Question

我想拼接数组的多个索引，这是我的代码示例：

    let arrayA = ["aa","bb","cc","dd","ee","ff","gg","hh"];
    let arrrayIndexSplice = [0,1,3];
    let test = null;
    
    for (let i of arrrayIndexSplice) {
      test = arrayA.splice(i,1);
    }
    console.log(arrayA);

预期结果是[“ cc”，“ ee”，“ ff”，“ gg”，“ hh”];

实际结果是[“ bb”，“ dd”，“ ee”，“ gg”，“ hh”]

Answer 1

拼接时，所有后续值的索引都会减少。因此，拼接出最后一个：

let arrrayIndexSplice = [0, 1, 3];
for (let i of arrrayIndexSplice.sort().reverse()) {
 arrayA.splice(i, 1);
}

或者只是过滤：

 arrayA = arrayA.filter((_, i) => !arrayIndexSplice.includes(i));

Answer 2

您应该从结尾开始，即- 让我们尝试关注

let arrayA=["aa","bb","cc","dd","ee","ff","gg","hh"];
let arrrayIndexSplice=[0,1,3];
for (let i=arrrayIndexSplice.length-1;i>=0;i--){arrayA.splice(arrrayIndexSplice[i],1)}

Answer 3

问题可能是您以升序遍历数组。因此，当您拼接第一个元素时，其他元素的索引也会改变。对数组进行负迭代可能会有所帮助。

let arrayA=["aa","bb","cc","dd","ee","ff","gg","hh"];
let arrrayIndexSplice=[0,1,3];
for (let i=arrrayIndexSplice.length-1; i > -1; i--){arrayA.splice(i,1)}

Answer 4

您可以过滤出所有要排除索引的元素。

let arrayA = ["aa", "bb", "cc", "dd", "ee", "ff", "gg", "hh"];
let arrrayIndexSplice = [0, 1, 3];
arrayA = arrayA.filter((item, index)=> !arrrayIndexSplice.includes(index));
console.log(arrayA);

Answer 5

我的想法是用特殊字符或字符串替换该索引中的项目，然后再次使用replace删除该特殊字符或字符串：

let arrayA = ["aa", "bb", "cc", "dd", "ee", "ff", "gg", "hh"];
let arrrayIndexSplice = [0, 1, 3];
let test = null;
arrayA.forEach((val, index) => {
  if(arrrayIndexSplice.includes(index)){
    arrayA[index] = '##';
  }
});
arrayA = arrayA.join(',').replace(/##,/g, '').split(',');

console.log(arrayA);

Answer 6

遍历for of循环时，您会不断更改数组本身，因此索引会在每个循环步骤向右移

您的示例

arrayA: ["aa", "bb", "cc", "dd", "ee", "ff", "gg", "hh"]
// start loop
// 1st iteration
// test = arrayA.splice(i,1);
test = 0
arrayA = ["bb", "cc", "dd", "ee", "ff", "gg", "hh"]  // index 0 removed

// 2nd iteration
// test = arrayA.splice(i,1);
test = 1
arrayA = ["bb", "dd", "ee", "ff", "gg", "hh"]  // index 1 removed

// 3rd iteration
// test = arrayA.splice(i,1);
test = 3
arrayA = ["bb", "dd", "ee", "gg", "hh"]  // index 3 removed

这有效：您需要获取当前arrrayIndexSplice元素的索引，以便可以将拼接起始索引减少该数量

console.clear()
let arrayA = ["aa", "bb", "cc", "dd", "ee", "ff", "gg", "hh"];
let arrrayIndexSplice = [0, 1, 3];
for (let k in arrrayIndexSplice) {
  if (arrrayIndexSplice.hasOwnProperty(k)) {
    arrayA.splice(arrrayIndexSplice[k]-k, 1)
  }
}

console.log(arrayA)

这也有效：它不适用于循环，但可以使用filter函数

let arr = ["aa", "bb", "cc", "dd", "ee", "ff", "gg", "hh"];
let indices = [0,1,3]

function filterIndices(arr, indices, swap = false) {
  return arr.filter((item, index, source) => {
    return swap ? !indices.includes(index) : indices.includes(index);
  })
}

newArr = filterIndices(arr, indices, true)
console.log(newArr)

Answer 7

您可以使用以下代码：

let arrayA = ["aa", "bb", "cc", "dd", "ee", "ff", "gg", "hh"];
let arrrayIndexSplice = [0, 1, 3];
for (let i of arrrayIndexSplice.length) {
  arrayA.splice(arrrayIndexSplice[i], 1)
}

使用for循环从数组中拼接多个元素

7 个答案: