对具有嵌套级别的对象数组进行分组

Question

我有一个表格模式数组对象，如下所示，最多 n 级。可以是任何父子的记录

var records = [
  { country: "USA", state: "FLORIDA", city: "city1" },
  { country: "USA", state: "FLORIDA", city: "city2" },
  { country: "USA", state: "FLORIDA", city:"city3" },
  { country: "USA", state: "ALASKA" },
  { country: "USA", state: "ALBAMA" },
]

var columns = ["country","state","city"]  // upto n column

我需要按以下格式对第 n 级进行分组，因为可以有 n 级关系，按以下格式分组记录

{
  sequencer: 1, value: 'USA', loop: [
    { sequencer: 1, value: 'FLORIDA', loop: [
      { sequencer: 1, value: 'city1' },
      { sequencer: 2, value: 'city2' },
      { sequencer: 3, value: 'city3' },
    ], },
    { sequencer: 2, value: 'ALASKA' },
    { sequencer: 3, value: 'ALBAMA' },
  ],
}

有人可以编写一个递归函数来对 n 级列对象进行分组吗？ 谢谢

Answer 1

您可以通过避免不需要的循环属性来对数据进行分组。

const
    data = [{ country: "USA", state: "FLORIDA", city: "city1" }, { country: "USA", state: "FLORIDA", city: "city2" }, { country: "USA", state: "FLORIDA", city: "city3" }, { country: "USA", state: "ALASKA" }, { country: "USA", state: "ALBAMA" }],
    keys = ['country', 'state', 'city'],
    result = data.reduce((loop, o) => {
        keys
            .map(k => o[k])
            .filter(Boolean)
            .reduce((r, value) => {
                let temp = (r.loop ??= []).find(q => q.value === value);
                if (!temp) r.loop.push(temp = { sequence: r.loop.length + 1, value });
                return temp;
            }, { loop });
        return loop;
    }, []);

console.log(result);

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Answer 2

我最近回复了similar question。您可以使用flatMap 和递归来编写您的transform -

const transform = ([ name, ...more ], { value, loop = [{}] }, r = {}) =>
  name === undefined
    ? [ r ]
    : loop.flatMap(t => transform(more, t, { [name]: value, ...r }))

const records =
  {sequencer:1,value:'USA',loop:[{sequencer:1,value:'FLORIDA',loop:[{sequencer:1,value:'city1'},{sequencer:2,value:'city2'},{sequencer:3,value:'city3'}]},{sequencer:2,value:'ALASKA'},{sequencer:3,value:'ALBAMA'}]}

const columns =
  ['country', 'state', 'city']

const result =
  transform(["country", "state", "city"], records)
  
console.log(JSON.stringify(result, null, 2))

---

function 语句等效于上面的 => 表达式 -

function transform
  ( [ name, ...more ]
  , { value, loop = [{}] }
  , r = {}
  )
  { if (name === undefined)
      return [ r ]
    else 
      return loop.flatMap
        ( t =>
            transform
              ( more
              , t
              , { [name]: value, ...r }
              )
        )
  }

阅读original Q&A 了解有关此技术的更多信息。