过滤数组以显示唯一值javascript

问题描述

我有一个如下所示的数组：

I have an array which looks like this:

["A", "A", "A", "A", "B", "B", "B", "C", "C", "D", "A", "A", "B", "B", "B"]

如何过滤它，因此，一次只有一个唯一值，如下所示：

How can I filter it, so there is only one unique value at a time, like this:

['A', 'B', 'C', 'D', 'A', 'B']

什么我试过了：



我尝试了两种不同的方法，我决定先使用for循环

What I have tried:

I have tried two different approaches, I decided to use a for loop first

function uniqueInOrder(arg) {
  var arr = [];
   arg.split('');
  for(var i = 0; i < arg.length; i++){
    if(arg[i] === arg[i+1]) {arr.push(arg[i])}
    
  }
 return arr
}

但据我所知，我需要以某种方式禁用它，如果它已经找到了副本，因为我得到了这个输出：

[A， A，A，B，B，C，A，B，B]



此外，我试过这个方法：

But as far as I understand, I would need to disable it somehow from pushing if it already found the duplicate, because I get this output:
["A", "A", "A", "B", "B", "C", "A", "B", "B"]

Also, I tried this method:

array.filter(function (value, index, self) { 
    return self.indexOf(value) === index;
})

但是这不允许重复

But that just doesn`t allow duplicates

推荐答案

if(arg[i] != arg[i+1]) {arr.push(arg[i])}

它应该是更好，即使有更多的错误。

it should be better, even if there is more bugs.

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