Group and count values inside array with nested object

Question

[{
    "id":35,
    "aktivis":{
        "id":450,
        "name":"A. Sri Nugroho Hadi",
        "gender": "male",
        "active_job":{
            "id":491,
            "id_aktivis":450,
            "id_company":20,
            "company":{
                "id":20,
                "name":"Company 1"
            }
        }
    }
},
{
    "id":36,
    "aktivis":{
        "id":451,
        "name":"Sri ",
        "gender": "female",
        "active_job":{
            "id":492,
            "id_aktivis":451,
            "id_company":21,
            "company":{
                "id":21,
                "name":"compnay 2"
            }
        }
    }
}]

so after my previous question and there is the anwser in Group and count values in an array

so it answer my question but then i get to the another problem that i didn't realize i did have. What if i want to show my array like this?

[{
        "id":20,
        "name":"Company 1"
        "employee_count":1
        "female":0
        "male":1
    },
    {
        "id":22,
        "name":"Company 2"
        "employee_count":1
        "female":1
        "male":0
}]

What have you tried so far?

dbramwell
– dbramwell

2020-04-26 09:34:05 +00:00
Commented Apr 26, 2020 at 9:34 — dbramwell
– dbramwell, Commented Apr 26, 2020 at 9:34

dellink · Accepted Answer · 2020-04-26 09:40:42Z

1

Object.values([YOUR ARRAY].reduce(function (accumulator, current) {
     const company = current.aktivis.active_job.company;
     accumulator[company.id] = accumulator[company.id] || {
       "id": company.id,
        "name": company.name,
        "employee_count": 0,
        "female": 0,
        "male": 0,
     };

     accumulator[company.id].employee_count += 1;
     if (current.aktivis.gender === 'male') {
       accumulator[company.id].male += 1;
     } else {
       accumulator[company.id].female += 1;
     }

     return accumulator;
   }, {}));

answered Apr 26, 2020 at 9:40

dellink

5493 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

jasper · Accepted Answer · 2020-04-26 09:35:27Z

Normally I'd create hashmap of company id to data points but since you ask for an array specifically we can also use the .findIndex method to see if a company exists.

const data = [{
    "id":35,
    "aktivis":{
        "id":450,
        "name":"A. Sri Nugroho Hadi",
        "gender": "male",
        "active_job":{
            "id":491,
            "id_aktivis":450,
            "id_company":20,
            "company":{
                "id":20,
                "name":"Company 1"
                }
            }
        }
    },
    {
    "id":36,
    "aktivis":{
        "id":451,
        "name":"Sri ",
        "gender": "female",
        "active_job":{
            "id":492,
            "id_aktivis":451,
            "id_company":21,
            "company":{
                "id":21,
                "name":"compnay 2"
                }
            }
        }
    }
]

let companies = [];

data.forEach(employee => {
    const sex = employee.aktivis.gender;
    const jobData = employee.aktivis.active_job;
    const companyIndex = companies.findIndex(c => c.id === jobData.company.id);
    const companyData = companies[companyIndex]
    if (!companyData) {
        // first time encountering this company need to create a new object
        companies.push({
            id: jobData.company.id,
            name: jobData.company.name,
            employee_count: 1,
            female: sex === "female" ? 1 : 0,
            male: sex === "male" ? 1 : 0,
        })
    } else {
        // company already exists in our array, just need to increment values
        companies[companyIndex][sex]++
        companies[companyIndex].employee_count++
    }
})

Collectives™ on Stack Overflow

Group and count values inside array with nested object

2 Answers 2

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related