Question

var a = [["Green","Medium"],["Green","Small"],["Medium","Red"],["Red","Small"]];

var b = [["Green","Medium"],["Green","Small"],["Medium","Red"]];

所以我的结果将[["Red","Small"]]

不是a.filter(x => !b.includes(x))无效，因为所有元素都是数组。我已经尝试过类似的事情

    var diff = [];
    a.forEach((res, i) => {
      b.forEach((res2, j) => {
        if (i === j && !_.isEqual(res, res2)) {
          diff.push(res);
        }
      });
    });
console.log(diff);

当不同元素位于最后位置时，此方法不起作用

Answer 1

在lodash中，您可以使用_.differenceWith()并提供_.isEqual()作为比较器来执行深度比较：

const c = _.differenceWith(a, b, _.isEqual);

完整摘要：

const a = [["Green","Medium"],["Green","Small"],["Medium","Red"],["Red","Small"]];
const b = [["Green","Medium"],["Green","Small"],["Medium","Red"]];

const c = _.differenceWith(a, b, _.isEqual);

console.log(c);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

Answer 2

对于不需要库的O(N)解决方案，我通过对每个子数组的内容进行字符串化，将map b转换为Set字符串，然后根据集合中是否包含filter的字符串项来a a：

var a = [["Green","Medium"],["Green","Small"],["Medium","Red"],["Red","Small"]];
var b = [["Green","Medium"],["Green","Small"],["Medium","Red"]];

const bSet = new Set(b.map(arr => JSON.stringify(arr)));
const aFiltered = a.filter(arr => !bSet.has(JSON.stringify(arr)));
console.log(aFiltered);

（{Set.has通常是O(1)，不像includes和indexOf这样的数组方法）

Answer 3

由于您已经在使用lodash，因此可以使用filter和every的组合来比较两个数组。

var a = [["Green","Medium"],["Green","Small"],["Medium","Red"],["Red","Small"]];

var b = [["Green","Medium"],["Green","Small"],["Medium","Red"]];


var unique = a.filter(a=> b.every(b=> !_.isEqual(a, b)));
console.log(unique)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.js"></script>

Answer 4

var a = [["Green","Medium"],["Green","Small"],["Medium","Red"],["Red","Small"]];

var b = [["Green","Medium"],["Green","Small"],["Medium","Red"]];

var c = b.map(ele=>JSON.stringify(ele));

var p = a.filter(ele=>!c.includes(JSON.stringify(ele)))

console.log(p)

元素为数组时两个数组之间的差异

4 个答案: