使用javascript中的值數組過濾深度嵌套的對象數組

Question

我想如何使用javascript中的值數組過濾深層嵌套的對象數組（動態）注意， Obj是動態的。

var result = getData(obj);
getData(obj){
  var getcn = obj.map(e=>e.cn);
  var tot = obj.filter(e=>getcn.includes(e.cn));
}

//input
var obj = [{
  "cn": "SG",
  "amt": "30"
},{
  "cn": "SG",
  "amt": "40"
},{
  "cn": "MY",
  "amt": "100"
},{
  "cn": "TH",
  "amt": "40"
}]

預期輸出：

[{
  "cn": "SG",
  "total": 2 // length of cn `SG`
},{
  "cn": "MY",
  "total": 1
},{
  "cn": "TH",
  "total": 1
}]

Answer 1

 var obj=[{cn:"SG",amt:"30"},{cn:"SG",amt:"40"},{cn:"MY",amt:"100"},{cn:"TH",amt:"40"}]; let res = obj.reduce((acc,cur) => { if(acc.some(obj => obj.cn === cur.cn)){ return acc.map(obj => obj.cn === cur.cn ? {cn: obj.cn, total: obj.total + 1} : obj) } return acc.concat({cn: cur.cn, total: 1}) },[]) console.log(res)

Answer 2

 var obj = [ { "cn": "SG", "amt": "30" }, { "cn": "SG", "amt": "40" }, { "cn": "MY", "amt": "100" }, { "cn": "TH", "amt": "40" } ] function getData(obj){ let frequency = {}, result = []; obj.map((item)=>{ if(item.cn in frequency) frequency[item.cn]++; else frequency[item.cn] = 1; }) for(let [key,value] of Object.entries(frequency)) result.push({cn:key,total:value}) return result; } console.log(getData(obj))

Answer 3

我不確定你所說的“obj 是動態的”是什么意思，但基本上，你需要遍歷一個值數組並將每個值添加到一個新數組中。 但是，如果該值已存在於這個新數組中，則只需將其總數增加 1。

//input
var input = [{
  "cn": "SG",
  "amt": "30"
},{
  "cn": "SG",
  "amt": "40"
},{
  "cn": "MY",
  "amt": "100"
},{
  "cn": "TH",
  "amt": "40"
}];

const getData = (data) => {
  const entryIndexByCn = {}; // store index of the value in the new array, 
                             // so we could then increase total by 1
                             // in case the value was already added

  return data.reduce((memo, entry) => {
    // get index of the value in the new array
    const entryIndex = entryIndexByCn[entry.cn];

    // if value is in the new array, increase total by 1
    if (entryIndex !== undefined) {
      memo[entryIndex].total += 1;
    } else { 
      // if not, record index of the value to
      // address it later if we meet the value again
      // to be able to increase its total
      entryIndexByCn[entry.cn] = memo.length;

      // and add new value to the new array
      memo.push({
        cn: entry.cn,
        total: 1
      });
    }

    return memo;
  }, []);
}