JavaScript 遍历二维数组并返回不匹配项 [英] JavaScript iterate through 2D array and return the mismatches

问题描述

我有两个二维数组，我想使用 JavaScript 比较它们，忽略匹配项，如果不匹配则将整行返回到一个新数组中.

I have two 2D arrays, I want to compare them using JavaScript, ignore the matches and if there are mismatches to return the entire row into a new array.

    var array1 = [ ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 [null,'ABC'],
                 ['6f93cfa0106f','xxx'],
                  ];

    var array2 = [ ['52a1fd0296fc','ABC'],
                   ['6f93cfa0106f','xxx'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','DEF'],
                   ['52a1fd0296fcasd','DEF'],  ];

我想获取这个输出，即存在于 array2 中而不存在于 array1 中的数组:

I want to take this output, the arrays that exists in array2 and NOT in array1:

array3 = [['52a1fd0296fcasd','DEF'],['52a1fd0296fc','ABC']]

有什么想法吗?

推荐答案

您可以使用数组的连接值作为键并过滤第二个数组的一次，同时指示已插入的项.

You could use the joined values of the array as key and filter the once of the second array, while indicating already inserted items.

var array1 = [ ['52a1fd0296fc','DEF'], ['52a1fd0296fc','DEF'], ['52a1fd0296fc','DEF'], ['52a1fd0296fc','DEF'], [null,'ABC'], ['6f93cfa0106f','xxx']],
    array2 = [ ['52a1fd0296fc','ABC'], ['6f93cfa0106f','xxx'], ['52a1fd0296fc','ABC'], ['52a1fd0296fc','ABC'], ['52a1fd0296fc','DEF'], ['52a1fd0296fcasd','DEF']],
    hash = Object.create(null),
    result;

array1.forEach(function (a) {
     hash[a.join('|')] = true;
});

result = array2.filter(function (a) {
    return !hash[a.join('|')] && (hash[a.join('|')] = true);
});

console.log(result);

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