使用递归函数向数组动态添加新维度

Question

我有一个这样的数组：0: "SuSPENSE"1: "Subcontractor Expense"2: "Data Entry"3: "Design"4: "Programming"5: "Subcontractor Expense - Other"6: "Total Subcontractor Expense"7: "Technology-Communication"8: "Licenses"9: "Domain Hosting Fee"10: "Internet"11: "Internet Servers"12: "Internet - Other"13: "Total Internet"14: "Telephone"15: "Call Center"16: "Cellular Phones"17: "Fax"18: "Telephone - Other"19: "Total Telephone"20: "Hardware/Software"21: "Computer Software"22: "Hardware/Software - Other"23: "Total Hardware/Software"24: "Technology-Communication - Other"25: "Total Technology-Communication"这是类别和子类别的列表。例如， 是一个子类别，以 结尾。与 和 相同。模板始终相同，类别以“（名称）”开头，以“总计（名称）”结尾。但每个类别可以有许多级别的子类别，就像树一样。我正在尝试使用递归函数将此数组解析为类似JSON的数组或多维数组，但我永远不知道最大深度是多少。我尝试使用以下方法执行以下操作："Subcontractor Expense""Total Subcontractor Expense""Internet""Total Internet"jsvar parsedArray = [];var items = getLineItems("Expense", "Total Expense"); //This is a function to return the mentioned arrayvar newArray = parseArray(items, "Expense");function parseArray(items, category){    var lineItems = [];    for(var i = 0; i < items.length; i++){        var inArray = $.inArray("Total " + items[i], items);        if(inArray !== -1){            parseArray(getLineItems(items[i], "Total " + items[i]), items[i]);        }        else {            lineItems.push(items[i]);        }    }    parsedArray[category] = lineItems;}但是这个递归函数永远不会比2级更深。有可能生成这样的东西吗？"SuSPENSE""Subcontractor Expense"    "Data Entry"    "Design"    "Programming""Subcontractor Expense - Other""Technology-Communication"    "Licenses"    "Domain Hosting Fee"    "Internet"        "Internet Servers"        "Internet - Other"    "Telephone"        "Call Center"        "Cellular Phones"        "Fax"        "Telephone - Other"    "Hardware/Software"        "Computer Software"        "Hardware/Software - Other"    "Technology-Communication - Other"

Answer 1

我仍然在猜测你的输出格式。这是一个递归解决方案，给出了我在评论中询问的格式：

[

    {name: "SuSPENSE"},

    {name: "Subcontractor Expense", children: [

        {name: "Data Entry"},

        {name: "Design"},

        {name: "Programming"},

        {name: "Subcontractor Expense - Other"}

    ]},

    {name: "Technology-Communication", children: [

       //...

    ]}

]

const restructure = (

  [s = undefined, ...ss], 

  index = s == undefined ? -1 : ss .indexOf ('Total ' + s)

) => 

  s == undefined

    ? []

    : index > -1

      ? [

          {name: s, children: restructure (ss .slice (0, index))}, 

          ... restructure (ss .slice (index + 1))

        ]

      : [{name: s}, ... restructure (ss)]

const data = ["SuSPENSE", "Subcontractor Expense", "Data Entry", "Design", "Programming", "Subcontractor Expense - Other", "Total Subcontractor Expense", "Technology-Communication", "Licenses", "Domain Hosting Fee", "Internet", "Internet Servers", "Internet - Other", "Total Internet", "Telephone", "Call Center", "Cellular Phones", "Fax", "Telephone - Other", "Total Telephone", "Hardware/Software", "Computer Software", "Hardware/Software - Other", "Total Hardware/Software", "Technology-Communication - Other", "Total Technology-Communication"]

console .log (

  restructure (data)

)

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

请注意，在其中一个递归情况下，我们调用 main 函数两次。一次用于嵌套数据，一次用于数组的其余部分。

另一种可能的结构，一个我不太喜欢的结构，但可能满足您的需求，看起来像这样：

[

    "SuSPENSE",

    {"Subcontractor Expense": [

        "Data Entry", "Design", "Programming", "Subcontractor Expense - Other"

    ]},

    { "Technology-Communication": [ 

        // ...

    ]}

]

这可以通过一个小的修改来实现：

const restructure = (

  [s = undefined, ...ss], 

  index = s == undefined ? -1 : ss .indexOf ('Total ' + s)

) => 

  s == undefined

    ? []

  : index > -1

    ? [

        {[s]: restructure (ss .slice (0, index))}, 

        ... restructure (ss .slice (index + 1))

      ]

    : [s, ... restructure (ss)]

更新

这个变体与第一个变体相同，但对于不习惯我的表达方式的人来说，可能看起来更熟悉：

const restructure = ([s = undefined, ...ss]) => {

  if (s == undefined) {return []}

  const index = ss .indexOf ('Total ' + s)

  return index < 0

    ? [{name: s}, ... restructure (ss)]

    : [

        {name: s, children: restructure (ss .slice (0, index))}, 

        ... restructure (ss .slice (index + 1))

      ]

}

Answer 2

您可以通过检查当前元素是否具有以单词开头，然后继续使用当前元素的文本的相应元素来执行此操作，如果是，则递增级别。当当前元素以单词总计开头时，您将递减级别。Total

const data = {"0":"SuSPENSE","1":"Subcontractor Expense","2":"Data Entry","3":"Design","4":"Programming","5":"Subcontractor Expense - Other","6":"Total Subcontractor Expense","7":"Technology-Communication","8":"Licenses","9":"Domain Hosting Fee","10":"Internet","11":"Internet Servers","12":"Internet - Other","13":"Total Internet","14":"Telephone","15":"Call Center","16":"Cellular Phones","17":"Fax","18":"Telephone - Other","19":"Total Telephone","20":"Hardware/Software","21":"Computer Software","22":"Hardware/Software - Other","23":"Total Hardware/Software","24":"Technology-Communication - Other","25":"Total Technology-Communication"}

function toNested(data) {

  const result = [];

  const levels = [result]

  let level = 0;

  const checkChildren = (string, data) => {

    return data.some(e => e === `Total ${string}`)

  }

  data.forEach((e, i) => {

    const object = { name: e, children: []}

    levels[level + 1] = object.children;

    if (e.startsWith('Total')) level--;

    else levels[level].push(object);

    if (checkChildren(e, data.slice(i))) level++;

  })

  return result;

}

const result = toNested(Object.values(data));

console.log(result)