Question

我正在尝试通过使用以下代码来检索第二个数组中的不包括的值：

function diffArray(arr1, arr2) {
  var newArr = [];
  for (let i of arr1) {
    if (arr2.includes(i)) {
      newArr.push(i)
    }
  }
  return newArr
}

console.log(
  diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5])
)

有什么办法可以使用另一种方法来做到这一点。我尝试了indexOf，但是我不想索引。

谢谢

Answer 1

您可以使用filter()：

let arr1 = [1, 2, 3, 5];
let arr2 = [1, 2, 3, 4, 5];

let result = arr2.filter(a2 => !arr1.includes(a2));
console.log(result);

Answer 2

if (!arr2.includes(i)) {
     newArr.push(i)
   }

！意味着不是

您也可以始终使用else，但这需要更多代码行：

if (arr2.includes(i)) {
     // newArr.push(i)
 }  else {
    newArr.push(i);
}

Answer 3

const a1 = [1, 2, 3, 4, 5];
const a2 = [1, 2, 3, 5];

function diffArray(arr1, arr2) {
  const frequencies = arr1.concat(arr2).reduce((frequencies, number) => {
    const frequency = frequencies[number];
    frequencies[number] = frequency ? frequency + 1 : 1;
    return frequencies;
  }, {});

  return Object.keys(frequencies).filter(number => frequencies[number] === 1);
}

如何将“ includes（）”方法反转为“ not-includes（）”并检索不包含的值？

3 个答案: