在 JavaScript 中简化对象的最佳方法是什么（最好使用 Jquery）

Question

I have an Object that looks like this:

var dataSource = [{
    date: new Date(1994, 2, 2),
    name: "a",
    l: 24.00,
    h: 25.00,
    o: 25.00,
    c: 24.875
}, {
    date: new Date(1994, 2, 2),
    name: "a",
    l: 23.625,
    h: 25.125,
    o: 24.00,
    c: 24.875
}, {
    date: new Date(1994, 2, 3),
    name: "a",
    l: 26.25,
    h: 28.25,
    o: 26.75,
    c: 27.00
}, {
    date: new Date(1994, 2, 4),
    name: "c",
    l: 26.50,
    h: 27.875,
    o: 26.875,
    c: 27.25
}, { 

and so on... I want to combine entries by their date, meaning if two datapoints have the same date and name, I want to add them together so the output would be:

var dataSource = [{
    date: new Date(1994, 2, 2),
    name: "a",
    l: 47.625,
    h: 50.125,
    o: 49.00,
    c: 49.75
}, {
    date: new Date(1994, 2, 3),
    name: "a",
    l: 26.25,
    h: 28.25,
    o: 26.75,
    c: 27.00
}, {
    date: new Date(1994, 2, 4),
    name: "c",
    l: 26.50,
    h: 27.875,
    o: 26.875,
    c: 27.25
}, { 

Right now the best way I can think of doing this would be a for loop that runs until the size of the object doesnt change any more. Is there a better way of doing this, possibly a jquery function similar to grep that could do this?

Answer 1

You can use a reduce() call grouping elements by a compound key made of the the date concatenated with the name ( o.date.valueOf() + o.name ), summing the relevant keys and then calling Object.values() on the result to return an array of the merged objects.

 const dataSource = [{ date: new Date(1994, 2, 2), name: "a", l: 24.00, h: 25.00, o: 25.00, c: 24.875 }, { date: new Date(1994, 2, 2), name: "a", l: 23.625, h: 25.125, o: 24.00, c: 24.875 }, { date: new Date(1994, 2, 3), name: "a", l: 26.25, h: 28.25, o: 26.75, c: 27.00 }, { date: new Date(1994, 2, 4), name: "c", l: 26.50, h: 27.875, o: 26.875, c: 27.25 }]; const sumKeys = (a, b) => ['l', 'h', 'o', 'c'].forEach(k => a[k] += b[k]), grouped = Object.values( dataSource.reduce((a, o) => { const entry = (a[o.date.valueOf() + o.name]??= { name: o.name, date: o.date.valueOf(), l: 0, h: 0, o: 0, c: 0 }); sumKeys(entry, o); return a; }, {})); console.log(grouped);

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or if you need to avoid logical nullish assignment (??=) for compatibility ...

 const dataSource = [{ date: new Date(1994, 2, 2), name: "a", l: 24.00, h: 25.00, o: 25.00, c: 24.875 }, { date: new Date(1994, 2, 2), name: "a", l: 23.625, h: 25.125, o: 24.00, c: 24.875 }, { date: new Date(1994, 2, 3), name: "a", l: 26.25, h: 28.25, o: 26.75, c: 27.00 }, { date: new Date(1994, 2, 4), name: "c", l: 26.50, h: 27.875, o: 26.875, c: 27.25 }]; const sumKeys = (a, b) => ['l', 'h', 'o', 'c'].forEach(k => a[k] += b[k]), grouped = Object.values( dataSource.reduce((a, o) => { const entry = (a[o.date.valueOf() + o.name] = a[o.date.valueOf() + o.name] || { name: o.name, date: o.date.valueOf(), l: 0, h: 0, o: 0, c: 0 }); sumKeys(entry, o); return a; }, {})); console.log(grouped);

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Answer 2

this way

 const data = [ { date: new Date(1994, 2, 2), name: "a", l: 24.00, h: 25.00, o: 25.00, c: 24.875 }, { date: new Date(1994, 2, 2), name: "a", l: 23.625, h: 25.125, o: 24.00, c: 24.875 }, { date: new Date(1994, 2, 3), name: "a", l: 26.25, h: 28.25, o: 26.75, c: 27.00 }, { date: new Date(1994, 2, 4), name: "c", l: 26.50, h: 27.875, o: 26.875, c: 27.25 } ] const Result = data.reduce((acc,c)=> { let same = acc.find(x=>x.date.getTime()===c.date.getTime() && x.name===c.name) if(.same) acc.push({...c}) else { same.l += c.l same.h += c.h same.o += c.o same.c += c,c } return acc }. []) console.log( Result )

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Answer 3

you can track the date and name in a mapper first

 var dataSource = [{ date: new Date(1994, 2, 2), name: "a", l: 24.00, h: 25.00, o: 25.00, c: 24.875 }, { date: new Date(1994, 2, 2), name: "a", l: 23.625, h: 25.125, o: 24.00, c: 24.875 }, { date: new Date(1994, 2, 3), name: "a", l: 26.25, h: 28.25, o: 26.75, c: 27.00 }, { date: new Date(1994, 2, 4), name: "c", l: 26.50, h: 27.875, o: 26.875, c: 27.25 }]; const mapper = dataSource.reduce((acc, cur) => { acc[cur.date] = acc[cur.date] || {}; acc[cur.date][cur.name] = acc[cur.date][cur.name] || {}; ["l", "h", "o", "c"].forEach(k => { acc[cur.date][cur.name][k] = (acc[cur.date][cur.name][k] || 0) + cur[k]; }); return acc; }, {}); const result = []; Object.keys(mapper).forEach(d => { Object.keys(mapper[d]).forEach(name => { const item = mapper[d][name]; result.push({ date: d, name, l: item.l, h: item.h, o: item.o, c: item.c, }); }); }); console.log(result);

Answer 4

  let arr = dataSource.reduce((cur, next) => {
    const flag = cur.some(ele => {
      return ele.date.valueOf() === next.date.valueOf() && ele.name === next.name
    })
    if(!flag) {
      cur.push(next)
    }
    return cur;
  }, [])
  console.log(arr)