Manipulating arrays to extract unique objects and count occurrences











up vote
-1
down vote

favorite












The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.




  • Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?

  • I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?





// Original array
const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
console.log(arr);

// Step 1: Extract meaningful information from original array
const arrOfArrOfObj = arr
.map(category => category.key1
.map(subCategory => (
{
id: subCategory.key2.id,
name: subCategory.key2.name
}
)
)
);
console.log(arrOfArrOfObj);

// Step 2: Make the array one dimensional
const arrOfObj = .concat(...arrOfArrOfObj);
console.log(arrOfObj);

// Step 3: Remove duplicates and count object occurrences.
let dedupedArrWithCount = ;
l = arrOfObj.length;
for (let i = 0; i < l; i++) {
let objExists = false;
for (let j = 0; j < dedupedArrWithCount.length; j++) {
// Two objects are identical if their ids are identical.
if (arrOfObj[i].id === dedupedArrWithCount[j].id) {
objExists = true;
dedupedArrWithCount[j].count += 1;
}
}
if (!objExists) {
dedupedArrWithCount.push({
id: arrOfObj[i].id,
name: arrOfObj[i].name,
count: 1
})
}
}

console.log(dedupedArrWithCount);












share|improve this question




















  • 1




    what does this data describe? will there always be two levels or might there ever be 3 or more?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 24 at 21:42










  • The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.
    – Jelefra
    Oct 24 at 21:48










  • so does that mean that there will only ever be two levels of keys?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 26 at 20:11










  • Yes that's the case.
    – Jelefra
    Oct 27 at 14:10















up vote
-1
down vote

favorite












The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.




  • Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?

  • I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?





// Original array
const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
console.log(arr);

// Step 1: Extract meaningful information from original array
const arrOfArrOfObj = arr
.map(category => category.key1
.map(subCategory => (
{
id: subCategory.key2.id,
name: subCategory.key2.name
}
)
)
);
console.log(arrOfArrOfObj);

// Step 2: Make the array one dimensional
const arrOfObj = .concat(...arrOfArrOfObj);
console.log(arrOfObj);

// Step 3: Remove duplicates and count object occurrences.
let dedupedArrWithCount = ;
l = arrOfObj.length;
for (let i = 0; i < l; i++) {
let objExists = false;
for (let j = 0; j < dedupedArrWithCount.length; j++) {
// Two objects are identical if their ids are identical.
if (arrOfObj[i].id === dedupedArrWithCount[j].id) {
objExists = true;
dedupedArrWithCount[j].count += 1;
}
}
if (!objExists) {
dedupedArrWithCount.push({
id: arrOfObj[i].id,
name: arrOfObj[i].name,
count: 1
})
}
}

console.log(dedupedArrWithCount);












share|improve this question




















  • 1




    what does this data describe? will there always be two levels or might there ever be 3 or more?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 24 at 21:42










  • The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.
    – Jelefra
    Oct 24 at 21:48










  • so does that mean that there will only ever be two levels of keys?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 26 at 20:11










  • Yes that's the case.
    – Jelefra
    Oct 27 at 14:10













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.




  • Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?

  • I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?





// Original array
const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
console.log(arr);

// Step 1: Extract meaningful information from original array
const arrOfArrOfObj = arr
.map(category => category.key1
.map(subCategory => (
{
id: subCategory.key2.id,
name: subCategory.key2.name
}
)
)
);
console.log(arrOfArrOfObj);

// Step 2: Make the array one dimensional
const arrOfObj = .concat(...arrOfArrOfObj);
console.log(arrOfObj);

// Step 3: Remove duplicates and count object occurrences.
let dedupedArrWithCount = ;
l = arrOfObj.length;
for (let i = 0; i < l; i++) {
let objExists = false;
for (let j = 0; j < dedupedArrWithCount.length; j++) {
// Two objects are identical if their ids are identical.
if (arrOfObj[i].id === dedupedArrWithCount[j].id) {
objExists = true;
dedupedArrWithCount[j].count += 1;
}
}
if (!objExists) {
dedupedArrWithCount.push({
id: arrOfObj[i].id,
name: arrOfObj[i].name,
count: 1
})
}
}

console.log(dedupedArrWithCount);












share|improve this question















The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.




  • Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?

  • I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?





// Original array
const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
console.log(arr);

// Step 1: Extract meaningful information from original array
const arrOfArrOfObj = arr
.map(category => category.key1
.map(subCategory => (
{
id: subCategory.key2.id,
name: subCategory.key2.name
}
)
)
);
console.log(arrOfArrOfObj);

// Step 2: Make the array one dimensional
const arrOfObj = .concat(...arrOfArrOfObj);
console.log(arrOfObj);

// Step 3: Remove duplicates and count object occurrences.
let dedupedArrWithCount = ;
l = arrOfObj.length;
for (let i = 0; i < l; i++) {
let objExists = false;
for (let j = 0; j < dedupedArrWithCount.length; j++) {
// Two objects are identical if their ids are identical.
if (arrOfObj[i].id === dedupedArrWithCount[j].id) {
objExists = true;
dedupedArrWithCount[j].count += 1;
}
}
if (!objExists) {
dedupedArrWithCount.push({
id: arrOfObj[i].id,
name: arrOfObj[i].name,
count: 1
})
}
}

console.log(dedupedArrWithCount);








// Original array
const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
console.log(arr);

// Step 1: Extract meaningful information from original array
const arrOfArrOfObj = arr
.map(category => category.key1
.map(subCategory => (
{
id: subCategory.key2.id,
name: subCategory.key2.name
}
)
)
);
console.log(arrOfArrOfObj);

