找出两个多维数组之间的差异

Question

我在找出两个多维数组之间的差异时遇到了一些麻烦。

以下是数组：

let arr1 = [["Tag2","TES"],["Tag3","TES"],["Fedex Ground","TES"],["Fedex Air","TES"],["AMER","TES"],["CA BC","TES"],["EMEA","TES"],["Express","TES"],["tag-5","TES"]]

let arr2 = [["Tag2","TES"],["Tag3","TES"],["Fedex Ground","TES"],["Fedex Air","TES"],["AMER","TES"],["CA BC","TES"],["EMEA","TES"],["testingTag","TES"],["Express","TES"],["tag-5","TES"], ["tag-6", "TES"]]

我希望结果是两个数组中都不存在的标签的二维数组，所以在本例中：

resultArr = [["testingTag","TES"],["tag-6","TES"]]

我已经能够成功地检索到数组中匹配的所有标记。

然后，我尝试使用所有匹配的标签遍历arr2，如果循环中的标签与arr2i中的标签不匹配，则将其推入数组，在推入数组之前，我将检查标签是否也已存在于正在推送的数组中，如果存在，则将其从数组中删除。但是我在这里解释的逻辑没有起作用，它只会让我更困惑。

我刚才尝试解释的逻辑的代码：

let existingTags = [], deletedTags = [];

    //Checks which tags are not to be deleted
    for(let i=0; i<newTags.length;i++){
        for(let j=0; j<oldTags.length;j++){
            if(newTags[i][0] == oldTags[j][0]){
                existingTags.push(newTags[i][0])
            }
        }
    }
    
    //Pushes to array the tags to be deleted 
    for(let i=0; i<existingTags.length;i++){
        for(let j=0; j<oldTags.length;j++){
            if(existingTags[i] != oldTags[j][0]){
                deletedTags.push(oldTags[j][0])
            }else{
                for(let k=0; k<deletedTags.length; k++){
                    if(oldTags[j][0] == deletedTags[k]){
                        deletedTags.splice(i,1)
                    }
                }
            }
        }
    }

有没有一种更简单、更有效的方法来找出两个多维数组之间的差异？

Answer 1

假设有一些字符(例如'_')不能出现在字符串中，您可以将您的问题转换为处理字符串数组，例如，如下所示：(我没有压缩代码，以保持一定的可读性)

set1 = new Set(arr1.map(x => x[0]+'_'+x[1]))
set2 = new Set(arr2.map(x => x[0]+'_'+x[1]))
diff1 = arr1.filter(x => !set2.has(x[0]+'_'+x[1]))
diff2 = arr2.filter(x => !set1.has(x[0]+'_'+x[1]))
res = diff1.concat(diff2)

Answer 2

您可以将一个对象作为键/值对，并通过检查存储的值与实际值来过滤第二个数组。

​

let array1 = [["Tag2", "TES"], ["Tag3", "TES"], ["Fedex Ground", "TES"], ["Fedex Air", "TES"], ["AMER", "TES"], ["CA BC", "TES"], ["EMEA", "TES"], ["Express", "TES"], ["tag-5", "TES"]],
    array2 = [["Tag2", "TES"], ["Tag3", "TES"], ["Fedex Ground", "TES"], ["Fedex Air", "TES"], ["AMER", "TES"], ["CA BC", "TES"], ["EMEA", "TES"], ["testingTag", "TES"], ["Express", "TES"], ["tag-5", "TES"], ["tag-6", "TES"]],
    values = Object.fromEntries(array1),
    result = array2.filter(([k, v]) => values[k] !== v);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

​