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合并基于子數組索引的數組數組作為鍵（NodeJS/Javascript）

JavaScript

慕妹3242003 2023-12-14 14:36:24

如何編寫代碼以以下方式合并我的列表？性能很重要。我想轉換以下數組："list": [ [ "marketing", "page_sections", "PageOne" ], [ "marketing", "page_sections", "PageTwo" ], [ "webapp", "page", "pageone" ], [ "webapp", "page", "pagetwo" ],改成下面的格式：[ { name: "marketing", path: "marketing/", children: [ { name: "page_sections", path: "marketing/page_sections", children: [ { name: "pageOne", path: "marketing/page_sections/pageOne", children: [] }, { name: "pageTwo", path: "marketing/page_sections/pageTwo", children: [] }, } ], }, { name: "webapp", path: "webapp/" children: [ { name: "page", path: "webapp/page/" children: [ { name: "pageone", path: "webapp/page/pageone" children: [] }, { name: "pagetwo", path: "webapp/page/pagetwo" children: [] }, } ] },]子數組的第一個索引是父級，第二個索引是父級的子級，第三個索引是第二個索引的子級（依此類推）。

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4 回答

MM們

TA貢獻1886條經驗獲得超2個贊

最短的方法是迭代嵌套名稱并查找具有相同名稱的對象。如果不存在，則創建一個新對象。將數組返回children為新級別。

此方法的特點Array#reduce是迭代外部數組data和所有內部數組。

const

? ? data = [["marketing", "page_sections", "PageOne"], ["marketing", "page_sections", "PageTwo"], ["webapp", "page", "pageone"], ["webapp", "page", "pagetwo"]],

? ? result = data.reduce((r, names) => {

? ? ? ? names.reduce((level, name, i, values) => {

? ? ? ? ? ? let temp = level.find(q => q.name === name),

? ? ? ? ? ? ? ? path = values.slice(0, i + 1).join('/') + (i ? '' : '/');

? ? ? ? ? ? if (!temp) level.push(temp = { name, path, children: [] });

? ? ? ? ? ? return temp.children;

? ? ? ? }, r);

? ? ? ? return r;

? ? }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

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嗶嗶one

TA貢獻1854條經驗獲得超8個贊

查看來源以及您的預期結果。

我要做的是循環list，然后在列表中進行另一個循環。將此與Array.find..混合

例如..

const data = {list:[

["marketing","page_sections","PageOne"],

["marketing","page_sections","PageTwo"],

["webapp","page","pageone"],

["webapp","page","pagetwo"]]};

function makeTree(src) {

const root = [];

for (const s of src) {

let r = root;

let path = '';

for (const name of s) {

path += `${name}/`;

let f = r.find(k => k.name === name);

if (!f) r.push(f = {name, path, children: []});

r = f.children;

}

return root;

}

console.log(makeTree(data.list));

.as-console-wrapper {

min-height: 100%;

}

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鴻蒙傳說

TA貢獻1865條經驗獲得超7個贊

您可以執行以下操作，

list= [

[

"marketing",

"page_sections",

"PageOne"

[

"marketing",

"page_sections",

"PageTwo"

[

"webapp",

"page",

"pageone"

[

"webapp",

"page",

"pagetwo"

];

getChildrenItem = (arr) => {

if(arr.length === 1) {

return { name: arr[0], children: []};

} else {

return { name: arr.splice(0,1)[0], children: [getChildrenItem([...arr])]};

}

merge = (srcArr, newObj) => {

const {name, children} = newObj;

let index = srcArr.findIndex(item => item.name === name);

if( index> -1) {

children.forEach(item => merge(srcArr[index].children, item))

return ;

} else {

srcArr.push(newObj);

return;

}

allObj = [];

list.forEach(item => {

let tempObj = getChildrenItem([...item]);

merge(allObj, tempObj);

});

console.log(allObj);

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翻過高山走不出你

TA貢獻1875條經驗獲得超3個贊

如果性能是一個問題，我認為這是最好的解決方案之一。

let list = [

["marketing", "page_sections", "PageOne"],

["marketing", "page_sections", "PageTwo"],

["webapp", "page", "pageone"],

["webapp", "page", "pagetwo"],

];

const dt = {};

const pushToOBJ = (Object, name) => {

if (Object[name]) return Object[name];

Object[name] = {

name,

children: {},

};

return Object[name];

};

for (let i = 0; i < list.length; i++) {

let subArray = list[i];

let st = pushToOBJ(dt, subArray[0]);

for (let j = 1; j < subArray.length; j++) {

st = pushToOBJ(st.children, subArray[j]);

}

let result = [];

const convertObjToChildArray = (obj) => {

if (obj === {}) return [];

let arr = Object.values(obj);

for (let i = 0; i < arr.length; i++) {

arr[i].children = convertObjToChildArray(arr[i].children);

}

return arr;

};

result = convertObjToChildArray(dt);

console.log(result);

沒有使用 JS find 函數，該函數已經具有 O(n) 復雜度。

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合并基于子數組索引的數組數組作為鍵（NodeJS/Javascript）

合并基于子數組索引的數組數組作為鍵（NodeJS/Javascript）

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