将2个数组的元素相互比较并返回计数JavaScript [英] Compare 2 array's elements against each other and return count JavaScript

问题描述

我有2个数组，我需要相互比较并返回相同的数量。

I have 2 arrays that I need to compare against each other and return the count of the same.

示例：将array1 [abcd]与array2进行比较[adce] 。返回为2,1，因为a和c处于相同位置且d处于错误位置。

Example: compare array1 [abcd] against array2 [adce]. Return would be 2,1 as both a and c are in same position and d is in the wrong position.

function () {
    var index = 0;
    for (index = 0; index < length; index++) {
        if (array1[index] == array2[index]) {
            count++
        } 
    }
    return count
}

我得到1的回报。我认为这是因为它们的长度相等，这就是为什么我得到1.所以我想我现在需要为循环添加另一个并让该循环去单独通过元素，但不知道如何做到这一点。我上面提到的可能完全错了，如果是的话，有人可以向我解释这个过程。

I'm getting a return of 1. I think that's because they equal in length that is why I'm getting a 1. So I'm thinking that I now need to put another for loop and have that loop go through the elements individually, but not sure how to do this. I could be completely wrong in what I have put above, if so, could someone explain the process to me please.

推荐答案

你获取 1 作为输出，因为 length 未在您的代码中定义

You get 1 as output because length is not defined in your code

var array1 = ['a','b','c','d'];
var array2 = ['a','d','c','e'];

var length = Math.min(array1.length,array2.length);
var countMatched = 0,countNotMatched = 0;

for(var index=0;index<length;index++)
{
  if(array1[index] == array2[index])
    countMatched++;
  else if(array2.indexOf(array1[index]) >= 0)
    countNotMatched++;
}
alert(countMatched );
alert(countNotMatched);

演示小提琴： http://jsfiddle.net/tCKE7/2/

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