如何将一些数据减少到一个新的对象数组中？

Question

更新我尝试了自己的解决方案，这就是我所做的也许可以做得更好

 const dbData = [{ studyId: 'X', siteId: 'A', day: '2000-01-01', status: 'PENDING_CALLCENTER', current: 5, total: 17, }, { studyId: 'X', siteId: 'A', day: '2000-01-01', status: 'PENDING_SITE', current: 3, total: 9, }, { studyId: 'Y', siteId: 'B', day: '2000-01-01', status: 'PENDING_SITE', current: 3, total: 9, }, { studyId: 'Y', siteId: 'B', day: '2000-01-01', status: 'PENDING_SITE', current: 3, total: 9, }, { studyId: 'Y', siteId: 'B', day: '2000-01-01', status: 'PENDING_CALLCENTER', current: 3, total: 9, }, ]; const reduced = dbData.reduce((acc, row) => { const { studyId, siteId, status, current, total } = row; const idx = acc.findIndex(x => studyId === x.studyId && siteId === x.siteId); const item = idx === -1 ? { studyId, siteId, currents: {}, totals: {} } : { ...acc[idx] }; item.currents[status] = item.currents[status] ? item.currents[status] + current : current; item.totals[status] = item.totals[status] ? item.totals[status] + total : total; if (idx === -1) { acc.push(item); } else { acc[idx] = item; } return acc; }, []); console.log(reduced);

我是使用 reduce 的新手，我在理解如何正确使用它时遇到了一些困难。

我想解决一个问题，我需要在具有不同形状的新 obj 数组中减少一些数据。

我有一些数据

[
      {
        studyId: 'X',
        siteId: 'A',
        day: '2000-01-01',
        status: 'PENDING_CALLCENTER',
        current: 5,
        total: 17,
      },
      {
        studyId: 'X',
        siteId: 'A',
        day: '2000-01-01',
        status: 'PENDING_SITE',
        current: 3,
        total: 9,
      },
    ];

我想把它减少到这个新数据

[
      {
        studyId: 'X',
        siteId: 'A',
        currents: {
          PENDING_CALLCENTER: 5,
          PENDING_SITE: 3,
        },
        totals: {
          PENDING_CALLCENTER: 17,
          PENDING_SITE: 9,
        },
      },
    ];

为了学习，我能够创建一个计算总和的减速器，并且我做到了，但上述问题我很难开始并想理解它。

Answer 1

你可以尝试这样的事情：

请注意 - 我没有考虑验证/异常处理。 这只是对数据使用 reduce 的一种方式。 您还可以找到一种减少代码行的可爱方法 - 随意:-)

const arr = [
  {
    studyId: 'X',
    siteId: 'A',
    day: '2000-01-01',
    status: 'PENDING_CALLCENTER',
    current: 5,
    total: 17,
  },
  {
    studyId: 'X',
    siteId: 'A',
    day: '2000-01-01',
    status: 'PENDING_SITE',
    current: 3,
    total: 9,
  },
];

const res = arr.reduce((accum, curr) => {
  const exist = accum.find(e => e.studyId === curr.studyId && e.siteId === curr.siteId);
  if (exist) {
    exist.currents[curr.status] += curr.current;
    exist.totals[curr.status] += curr.total;
  } else {
    const elem = {
      studyId: curr.studyId,
      siteId: curr.siteId,
      currents: {
        PENDING_CALLCENTER: 0,
        PENDING_SITE: 0,
      },
      totals: {
        PENDING_CALLCENTER: 0,
        PENDING_SITE: 0,
      }
    }
    elem.currents[curr.status] = curr.current;
    elem.totals[curr.status] = curr.total;
    accum.push(elem);
  }
  return accum;
}, []);

console.log(res);

输出：

[
  {
    studyId: 'X',
    siteId: 'A',
    currents: { PENDING_CALLCENTER: 5, PENDING_SITE: 3 },
    totals: { PENDING_CALLCENTER: 17, PENDING_SITE: 9 }
  }
]

Answer 2

下面的currents功能只是合并了相同 ID 的对象的当前值和totals值。 因此，对于由 OP 提供的重复项（如Y / B的重复项），相同属性的数值将被覆盖但不会相加。

 function mergeCurrentsAndTotalsOfSameIds({ lookup = new Map, result }, item) { const { studyId, siteId, status, current, total } = item; const mergerKey = [studyId, '_###_', siteId].join(''); let merger = lookup.get(mergerKey); if (!merger) { merger = { studyId, siteId, currents: {}, totals: {} }; lookup.set(mergerKey, merger); result.push(merger); } merger.currents[status] = current; merger.totals[status] = total; return { lookup, result }; } const dbData = [{ studyId: 'X', siteId: 'A', day: '2000-01-01', status: 'PENDING_CALLCENTER', current: 5, total: 17, }, { studyId: 'X', siteId: 'A', day: '2000-01-01', status: 'PENDING_SITE', current: 3, total: 9, }, { studyId: 'Y', siteId: 'B', day: '2000-01-01', status: 'PENDING_SITE', current: 3, total: 9, }, { studyId: 'Y', siteId: 'B', day: '2000-01-01', status: 'PENDING_SITE', current: 3, total: 9, }, { studyId: 'Y', siteId: 'B', day: '2000-01-01', status: 'PENDING_CALLCENTER', current: 3, total: 9, }]; const { result: mergedData } = dbData .reduce(mergeCurrentsAndTotalsOfSameIds, {result: [] }) console.log({ mergedData, dbData });

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