從 NESTED 對象數組中刪除重復項

Question

我有一個像下面這樣的對象數組，其中有另一個對象數組，這些對象具有重復的verticalName值，需要刪除這些值才能在頁面上顯示。

 "landingPages": [
        {
          "programmes": [
            {
              "progName": "Programme 1",
              "verticals": [
                {
                  "id": "8",
                  "verticalName": "Law and Criminology"
                }
              ]
            }
          ]
        },
        {
          "programmes": [
            {
              "progName": "Programme 2",
              "verticals": [
                {
                  "id": "1",
                  "verticalName": "Psychology and Sociology"
                }
              ]
            }
          ]
        },
        {
          "programmes": [
            {
              "progName": "Programme 3",
              "verticals": [
                {
                  "id": "3",
                  "verticalName": "Computing and IT"
                }
              ]
            }
          ]
        },
        {
          "programmes": [
            {
              "progName": "Programme 4",
              "verticals": [
                {
                  "id": "1",
                  "verticalName": "Psychology and Sociology"
                }
              ]
            },
            {
              "progName": "Programme 5",
              "verticals": [
                {
                  "id": "3",
                  "verticalName": "Computing and IT"
                }
              ]
            },
            {
              "progName": "Programme 6",
              "verticals": [
                {
                  "id": "2",
                  "verticalName": "Business and Management"
                }
              ]
            },
            {
              "progName": "Programme 7",
              "verticals": [
                {
                  "id": "3",
                  "verticalName": "Computing and IT"
                }
              ]
            },
            {
              "progName": "Programme 8",
              "verticals": [
                {
                  "id": "3",
                  "verticalName": "Computing and IT"
                }
              ]
            }
          ]
        }
      ]

我嘗試了幾種解決方案，其中一些不起作用，但一個起作用，我在下面復制/粘貼。 我已經設法將它們全部存儲在一個對象數組中，只是代碼對我來說看起來很糟糕。 我正在嘗試找到一個更清潔的解決方案來解決這個問題。 這是我的解決方案，我想看看它與您的解決方案相比如何，如果您能給我一些反饋，那將是驚人的。

    let known = {}
    let noduplicates = programmes.map((subarray) => {
        const newarrays = []
        subarray.verticals.forEach((item) => {
            if (
                !known.hasOwnProperty(item.verticalName) &&
                (known[item.verticalName] = true)
            ) {
                newararys.push(item)
            } else {
                console.log('test')
            }
        })
        console.log(newarrays)
        return newarrays
    })
    const fil = filtered.filter((item) => item.length)
    const singleArr = fil.reduce((a, b) => a.concat(b), [])
    console.log(singleArr)

我想要得到的是這樣的東西，基本上，只是在垂直數組中找到的所有對象，但是 - 沒有重復：

[
   {id: "1", verticalName: "Psychology and Sociology"}, 
   {id: "3", verticalName: "Computing and IT"}, 
   {id: "2", verticalName: "Business and Management"}
]

大家加油！

Answer 1

你可以通過純 ES6 函數來實現，我會分解，但你可以在一行中擁有它：）

首先，您需要檢索所有verticalNames

const allVerticalNames = landingPages.map(lp => lp.programmes.map(p => p.verticals))

你會有很多 arrays 或 arrays，所以我們需要將所有這些都展平

const flattened = allVerticalNames.flat(2)

此時您擁有所有的verticalNames，但沒有uniq。 為此，我們應該將其轉換為 Set，Map 或更簡單的 Object

const keyValuePairs = flattened.map(v => [v.id, v.verticalName]); // We transform to a pair key / value
const uniq = Object.fromEntries(keyValuePairs)

此時你有一個 object 像{1: "Psychology and Sociology", 2: "Business and Management", 3: "Computing and IT", 8: "Law and Criminology"}所以如果你想讓它回到例如，您將其移回數組！

const final = Object.entries(uniq).map(([id, verticalName]) => ({id, verticalName}))

就是這樣！

全部在一條線上？

const final = Object.entries(Object.fromEntries(landingPages.map(lp => lp.programmes.map(p => p.verticals)).flat(2).map(v => [v.id, v.verticalName]))).map(([id, verticalName]) => ({id, verticalName}))

