Question

如果它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>

Answer 1

array1 = array1.filter(function(val) {
  return array2.indexOf(val) == -1;
});

或者，随着ES6的可用性：

array1 = array1.filter(val => !array2.includes(val));

filter() reference here

indexOf() reference here

includes() reference here

Answer 2

使用$SOLR_HOME/logs

Array.splice()

Answer 3

由于超出我的原因，诀窍是将外循环向下循环（i--）并将内循环向上循环（j ++）。

见下面的例子：

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)
  }

我怎么知道这个？见下文：

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);
    }
  }
}

或

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));

就此而言，是否有人能够解释为什么外环需要向下循环（ - ）？

祝你好运！

Answer 4

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' ]

Answer 5

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>

Answer 6

这是我的解决方法

array1 = array1.filter(function(val) { return array2.indexOf(val.toString()) == -1; });

Answer 7

这是我的解决方案，用于删除ES6中的重复项。

let foundDuplicate = false;
existingOptions.some(existingItem => {
  result = result.filter(item => {
    if (existingItem.value !== item.value) {
      return item;
    } else {
      foundDuplicate = true;
    }
  });
  return foundDuplicate;
});

之所以使用这种方法，是因为在我的情况下，我有对象数组，而indexOf却有问题。

Answer 8

arr1 = arr1.filter（val =>！arr2.includes（val））;

比较两个Javascript数组并删除重复项

8 个答案: