Manipulating information from one format to another

Question

Manipulating information from one format to another

I am currently tackling the task of calculating scores based on departments within groups. For simplicity, I will focus on just one group as an example.

Here is the data structure that I have:

const data = [{
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 80,
            "count": 1,
            "department": "Engineering",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 40,
            "count": 1,
            "department": "Executive",
            "group": "Group one"
        },
        ... (additional data entries)
    ];

My goal is to construct a data structure for a heatmap:

const heatmapData = [
{
  row: 'Group one',
  columns: [
    { 
      name: 'Engineering', 
      averageScore: 70,  
    },
    { 
      name: 'Supporting Department', 
      averageScore: 100,  
    },
    .... (more unique departments)
  ]
}
]

I am struggling to find a straightforward solution for grouping data and performing calculations. Any assistance or guidance would be greatly appreciated. Thank you!

javascript typescript

Answer 1

Answer №1

To organize your data, group it by the group attribute. Then, for each group, further categorize it based on the department and calculate the total score and count. Once you have this structure, you can compute the average score for each department.

const data = [{ "id": "cklt7ln1k0922o0sabjkk74m9", "score": 80, "count": 1, "department": "Engineering", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score": 40, "count": 1, "department": "Executive", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score": 40, "count": 1, "department": "OOO Boost", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score": 60, "count": 1, "department": "Engineering", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score...
// Code continues with data manipulation
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

To organize your data, group it by the group attribute. Then, for each group, further categorize it based on the department and calculate the total score and count. Once you have this structure, you can compute the average score for each department.

const data = [{ "id": "cklt7ln1k0922o0sabjkk74m9", "score": 80, "count": 1, "department": "Engineering", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score": 40, "count": 1, "department": "Executive", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score": 40, "count": 1, "department": "OOO Boost", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score": 60, "count": 1, "department": "Engineering", "group": "Group one" }, { "id": "cklt7ln1k0922o0sabjkk74m9", "score...
// Code continues with data manipulation
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

Answer №2

To organize elements into a dictionary, you can use the following code snippet:

let organizedData = data.reduce((accumulator, value) => {
  if(!accumulator[value.category]) accumulator[value.category] = {};
  if(!accumulator[value.category][value.section]) accumulator[value.category][value.section] = {total: 0, count: 0, avgScore: 0};
  accumulator[value.category][value.section] = {
    total: accumulator[value.category][value.section].total + value.score,
    count: accumulator[value.category][value.section].count + 1,
    avgScore: (accumulator[value.category][value.section].total + value.score) / (accumulator[value.category][value.section].count + 1)
  }
return accumulator
}, {});

Next step is to transform it into the requested format:

Object.keys(organizedData).map(category => {
  return {
    category: category,
    sections: Object.keys(organizedData[category]).map(section => {return {name: section, avgScore: organizedData[category][section].avgScore}})
  }
})

Answer 4

To organize elements into a dictionary, you can use the following code snippet:

let organizedData = data.reduce((accumulator, value) => {
  if(!accumulator[value.category]) accumulator[value.category] = {};
  if(!accumulator[value.category][value.section]) accumulator[value.category][value.section] = {total: 0, count: 0, avgScore: 0};
  accumulator[value.category][value.section] = {
    total: accumulator[value.category][value.section].total + value.score,
    count: accumulator[value.category][value.section].count + 1,
    avgScore: (accumulator[value.category][value.section].total + value.score) / (accumulator[value.category][value.section].count + 1)
  }
return accumulator
}, {});

Next step is to transform it into the requested format:

Object.keys(organizedData).map(category => {
  return {
    category: category,
    sections: Object.keys(organizedData[category]).map(section => {return {name: section, avgScore: organizedData[category][section].avgScore}})
  }
})

Answer 5

Answer №3

Let me break down how this solution functions.. Considering that objects act more as pointers than values, I leveraged this concept in my approach. I utilized an array to store data (where new elements are added), along with 2 Objects to manage values (modifying these Objects directly impacted the "values" in the array). By populating the array with values linked to these Objects, I was able to structure the data according to your specified format.

UPDATED ONCE MORE.. I incorporated the logic from your calculation examples >:D

function group(array){
  let arr=[]; let groups={}; let departments={}
  array.forEach(a=>{
    if(!groups[a.group]){ //defining each group
      groups[a.group]={row:a.group,columns:[]}
      arr.push(groups[a.group])
    }
    if(!departments[a.department]){ //establishing each department
      departments[a.department]=[{name:a.department,averageScore:a.score*a.count},a.count] //score * count and count(based on your calculation examples)
      groups[a.group].columns.push(departments[a.department][0])
    }
    else{ //updating existing departments
      departments[a.department][1]+=a.count //count
      departments[a.department][0].averageScore+=a.score*a.count //score * count
      //(based on your calculation examples)
    }
  })
  Object.keys(departments).forEach(a=>departments[a][0].averageScore/=departments[a][1])
  return arr
}

console.log(group(data))

<script>
//to retain unnecessary space in the answer
window.data = [{
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 80,
            "count": 1,
            "department": "Engineering",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 40,
            "count": 1,
            "department": "Executive",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 40,
            "count": 1,
            "department": "OOO Boost",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 60,
            "count": 1,
            "department": "Engineering",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 100,
            "count": 2,
            "department": "Supporting Department",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 20,
            "count": 1,
            "department": "Designers",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 20,
            "count": 1,
            "department": "Executive",
            "group": "Group one"
        }
    ];
</script>

Answer 6

Let me break down how this solution functions.. Considering that objects act more as pointers than values, I leveraged this concept in my approach. I utilized an array to store data (where new elements are added), along with 2 Objects to manage values (modifying these Objects directly impacted the "values" in the array). By populating the array with values linked to these Objects, I was able to structure the data according to your specified format.

UPDATED ONCE MORE.. I incorporated the logic from your calculation examples >:D

function group(array){
  let arr=[]; let groups={}; let departments={}
  array.forEach(a=>{
    if(!groups[a.group]){ //defining each group
      groups[a.group]={row:a.group,columns:[]}
      arr.push(groups[a.group])
    }
    if(!departments[a.department]){ //establishing each department
      departments[a.department]=[{name:a.department,averageScore:a.score*a.count},a.count] //score * count and count(based on your calculation examples)
      groups[a.group].columns.push(departments[a.department][0])
    }
    else{ //updating existing departments
      departments[a.department][1]+=a.count //count
      departments[a.department][0].averageScore+=a.score*a.count //score * count
      //(based on your calculation examples)
    }
  })
  Object.keys(departments).forEach(a=>departments[a][0].averageScore/=departments[a][1])
  return arr
}

console.log(group(data))

<script>
//to retain unnecessary space in the answer
window.data = [{
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 80,
            "count": 1,
            "department": "Engineering",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 40,
            "count": 1,
            "department": "Executive",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 40,
            "count": 1,
            "department": "OOO Boost",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 60,
            "count": 1,
            "department": "Engineering",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 100,
            "count": 2,
            "department": "Supporting Department",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 20,
            "count": 1,
            "department": "Designers",
            "group": "Group one"
        },
        {
            "id": "cklt7ln1k0922o0sabjkk74m9",
            "score": 20,
            "count": 1,
            "department": "Executive",
            "group": "Group one"
        }
    ];
</script>

Manipulating information from one format to another

Answer №1

Answer №2

Answer №3

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