Transforming a pair of lists into JSON format

Question

Transforming a pair of lists into JSON format

I'm currently working on converting two lists into JSON format.

Let's take a look at an example:

list1 = ['a','b','a']
list2 = ['q','r','s']

The conversion should result in:

[{
    "name": "g",
    "children": [{
            "name": "a",
            "children": [{
                "name": "q"
            }, {
                "name": "s"
            }]
        },
        {
            "name": "b",
            "children": [{
                "name": "r"
            }]
        }
    ]
}]

I have come close with the following code:

list1 = ['a','b','a']
list2 = ['q','r','s']

nameDict = {}
childrenDict = {}

list1 = list1.map(x => {
  return({name: x});
});
console.log(list1);

list2 = list2.map(x => {
  return({children: x});
});
console.log(list2);

var combined = list1.map(function(e, i) {
  return [e, list2[i]];
});

console.log(JSON.stringify(combined))

However, the output is currently:

[[{"name":"a"},{"children":"q"}],
[{"name":"b"},{"children":"r"}],
[{"name":"a"},{"children":"s"}]]

Any suggestions on how to combine these elements into the desired format?

[{
    "name": "g",
    "children": [{
            "name": "a",
            "children": [{
                "name": "q"
            }, {
                "name": "s"
            }]
        },
        {
            "name": "b",
            "children": [{
                "name": "r"
            }]
        }
    ]
}]

javascript json

Answer 1

Answer №1

Important Note: Due to the unknown origin of the letter g, I will focus solely on creating the root children array.

Given that both arrays are of equal length, a simple for loop can be used along with the array index to manipulate both arrays simultaneously. The approach involves constructing an array and checking during each iteration if the "child" already exists. If it does not, it is created.

l1 = ['a','b','a']
l2 = ['q','r','s']

let gChildren = []
for(let i = 0; i < l1.length; i++){
  let group = gChildren.find(c => c.name === l1[i])
  if(!group){
    group = { name: l1[i], children: [] }
    gChildren.push(group)
  }
  
  group.children.push({ name: l2[i] })
}

console.log(gChildren)

Answer 2

Important Note: Due to the unknown origin of the letter g, I will focus solely on creating the root children array.

Given that both arrays are of equal length, a simple for loop can be used along with the array index to manipulate both arrays simultaneously. The approach involves constructing an array and checking during each iteration if the "child" already exists. If it does not, it is created.

l1 = ['a','b','a']
l2 = ['q','r','s']

let gChildren = []
for(let i = 0; i < l1.length; i++){
  let group = gChildren.find(c => c.name === l1[i])
  if(!group){
    group = { name: l1[i], children: [] }
    gChildren.push(group)
  }
  
  group.children.push({ name: l2[i] })
}

console.log(gChildren)

Answer 3

Answer №2

If you are looking to achieve the desired outcome while preserving your existing structure, the following code snippet will do just that:

let data1 = ["x","y","z"];
let data2 = ["1","2","3"];

let result = [{name: "group", elements: []}];

for (let i=0;i < data1.length;i++) {
        let match = false;
    for (let j=0;j < result[0].elements.length;j++) {
        if (result[0].elements[j].name === data1[i]) {
            result[0].elements[j].elements.push({name: data2[i]});
        match = true;
        }
      }
      if (match === false) {
            result[0].elements.push({name: data1[i], elements: [{name: data2[i]}]});
        }

}

console.log(JSON.stringify(result));

Answer 4

If you are looking to achieve the desired outcome while preserving your existing structure, the following code snippet will do just that:

let data1 = ["x","y","z"];
let data2 = ["1","2","3"];

let result = [{name: "group", elements: []}];

for (let i=0;i < data1.length;i++) {
        let match = false;
    for (let j=0;j < result[0].elements.length;j++) {
        if (result[0].elements[j].name === data1[i]) {
            result[0].elements[j].elements.push({name: data2[i]});
        match = true;
        }
      }
      if (match === false) {
            result[0].elements.push({name: data1[i], elements: [{name: data2[i]}]});
        }

}

console.log(JSON.stringify(result));

Answer 5

Answer №3

Using Array.prototype.reduce is a great strategy when you need to loop through an array and consolidate all values into a single output.

list1.reduce((accumulator, value, index) => {
  const correspondingValue = list2[index]
  const obj = accumulator.find(item => item.name === value)
  if (obj) {
    obj.children.push({name: correspondingValue})
  } else {
    accumulator.push({
      name: value,
      children: [{name: correspondingValue}]
    })
  }

  return accumulator
}, [])

This code efficiently adds children to an existing item or creates a new item if needed during each loop iteration.

The variable g is not included in this code snippet as its purpose is unknown. However, you can integrate the resulting array from the reduce function into another object or array as required.

