Question

我有两个2D数组，我想用JavaScript比较它们，忽略匹配，如果不匹配则将整行返回到一个新数组。

    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'],  ];

我想取这个输出，array2中存在的数组而不是array1中的数组：

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

请问好吗？

Answer 1

只需遍历两个数组：

＆＃13;

var array1 = [ ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 [null,'ABC'],
                 ['6f93cfa0106f','xxx'],
                 ['52a1fd0296fc','ABC'] ];

var array2 = [ ['52a1fd0296fc','ABC'],
                   ['6f93cfa0106f','xxx'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','DEF'] ];
                   
var array3 = [];
for(var i = 0; i<array1.length; ++i) {
	var a = array1[i];
	var found = false;
	for(var j = 0; j<array2.length; ++j) {
		var b = array2[j];
		if(a[0] == b[0] && a[1] == b[1]) {
			found = true;
			break;
		}
	}
	if(!found) {
		array3.push(a);
	}
}

console.dir(array3);

＆＃13;

我假设你想要array1而不是array2，而不是相反。

Answer 2

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目。

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);

Answer 3

这是我的目标，可能更紧凑（并使用ES6）

此代码从主数组中删除重复的数组。

var array1 = [ ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 [null,'ABC'],
                 ['6f93cfa0106f','xxx'],
                 ['52a1fd0296fc','ABC'] ];

let array3 = []

array1.forEach( a1 => {
	if(!array3.find(a2 => a2[0]===a1[0] && a2[1]===a1[1])) array3.push(a1)
})

console.log(array3)

JavaScript迭代2D数组并返回不匹配

3 个答案: