按多个键分组的逻辑答案

【问题标题】：Logic for Group by multiple keys按多个键分组的逻辑
【发布时间】：2020-09-29 09:13:31
【问题描述】：

输入

[{"DefectsCount":1,"Release":"5.7.1.0","SWID":"23","AgeBin":"0-20","BinStart":0,"ReleaseSWID":"5.6.1.1.23"},{"DefectsCount":1,"Release":"5.7.1.0","SWID":"172","AgeBin":"0-20","BinStart":0,"ReleaseSWID":"5.7.1.0.172"},{"DefectsCount":1,"Release":"5.6.1.2","SWID":null,"AgeBin":"0-21","BinStart":0}]

输出

[
{"DefectsCount":1,"AgeBin":"0-21","Release":"5.6.1.2"},

{"DefectsCount":2,"AgeBin":"0-20","Release":"5.7.1.0"},

]

需要按 AgeBin 和 Release 分组并获取它的计数

项目中不允许Loadash，请在for循环中说一下怎么做

【问题讨论】：

标签： javascript for-loop group-by count logic

【解决方案1】：

解决此问题的另一种方法是从AgeBin 和Release 值中创建一个键，并使用reduce 通过该键对DefectsCount 求和。然后可以使用map 将该对象拆分为结果数组：

const data = [{
    DefectsCount: 1,
    Release: "5.7.1.0",
    SWID: "23",
    AgeBin: "0-20",
    BinStart: 0,
    ReleaseSWID: "5.6.1.1.23",
  },
  {
    DefectsCount: 1,
    Release: "5.7.1.0",
    SWID: "172",
    AgeBin: "0-20",
    BinStart: 0,
    ReleaseSWID: "5.7.1.0.172",
  },
  {
    DefectsCount: 1,
    Release: "5.6.1.2",
    SWID: null,
    AgeBin: "0-21",
    BinStart: 0,
  },
]

const out = Object.entries(data.reduce((c, o) => {
    const group = o.AgeBin + '#' + o.Release;
    c[group] = (c[group] || 0) + o.DefectsCount;
    return c;
  }, {}))
  .map(a => {
    [AgeBin, Release] = a[0].split('#');
    DefectsCount = a[1];
    return {
      AgeBin,
      Release,
      DefectsCount
    }
  });

console.log(out)

【讨论】：

【解决方案2】：

您可以使用reduce 和findIndex 来累积最终结果。每次迭代，检查累积结果中是否已经有 AgeBin 和 Release 的元素，然后添加或修改该元素

data.reduce((acc, el) => {
  const groupIndex = acc.findIndex(
    (g) => g.AgeBin === el.AgeBin && g.Release === el.Release
  )
  if (groupIndex >= 0) {
    acc[groupIndex] = {
      ...acc[groupIndex],
      DefectsCount: acc[groupIndex].DefectsCount + el.DefectsCount,
    }
  } else {
    acc.push({
      DefectsCount: el.DefectsCount,
      AgeBin: el.AgeBin,
      Release: el.Release,
    })
  }

  return acc
}, [])

可运行示例

const data = [
  {
    DefectsCount: 1,
    Release: "5.7.1.0",
    SWID: "23",
    AgeBin: "0-20",
    BinStart: 0,
    ReleaseSWID: "5.6.1.1.23",
  },
  {
    DefectsCount: 1,
    Release: "5.7.1.0",
    SWID: "172",
    AgeBin: "0-20",
    BinStart: 0,
    ReleaseSWID: "5.7.1.0.172",
  },
  {
    DefectsCount: 1,
    Release: "5.6.1.2",
    SWID: null,
    AgeBin: "0-21",
    BinStart: 0,
  },
]

const res = data.reduce((acc, el) => {
  const groupIndex = acc.findIndex(
    (g) => g.AgeBin === el.AgeBin && g.Release === el.Release
  )
  if (groupIndex >= 0) {
    acc[groupIndex] = {
      ...acc[groupIndex],
      DefectsCount: acc[groupIndex].DefectsCount + el.DefectsCount,
    }
  } else {
    acc.push({
      DefectsCount: el.DefectsCount,
      AgeBin: el.AgeBin,
      Release: el.Release,
    })
  }

  return acc
}, [])

console.log(res)

【讨论】：