根据特定性别数据拆分组合列表用于 ID

Question

这是一个组合列表，具有唯一的 ageCode。

    [
      {
        "ageCode": 1,
        "ageDesc": "0-4",
        "qM": 358,
        "sM": 158,
        "qF": 328,
        "sF": 258
      },
      {
        "ageCode": 3,
        "ageDesc": "15-59",
        "qM": 525,
        "sM": 125
      },
      {
        "ageCode": 4,
        "ageDesc": "60+",
        "qF": 458,
        "sF": 358
      }
    ]

需要根据年龄代码的 M 和 F 值是否存在，将组合列表拆分为单独的对象。转换后的列表如下所示。如果没有 "M" 日期，则不会有 "gender" 的对象："M" 同样适用于 "F"

    [
      {
        "ageCode": 1,
        "ageDesc": "0-4",
        "q": 358,
        "s": 158,
        "gender": "M"
      },
      {
        "ageCode": 1,
        "ageDesc": "0-4",
        "q": 328,
        "s": 258,
        "gender": "F"
      },
      {
        "ageCode": 3,
        "ageDesc": "15-59",
        "q": 525,
        "s": 125,
        "gender": "M"
      },
      {
        "agCode": 4,
        "ageDesc": "60+",
        "q": 458,
        "s": 358,
        "gender": "F"
      }
    ]

尝试过的解决方案：

      for(let item  of this.ageData) {
            if (this.ageData.find((i) => { i.agCode=== item.agCode})){

    //}

        }

这里需要重复和多次for循环等问题，有什么有效的方法可以实现吗

Answer 1

尝试下一个代码

const combined = [
  {
    "ageCode": 1,
    "ageDesc": "0-4",
    "qM": 358,
    "sM": 158,
    "qF": 328,
    "sF": 258
  },
  {
    "ageCode": 3,
    "ageDesc": "15-59",
    "qM": 525,
    "sM": 125
  },
  {
    "ageCode": 4,
    "ageDesc": "60+",
    "qF": 458,
    "sF": 358
  }
];

const original = combined.reduce((result, item) => {
  if (item.qM !== undefined) {
    result.push({
      "ageCode": item.ageCode,
      "ageDesc": item.ageDesc,
      "q": item.qM,
      "s": item.sM,
      "gender": "M"
    });
  }
  if (item.qF !== undefined) {
    result.push({
      "ageCode": item.ageCode,
      "ageDesc": item.ageDesc,
      "q": item.qF,
      "s": item.sF,
      "gender": "F"
    });
  }
  return result;
}, []);

console.log(original);

根据特定性别数据拆分组合列表用于 ID

Split a combined List based on particular gender data is present for ID

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