使用for循環從數組中拼接多個元素

Question

我想拼接數組的多個索引，這是我的代碼示例：

  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); 

預期結果為[“ cc”，“ ee”，“ ff”，“ gg”，“ hh”]；

實際結果是[“ bb”，“ dd”，“ ee”，“ gg”，“ hh”]

Answer 1

拼接時，所有后續值的索引都會減少。 因此，拼接出最后一個：

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

或者只是過濾器：

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

Answer 2

您應該從結尾開始，我-讓我們嘗試跟隨

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

問題可能是您以升序遍歷數組。 因此，當您拼接第一個元素時，其他元素的索引也會改變。 數組上的負迭代可能會有所幫助。

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

您可以過濾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換該索引中的項目，然后再次使用replace刪除該特殊字符或字符串：

 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

在遍歷for of循環時，您會不斷更改數組本身，因此索引會在每個循環步移到右側

你的例子

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

這有效：您需要獲取當前arrrayIndexSplice元素的索引，因此可以將拼接起始索引減少該數量

 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) 

這也起作用：它不適用於循環，但可以使用filter功能

 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

您可以使用以下代碼：

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)
}