Answer 2

您可以將set用於此用例，它允許您存儲任何類型的唯一值。

 const data = [ { "programmes": [ { "progName": "Programme 1", "verticals": [ { "id": "8", "verticalName": "Law and Criminology" } ] } ] }, { "programmes": [ { "progName": "Programme 2", "verticals": [ { "id": "1", "verticalName": "Psychology and Sociology" } ] } ] }, { "programmes": [ { "progName": "Programme 3", "verticals": [ { "id": "3", "verticalName": "Computing and IT" } ] } ] }, { "programmes": [ { "progName": "Programme 4", "verticals": [ { "id": "1", "verticalName": "Psychology and Sociology" } ] }, { "progName": "Programme 5", "verticals": [ { "id": "3", "verticalName": "Computing and IT" } ] }, { "progName": "Programme 6", "verticals": [ { "id": "2", "verticalName": "Business and Management" } ] }, { "progName": "Programme 7", "verticals": [ { "id": "3", "verticalName": "Computing and IT" } ] }, { "progName": "Programme 8", "verticals": [ { "id": "3", "verticalName": "Computing and IT" } ] } ] } ]; const seen = new Set(); const filteredArr = data.filter(el => { const verticalName = el.programmes[0].verticals[0].verticalName; const duplicate = seen.has(verticalName); seen.add(verticalName); return;duplicate; }). console;log(filteredArr);

Answer 3

如果我正確理解了這個問題，我們需要從嵌套programmes數組中刪除verticalNames 。

 const landingPages = [ { programmes: [ { progName: 'Programme 1', verticals: [ { id: '8', verticalName: 'Law and Criminology', }, ], }, ], }, { programmes: [ { progName: 'Programme 2', verticals: [ { id: '1', verticalName: 'Psychology and Sociology', }, ], }, ], }, { programmes: [ { progName: 'Programme 3', verticals: [ { id: '3', verticalName: 'Computing and IT', }, ], }, ], }, { programmes: [ { progName: 'Programme 4', verticals: [ { id: '1', verticalName: 'Psychology and Sociology', }, ], }, { progName: 'Programme 5', verticals: [ { id: '3', verticalName: 'Computing and IT', }, ], }, { progName: 'Programme 6', verticals: [ { id: '2', verticalName: 'Business and Management', }, ], }, { progName: 'Programme 7', verticals: [ { id: '3', verticalName: 'Computing and IT', }, ], }, { progName: 'Programme 8', verticals: [ { id: '3', verticalName: 'Computing and IT', }, ], }, ], }, ] let page = landingPages.find(page => page.programmes.length > 1); const uniqueprogrammes = new Set(); const uniqueverticals = page.programmes.map(p => { const vn = p.verticals[0].verticalName; if (.uniqueprogrammes.has(vn)) { uniqueprogrammes;add(vn). return p;verticals[0]. } });filter(v =>.;v); console.log(uniqueverticals);

更新：

根據問題中的預期結果更新了代碼片段。

這個想法是首先找到嵌套的程序數組，這是通過根據.length過濾landingPages object 來完成的

let page = landingPages.find(page => page.programmes.length > 1);

然后，我們創建一個集合來跟蹤這個嵌套數組中的所有唯一程序，

const uniqueprogrammes = new Set();

接下來，我們通過嵌套數組map找到所有不屬於uniqueprogrammes集的垂直，

const uniqueverticals = page.programmes.map(p => {
  const vn = p.verticals[0].verticalName;
  if (!uniqueprogrammes.has(vn)) {
    uniqueprogrammes.add(vn);
    return p.verticals[0];
  }
})

但是，這會給我們一個 output 如下所示，

[
  { id: '1', verticalName: 'Psychology and Sociology' },
  { id: '3', verticalName: 'Computing and IT' },
  { id: '2', verticalName: 'Business and Management' },
  undefined,
  undefined
]

所以（最后），我們需要filter掉虛假值，

.filter(v => !!v);

Answer 4

您可以使用loadash來奉承然后刪除重復項。

var arrayWithDuplicatesRemoved= .flattenDeep(landingPages);

arrayWithDuplicatesRemoved = arrayWithDuplicatesRemoved .uniqBy(data, function (e) {return e.verticalName;});