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
asked Nov 16 '18 at 11:46
OmarOmar
1114
|
show 2 more comments
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
asked Nov 16 '18 at 11:46
OmarOmar
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
|
show 2 more comments
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
asked Nov 16 '18 at 11:46
OmarOmar
1114
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
asked Nov 16 '18 at 11:46
OmarOmar
1114
asked Nov 16 '18 at 11:46
OmarOmar
1114
edited Nov 17 '18 at 20:29
Omar
asked Nov 16 '18 at 11:46
OmarOmar
1114
asked Nov 16 '18 at 11:46
OmarOmar
1114
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
|
show 2 more comments
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
|
show 2 more comments
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[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; answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
This works perfectly, Thanks
– Omar
Nov 16 '18 at 12:36
add a comment |
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)answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
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
add a comment |
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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[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; answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
This works perfectly, Thanks
– Omar
Nov 16 '18 at 12:36
add a comment |
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[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; answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
This works perfectly, Thanks
– Omar
Nov 16 '18 at 12:36
add a comment |
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[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; answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
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[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[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[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; answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
answered Nov 16 '18 at 12:22
Nina ScholzNina Scholz
196k15108179
196k15108179
This works perfectly, Thanks
– Omar
Nov 16 '18 at 12:36
add a comment |
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
add a comment |
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)answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
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
add a comment |
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)answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
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
add a comment |
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)answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
4,8341533
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)answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
4,8341533
edited Nov 16 '18 at 13:30
answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
4,8341533
answered Nov 16 '18 at 12:21
Nurbol AlpysbayevNurbol Alpysbayev
4,8341533
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
add a comment |
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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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