两个数组之间的打字稿差异 [英] Typescript difference between two arrays

问题描述

有没有办法在TypeScrpit中从list_b返回list_a的缺失值?

Is there a way to return the missing values of list_a from list_b in TypeScrpit?

例如:

var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd', 'z'];

结果值为

['e', 'f', 'g'].

预先感谢

推荐答案

可能有很多方法，例如使用

There are probably a lot of ways, for example using the Array.prototype.filter():

var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd'];

let missing = a1.filter(item => a2.indexOf(item) < 0);
console.log(missing); // ["e", "f", "g"]

( filter函数在a1的元素上运行，并将其缩小(但在新数组中)为a1中的元素(因为我们正在对其元素进行迭代)而在a2.

The filter function runs over the elements of a1 and it reduce it (but in a new array) to elements who are in a1 (because we're iterating over it's elements) and are missing in a2.

元素不会包含在结果数组(missing)中，因为过滤器函数不会遍历a2元素:

Elements in a2 which are missing in a1 won't be included in the result array (missing) as the filter function doesn't iterate over the a2 elements:

var a1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var a2 = ['a', 'b', 'c', 'd', 'z', 'hey', 'there'];

let missing = a1.filter(item => a2.indexOf(item) < 0);
console.log(missing); // still ["e", "f", "g"]

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