Answer 6

Using Array.prototype.reduce is a great strategy when you need to loop through an array and consolidate all values into a single output.

list1.reduce((accumulator, value, index) => {
  const correspondingValue = list2[index]
  const obj = accumulator.find(item => item.name === value)
  if (obj) {
    obj.children.push({name: correspondingValue})
  } else {
    accumulator.push({
      name: value,
      children: [{name: correspondingValue}]
    })
  }

  return accumulator
}, [])

This code efficiently adds children to an existing item or creates a new item if needed during each loop iteration.

The variable g is not included in this code snippet as its purpose is unknown. However, you can integrate the resulting array from the reduce function into another object or array as required.

Answer 7

Answer №4

If you want to rearrange the array and use the data as a path to the final child object, here is a way to do it:

arr1 = ['apple', 'banana', 'apple']
arr2 = ['queen', 'rook', 'knight']

After rearranging, it would look like this:

[
    ['apple', 'queen'], // apple -> queen
    ['banana', 'rook'], // banana -> rook
    ['apple', 'knight'] // apple -> knight
]

This arrangement can be used effectively with the reduce method.

One advantage of this method is that it can handle longer paths to the final children, such as:

[
    ['apple', 'x', 'y', 'z'],
    ...
]

When used with reduce, it returns a nested object with the specified relationship between the elements.

const
    rearrange = array => array.reduce((result, current) => current.map((value, index) => [...(result[index] || []), value]), []);

var arr1 = ['apple', 'banana', 'apple'],
    arr2 = ['queen', 'rook', 'knight'],
    finalResult = rearrange([arr1, arr2]).reduce((result, current) => {
        current.reduce((prev, element) => {
            var temp = (prev.children = prev.children || []).find(obj => obj.name === element);
            if (!temp) prev.children.push(temp = { name: element });
            return temp
        }, result);
        return result;
    }, { name: 'start' });

console.log(finalResult);

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

Answer 8

If you want to rearrange the array and use the data as a path to the final child object, here is a way to do it:

arr1 = ['apple', 'banana', 'apple']
arr2 = ['queen', 'rook', 'knight']

After rearranging, it would look like this:

[
    ['apple', 'queen'], // apple -> queen
    ['banana', 'rook'], // banana -> rook
    ['apple', 'knight'] // apple -> knight
]

This arrangement can be used effectively with the reduce method.

One advantage of this method is that it can handle longer paths to the final children, such as:

[
    ['apple', 'x', 'y', 'z'],
    ...
]

When used with reduce, it returns a nested object with the specified relationship between the elements.

const
    rearrange = array => array.reduce((result, current) => current.map((value, index) => [...(result[index] || []), value]), []);

var arr1 = ['apple', 'banana', 'apple'],
    arr2 = ['queen', 'rook', 'knight'],
    finalResult = rearrange([arr1, arr2]).reduce((result, current) => {
        current.reduce((prev, element) => {
            var temp = (prev.children = prev.children || []).find(obj => obj.name === element);
            if (!temp) prev.children.push(temp = { name: element });
            return temp
        }, result);
        return result;
    }, { name: 'start' });

console.log(finalResult);

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

Answer 9

Answer №5

A more concise version using distinct keys filter:

var list1 = ['a','b','a'], list2 = ['q','r','s']

var children = [...new Set(list1)].map(key => ({ name: key, children: 
        list2.filter((value, index) => list1[index] == key).map(value => ({ name: value })) }))

console.log( [{ name: 'group', children }] )

Alternatively, a more efficient approach with intermediate object groups:

var list1 = ['a','b','a'], list2 = ['q','r','s']

var children = Object.entries(
                       list1.reduce((obj, key, index) => ((obj[key] = obj[key] || []).push(list2[index]), obj), {})
                     ).map(([key, values]) => ({ name: key, children: values.map(value => ({ name: value })) }))

console.log( [{ name: 'group', children }] )

Answer 10

A more concise version using distinct keys filter:

var list1 = ['a','b','a'], list2 = ['q','r','s']

var children = [...new Set(list1)].map(key => ({ name: key, children: 
        list2.filter((value, index) => list1[index] == key).map(value => ({ name: value })) }))

console.log( [{ name: 'group', children }] )

Alternatively, a more efficient approach with intermediate object groups:

var list1 = ['a','b','a'], list2 = ['q','r','s']

var children = Object.entries(
                       list1.reduce((obj, key, index) => ((obj[key] = obj[key] || []).push(list2[index]), obj), {})
                     ).map(([key, values]) => ({ name: key, children: values.map(value => ({ name: value })) }))

console.log( [{ name: 'group', children }] )

Transforming a pair of lists into JSON format

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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