How to create a tree-like structure from a JavaScript Object





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1















This is the Object sample I'm given:



Input :




{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }




I want to transform it to a JSON format like this: 



{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}


Any Ideas/Help would be appreciated






javascript json object







share|improve this question




















  • 2





    Can you show us what your attempt looked like?

    – SpoonMeiser
    Nov 16 '18 at 11:47











  • Wait a minute..

    – Nurbol Alpysbayev
    Nov 16 '18 at 11:57











  • I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

    – Omar
    Nov 16 '18 at 12:12













  • I almost finished it... it's taking too long.. :)

    – Nurbol Alpysbayev
    Nov 16 '18 at 12:12











  • I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

    – Patrik Valkovič
    Nov 16 '18 at 12:13




















1















This is the Object sample I'm given:



Input :




{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }




I want to transform it to a JSON format like this: 



{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}


Any Ideas/Help would be appreciated






javascript json object







share|improve this question




















  • 2





    Can you show us what your attempt looked like?

    – SpoonMeiser
    Nov 16 '18 at 11:47











  • Wait a minute..

    – Nurbol Alpysbayev
    Nov 16 '18 at 11:57











  • I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

    – Omar
    Nov 16 '18 at 12:12













  • I almost finished it... it's taking too long.. :)

    – Nurbol Alpysbayev
    Nov 16 '18 at 12:12











  • I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

    – Patrik Valkovič
    Nov 16 '18 at 12:13
















1












1








1


1






This is the Object sample I'm given:



Input :




{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }




I want to transform it to a JSON format like this: 



{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}


Any Ideas/Help would be appreciated






javascript json object







share|improve this question
















This is the Object sample I'm given:



Input :




{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }




I want to transform it to a JSON format like this: 



{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}


Any Ideas/Help would be appreciated






javascript json object




javascript json object






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 17 '18 at 20:29







Omar

















asked Nov 16 '18 at 11:46









OmarOmar

1114




1114








  • 2





    Can you show us what your attempt looked like?

    – SpoonMeiser
    Nov 16 '18 at 11:47











  • Wait a minute..

    – Nurbol Alpysbayev
    Nov 16 '18 at 11:57











  • I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

    – Omar
    Nov 16 '18 at 12:12













  • I almost finished it... it's taking too long.. :)

    – Nurbol Alpysbayev
    Nov 16 '18 at 12:12











  • I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

    – Patrik Valkovič
    Nov 16 '18 at 12:13
















  • 2





    Can you show us what your attempt looked like?

    – SpoonMeiser
    Nov 16 '18 at 11:47











  • Wait a minute..

    – Nurbol Alpysbayev
    Nov 16 '18 at 11:57











  • I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

    – Omar
    Nov 16 '18 at 12:12













  • I almost finished it... it's taking too long.. :)

    – Nurbol Alpysbayev
    Nov 16 '18 at 12:12











  • I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

    – Patrik Valkovič
    Nov 16 '18 at 12:13










2




2





Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47





Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47













Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57





Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57













I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12







I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12















I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12





I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12













I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13







I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13














2 Answers
2






active

oldest

votes


















1














You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.






var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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








share|improve this answer
























  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36



















0














OMG this seemingly simple stuff took so long






const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)








share|improve this answer


























  • you could add your answer as a code snipet

    – Dhaval Pankhaniya
    Nov 16 '18 at 12:27











  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36











  • @Hussler you could at least upvote it if it works...

    – Nurbol Alpysbayev
    Nov 16 '18 at 13:26











  • @NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

    – Omar
    Nov 16 '18 at 13:50












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.






var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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








share|improve this answer
























  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36
















1














You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.






var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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








share|improve this answer
























  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36














1












1








1







You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.






var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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








share|improve this answer













You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.






var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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








var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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





var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

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






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 16 '18 at 12:22









Nina ScholzNina Scholz

196k15108179




196k15108179













  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36



















  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36

















This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36





This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36













0














OMG this seemingly simple stuff took so long






const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)








share|improve this answer


























  • you could add your answer as a code snipet

    – Dhaval Pankhaniya
    Nov 16 '18 at 12:27











  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36











  • @Hussler you could at least upvote it if it works...

    – Nurbol Alpysbayev
    Nov 16 '18 at 13:26











  • @NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

    – Omar
    Nov 16 '18 at 13:50
















0














OMG this seemingly simple stuff took so long






const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)








share|improve this answer


























  • you could add your answer as a code snipet

    – Dhaval Pankhaniya
    Nov 16 '18 at 12:27











  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36











  • @Hussler you could at least upvote it if it works...

    – Nurbol Alpysbayev
    Nov 16 '18 at 13:26











  • @NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

    – Omar
    Nov 16 '18 at 13:50














0












0








0







OMG this seemingly simple stuff took so long






const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)








share|improve this answer















OMG this seemingly simple stuff took so long






const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)








const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)





const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 16 '18 at 13:30

























answered Nov 16 '18 at 12:21









Nurbol AlpysbayevNurbol Alpysbayev

4,8341533




4,8341533













  • you could add your answer as a code snipet

    – Dhaval Pankhaniya
    Nov 16 '18 at 12:27











  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36











  • @Hussler you could at least upvote it if it works...

    – Nurbol Alpysbayev
    Nov 16 '18 at 13:26











  • @NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

    – Omar
    Nov 16 '18 at 13:50



















  • you could add your answer as a code snipet

    – Dhaval Pankhaniya
    Nov 16 '18 at 12:27











  • This works perfectly, Thanks

    – Omar
    Nov 16 '18 at 12:36











  • @Hussler you could at least upvote it if it works...

    – Nurbol Alpysbayev
    Nov 16 '18 at 13:26











  • @NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

    – Omar
    Nov 16 '18 at 13:50

















you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27





you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27













This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36





This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36













@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26





@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26













@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50





@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50


















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