JS Loop through array based on key values and add values

Question

I have the following object inside an array:-

[
 {"score": 5, "question": 0, "weight": 2},
 {"score": 4, "question": 1, "weight": 2},
 {"score": 3, "question": 0, "weight": 4},
 {"score": 4, "question": 1, "weight": 4},
 {"score": 2, "question": 2, "weight": 4},
 {"score": 8, "question": 0, "weight": 2}
]

I am trying to loop through the array so I have the following output, so I am able to run some math against the results:-

[
  [
    {"score": 5, "question": 0, "weight": 2},
    {"score": 4, "question": 1, "weight": 2}
  ],
  [
    {"score": 3, "question": 0, "weight": 4},
    {"score": 4, "question": 1, "weight": 4},
    {"score": 2, "question": 2, "weight": 4}
  ],
  [
    {"score": 8, "question": 0, "weight": 2}
  ]
];

Is there a dynamic way I am able to get array1 to look like array2?

I am using flat JS for this please no jQuery answers.

Thanks in advance.

** Note **

Sometimes each section will have more or less values, this is why I require it to be dynamic.

@Slai, no, because it should group same weights, not collect some counts. — Nina Scholz
– Nina Scholz, Commented Mar 11, 2018 at 14:38
It's not grouping by weights as they could all be group 2 for example. It needs to loop through the questions and if it resets back to 0 start a new array inside the larger array. The main identifier is the question id. — Tony Hensler
– Tony Hensler, Commented Mar 11, 2018 at 14:42
@TonyHensler it should group same weights then why group1 and group3 are different as for both group weight is 2. — yajiv
– yajiv, Commented Mar 11, 2018 at 14:46
@yajiv it's splitting by sequences that start with "question": 0 — Slai
– Slai, Commented Mar 11, 2018 at 14:53

Nenad Vracar · Accepted Answer · 2018-03-11 14:44:56Z

2

You can do this with reduce() method you just need to keep track of current index for final array.

const data =[
  {score: 5, question: 0, weight: 2},
  {score: 4, question: 1, weight: 2},
  {score: 3, question: 0, weight: 4},
  {score: 4, question: 1, weight: 4},
  {score: 2, question: 2, weight: 4},
  {score: 8, question: 0, weight: 2}
]

const result = data.reduce(function(r, e, i) {
  if(i == 0) r = {values: [], counter: 0}
  if(e.question == 0 && i != 0) r.counter++
  if(!r.values[r.counter]) r.values[r.counter] = [e]
  else r.values[r.counter].push(e)
  return r;
}, {}).values

console.log(result)

answered Mar 11, 2018 at 14:44

Nenad Vracar

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Comments

Nina Scholz · Accepted Answer · 2018-03-11 14:47:06Z

1

You could check weight and if different, then take a new group.

var data = [{ score: 5, question: 0, weight: 2 }, { score: 4, question: 1, weight: 2 }, { score: 3, question: 0, weight: 4 }, { score: 4, question: 1, weight: 4 }, { score: 2, question: 2, weight: 4 }, { score: 8, question: 0, weight: 2 }],
    grouped = data.reduce(function (r, o, i, a) {
        if ((a[i - 1] || {}).weight !== o.weight) {
            r.push([]);
        }
        r[r.length - 1].push(o);
        return r;
    }, []);

console.log(grouped);

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answered Mar 11, 2018 at 14:47

Nina Scholz

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Comments

Ori Drori · Accepted Answer · 2018-03-11 14:47:51Z

0

You can use Array.reduce() to convert the 1st to the 2nd array. Whenever the current object weight doesn't match lastWeight, add another subarray. Always push the current item to the last subarray:

const arr = [
  {score: 5, question: 0, weight: 2},
  {score: 4, question: 1, weight: 2},
  {score: 3, question: 0, weight: 4},
  {score: 4, question: 1, weight: 4},
  {score: 2, question: 2, weight: 4},
  {score: 8, question: 0, weight: 2}
];

let lastWeight = null;
const result = arr.reduce((r, o) => {
  if(lastWeight === null || o.weight !== lastWeight) {
    lastWeight = o.weight;
    r.push([]);
  }
  r[r.length - 1].push(o);
  
  return r;
}, []);

console.log(result);

answered Mar 11, 2018 at 14:47

Ori Drori

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Comments

Igor · Accepted Answer · 2018-03-11 14:48:15Z

0

var oldA = [
 {"score": 5, "question": 0, "weight": 2},
 {"score": 4, "question": 1, "weight": 2},
 {"score": 3, "question": 0, "weight": 4},
 {"score": 4, "question": 1, "weight": 4},
 {"score": 2, "question": 2, "weight": 4},
 {"score": 8, "question": 0, "weight": 2}
];

var newA = [];
var prevW = 0;
var prevA;

for (var i = 0; i < oldA.length; i++) {
  if (oldA[i].weight != prevW) {
    prevA = [];
    newA.push(prevA);
    prevW = oldA[i].weight;
  }
  prevA.push(oldA[i]);
}

console.log(newA);

answered Mar 11, 2018 at 14:48

Igor

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Comments

Hassan Imam · Accepted Answer · 2018-03-11 14:52:15Z

0

You can use array#reduce to group your array. Check if the value of question is 0 then push a new array and add the object to it.

var data = [ {"score": 5, "question": 0, "weight": 2}, {"score": 4, "question": 1, "weight": 2}, {"score": 3, "question": 0, "weight": 4}, {"score": 4, "question": 1, "weight": 4}, {"score": 2, "question": 2, "weight": 4}, {"score": 8, "question": 0, "weight": 2}],
    result = data.reduce((r,o) => {
      if(o.question == 0)
        r.push([]);
      r[r.length - 1].push(o);
      return r;
    },[]);
console.log(result);

answered Mar 11, 2018 at 14:52

Hassan Imam

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Comments

connexo · Accepted Answer · 2018-03-11 14:52:56Z

0

let scores = [
 {"score": 5, "question": 0, "weight": 2},
 {"score": 4, "question": 1, "weight": 2},
 {"score": 3, "question": 0, "weight": 4},
 {"score": 4, "question": 1, "weight": 4},
 {"score": 2, "question": 2, "weight": 4},
 {"score": 8, "question": 0, "weight": 2}
]

let groupedScores = [], group = [];

scores.forEach((entry) => {
  if(entry.question === 0) { 
    if (group.length) { 
      groupedScores.push(group); 
    }
    group = []; 
  }
  group.push(entry);
})
groupedScores.push(group)

console.log(groupedScores)

answered Mar 11, 2018 at 14:52

connexo

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Comments

Slai · Accepted Answer · 2018-03-11 15:22:59Z

0

A bit simplified if splitting by just question: 0 :

data = [ { score: 5, question: 0, weight: 2 }, { score: 4, question: 1, weight: 2 },
         { score: 3, question: 0, weight: 4 }, { score: 4, question: 1, weight: 4 },
         { score: 2, question: 2, weight: 4 }, { score: 8, question: 0, weight: 2 } ]

result = data.reduce((r, v) => (v.question ? r[r.length-1].push(v) : r.push([v]), r), [])

console.log( result )

answered Mar 11, 2018 at 15:22

Slai

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