JS拆分数组以生成子数组并找到所有数组的联合集。

Question

我有以下格式的数据：

要素1：“欧洲>德国>柏林>主办事处”(反对)

要素2：“美利坚合众国>美国>总部”(反对)

要素3：“美国>美国>纽约>主要办公室”(对象)

要素4："America > United States > NY > Sub Office“(Object)

要素5：“联合王国>英格兰>伦敦>主办事处”(对象)

首先，我尝试根据">“字符拆分数组，并以以下格式创建一个包含所有元素的树(vanilla /es6)：

"Europe": {
  "Germany": {
    "Berlin": {
      "Main Office": "Object"
    }
  }
},

我的尝试：

Array.prototype.slice.call(elements).forEach((el) => {

    let 
    stringIHave = el.name, // "Europe > Germany > Berlin > Main Office" 
    keywordsArr = stringIHave.split( " > " ),
    arrayTree =  new Array();

    Array.prototype.slice.call(keywordsArr).forEach((subEl) => {
        //arrayTree.push(el.toString()); // Result: {"Europe", "Germany", "Berlin", "Main Office" }  

        // What I'm Seeking:  
        // "Europe": { "Germany": { "Berlin": { "Main Office": "Object" } } },

    });

});

数据格式化后，我们需要找到所有数组的联合(或一个数组)，如下所示：

[
"Europe": {
  "Germany": {
    "Berlin": {
      "Main Office": "Object"
    }
  }
},

"America": {
  "United States": {
    "Headquarters",
    "NY": {
      "Main Office": "Object",
      "Sub Office": "Object"
    }
  }
},

"United Kingdom": {
  "England": {
    "London": {
      "Main Office": "Object"
    }
  }
}
]

我一直试图使用Array.from(new Set(masterArray))来查找所有数组的联合，但没有骰子！

Answer 1

可以对数组中的每个字符串使用递归函数来构造嵌套对象。

​

const data = [
  'Element 1: Europe > Germany > Berlin > Main Office',

  'Element 2: America > United States > Headquarters',

  'Element 3: America > United States > NY > Main Office',

  'Element 4: America > United States > NY > Sub Office',

  'Element 5: United Kingdom > England > London > Main Office'
]
 
 //remove the string before ':' on all array elements.
 let data1 = data.map(str => str.slice( str.indexOf(':') +1, -1 ));
 
 let arrayTree = data1.map(str => convertToTree(str)) ;
 
 console.log(arrayTree);

 function convertToTree(str) {
     let places = str.split('>');
   
     if(places.length === 1) {
         return ({"Main Office": "Object"});
     }
   
     //the first part of the splited string should be the property and the remaining part will be converted 
     //back to string and sent to the recursive function until the result of the split becomes a single element

     return ({ [places[0]]:convertToTree(places.slice(1).join('>'))  });
 }

​