What is the best way to simplify Objects in JavaScript (perferably using 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?

How does this have anything to do with jQuery? jQuery is for DOM manipulation, not general js stuff. — SuperStormer
– SuperStormer, Commented Jan 14, 2021 at 2:16
@SuperStormer I don't think so, They have lots of methods such as Ajax which is just an xhr call, but less code to write. — Luke Vanzweden
– Luke Vanzweden, Commented Jan 14, 2021 at 2:26
This is more of lodash/underscore's domain, if any at all. Also, Zawinski's law. — SuperStormer
– SuperStormer, Commented Jan 14, 2021 at 22:10

pilchard · Accepted Answer · 2021-01-14 02:54:27Z

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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D. Seah · Accepted Answer · 2021-01-14 02:15:00Z

0

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);

answered Jan 14, 2021 at 2:15

D. Seah

4,7121 gold badge15 silver badges27 bronze badges

1 Comment

Luke Vanzweden Over a year ago

This answer works, but the other answer is less code and faster, so I will use it instead

linchong · Accepted Answer · 2021-01-14 02:37:09Z

0

  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)

answered Jan 14, 2021 at 2:37

linchong

5032 silver badges4 bronze badges

2 Comments

pilchard Over a year ago

This just skips over matches, it doesn't sum them.

Borja Over a year ago

While this code may provide a solution to the question, it's better to add context as to why/how it works. This can help future users learn, and apply that knowledge to their own code. You are also likely to have positive feedback from users in the form of upvotes, when the code is explained.

Mister Jojo · Accepted Answer · 2021-01-14 03:10:03Z

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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