
for (var j = 0; j < array2.length; j++) {
if (array1[i] !== array2[j]) {
if (SecondarrayIndexcount === (array2.length - 1)) {
array3.push(array1[i]);
SecondarrayIndexcount = 0;
break;
}
SecondarrayIndexcount++;
}
}
}
for (var i in array3) {
alert(array3[i]);
}
}
</script>
