查找在Javascript数组中仅出现一次的项目 [英] Finding items that appear only one time in a Javascript array

问题描述

我试图找到在Javascript数组中仅出现一次的项目.在以下数组中:

I'm trying to find the items that appear only one time in a Javascript array. In the following array:

['txfa2','txfa9','txfa2','txfa1','txfa3','txfa4','txfa8','txfa9','txfa2','txfa8']

结果应为:

['txfa1','txfa3','txfa4']

我目前在jQuery和.sort()中使用.each()函数.是否有更聪明或更好的方法来做到这一点?您是否知道可以用更少的代码行做到这一点的jQuery插件.

I'm currently using the .each() function in jQuery and .sort(). Is there a smarter or better way on doing this? Do you know of any jQuery plugin that can do this in fewer lines of code.

<script src="../../js/jq.js"></script>
<script>
var items = ['txfa2', 'txfa9', 'txfa2', 'txfa1', 'txfa3', 'txfa4', 'txfa8', 'txfa9', 'txfa2', 'txfa8'];
var nopairs = []; //should contain only txfa1, txfa3, txfa4
var haspair = '';//contains the item which has pairs
var haspair_ctr = 0;
var nopair_ctr = 0;

var arranged = items.sort();

$(arranged).each(function(index){
    if(index != arranged.length){
        if(arranged[index] != arranged[index + 1] && arranged[index] != haspair){
            nopairs[nopair_ctr] = arranged[index];
            nopair_ctr++;

        }else{
            haspair  = arranged[index];

        }   
    }

});
console.log(nopairs);

</script>

推荐答案

一种简洁的方法:

function singles(array) {
  for (var index = 0, single = []; index < array.length; index++) {
    if (array.indexOf(array[index], array.indexOf(array[index]) + 1) == -1)
      single.push(array[index]);
  };
  return single;
};

var items = ['txfa2', 'txfa9', 'txfa2', 'txfa1', 'txfa3', 'txfa4', 'txfa8', 'txfa9', 'txfa2', 'txfa8'];

console.log(singles( items ))

演示: http://jsfiddle.net/ThinkingStiff/C849F/

以下是该问题(实际上有效)的非重复答案的性能，表明该方法(蓝色)可与其他方法媲美.

Here's the performance of the non-duplicate answers to this question (that actually work) showing this method (blue) is comparable with the other methods.

性能: http://jsperf.com/find-array-singles/3

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