是否有一个loadash函数可以比较两个数组，并且仅当arr1中存在所有来自arr2的值时才返回

Question

我必须比较两个数组，如果只有数组1包含数组2的所有值，则返回true。对此合适的loadash函数是什么？let arr1 = [1, 2, 3, 4]let arr2 = [1, 2] let arr3 = [1, 5] 比较arr1和arr2应该返回true比较arr1和arr3应该返回false

Answer 1

您可以使用every：

let arr1 = [1, 2, 3, 4];

let arr2 = [1, 2];

let arr3 = [1, 5];

const allElements = (a1, a2) => a2.every(e => a1.includes(e));

console.log(allElements(arr1, arr2));

console.log(allElements(arr1, arr3));

Answer 2

仅出于完整性考虑，您可以选择Set并与进行比较Set#has。

let arr1 = [1, 2, 3, 4],

    arr2 = [1, 2],

    arr3 = [1, 5],

    base = new Set(arr1);

console.log(arr2.every(Set.prototype.has, base));

console.log(arr3.every(Set.prototype.has, base));

Answer 3

这里不需要loadash，只需将本机JavaScriptArray#every方法与Array#includesmethod一起使用

function compareArray(arr1, arr2) {

  return arr2.every(v => arr1.includes(v))

}

let arr1 = [1, 2, 3, 4]

let arr2 = [1, 2]

let arr3 = [1, 5]

console.log(compareArray(arr1, arr2))

console.log(compareArray(arr1, arr3))