// Step 2: Make the array one dimensional
const arrOfObj = .concat(...arrOfArrOfObj);
console.log(arrOfObj);

// Step 3: Remove duplicates and count object occurrences.
let dedupedArrWithCount = ;
l = arrOfObj.length;
for (let i = 0; i < l; i++) {
let objExists = false;
for (let j = 0; j < dedupedArrWithCount.length; j++) {
// Two objects are identical if their ids are identical.
if (arrOfObj[i].id === dedupedArrWithCount[j].id) {
objExists = true;
dedupedArrWithCount[j].count += 1;
}
}
if (!objExists) {
dedupedArrWithCount.push({
id: arrOfObj[i].id,
name: arrOfObj[i].name,
count: 1
})
}
}

console.log(dedupedArrWithCount);





// Original array
const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
console.log(arr);

// Step 1: Extract meaningful information from original array
const arrOfArrOfObj = arr
.map(category => category.key1
.map(subCategory => (
{
id: subCategory.key2.id,
name: subCategory.key2.name
}
)
)
);
console.log(arrOfArrOfObj);

// Step 2: Make the array one dimensional
const arrOfObj = .concat(...arrOfArrOfObj);
console.log(arrOfObj);

// Step 3: Remove duplicates and count object occurrences.
let dedupedArrWithCount = ;
l = arrOfObj.length;
for (let i = 0; i < l; i++) {
let objExists = false;
for (let j = 0; j < dedupedArrWithCount.length; j++) {
// Two objects are identical if their ids are identical.
if (arrOfObj[i].id === dedupedArrWithCount[j].id) {
objExists = true;
dedupedArrWithCount[j].count += 1;
}
}
if (!objExists) {
dedupedArrWithCount.push({
id: arrOfObj[i].id,
name: arrOfObj[i].name,
count: 1
})
}
}

console.log(dedupedArrWithCount);






javascript beginner array functional-programming ecmascript-6






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edited Nov 1 at 16:38









Sᴀᴍ Onᴇᴌᴀ

7,71061748




7,71061748










asked Oct 24 at 21:24









Jelefra

12




12








  • 1




    what does this data describe? will there always be two levels or might there ever be 3 or more?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 24 at 21:42










  • The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.
    – Jelefra
    Oct 24 at 21:48










  • so does that mean that there will only ever be two levels of keys?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 26 at 20:11










  • Yes that's the case.
    – Jelefra
    Oct 27 at 14:10














  • 1




    what does this data describe? will there always be two levels or might there ever be 3 or more?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 24 at 21:42










  • The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.
    – Jelefra
    Oct 24 at 21:48










  • so does that mean that there will only ever be two levels of keys?
    – Sᴀᴍ Onᴇᴌᴀ
    Oct 26 at 20:11










  • Yes that's the case.
    – Jelefra
    Oct 27 at 14:10








1




1




what does this data describe? will there always be two levels or might there ever be 3 or more?
– Sᴀᴍ Onᴇᴌᴀ
Oct 24 at 21:42




what does this data describe? will there always be two levels or might there ever be 3 or more?
– Sᴀᴍ Onᴇᴌᴀ
Oct 24 at 21:42












The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.
– Jelefra
Oct 24 at 21:48




The data describes cycling activities (key1) and chunks of them (key2). In total there are around 1,000 key1 objects and arrays under key1 are 1-100 long.
– Jelefra
Oct 24 at 21:48












so does that mean that there will only ever be two levels of keys?
– Sᴀᴍ Onᴇᴌᴀ
Oct 26 at 20:11




so does that mean that there will only ever be two levels of keys?
– Sᴀᴍ Onᴇᴌᴀ
Oct 26 at 20:11












Yes that's the case.
– Jelefra
Oct 27 at 14:10




Yes that's the case.
– Jelefra
Oct 27 at 14:10










1 Answer
1






active

oldest

votes

















up vote
0
down vote













Responding to your questions




“Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?”




While I like the benefits of functional approaches, they are often slower because of the extra function calls.



Steps 1 and 2 could be simplified by using two for...of loops (since ecmascript-6 features like const are also used). Also, instead of constructing a new object to return in the nested map callback, you can just return subCategory.key2 (though perhaps you simplified the data and the original data contains more properties that aren't needed in the end).



const arrOfObj = ;
for (let category of arr) {
for (let subCategory of category.key1) {
arrOfObj.push(subCategory.key2);
}
}


This generally works faster, at least for the small dataset supplied - see this jsPerf test for comparison




“I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?”




I was thinking of using JSON.stringify() to make an array of counts in step 1 and then use JSON.parse() in step 3 for lookups in constant time, but apparently that was slower, possibly because of the original data set which only has 1 collision. Maybe for a larger dataset that would be faster. See this jsPerf for a comparison.






const arr = [
{key1: [{key2: {id: 1, name: 'a'}}]},
{key1: [{key2: {id: 2, name: 'b'}}]},
{key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
];
const counts = {};
for (let category of arr) {
for (let subCategory of category.key1) {
const countKey = JSON.stringify(subCategory.key2)
counts[countKey] = (counts[countKey] || 0) + 1;
}
}
const dedupedArrWithCount = ;
for (let key in counts) {
const obj = JSON.parse(key);
dedupedArrWithCount.push(Object.assign(obj, {
count: counts[key]
}));
}
console.log(dedupedArrWithCount);







Other feedback



The variable l is declared without any keyword:




let dedupedArrWithCount = ;
l = arrOfObj.length;



Unless this code is wrapped in an IIFE, l becomes a global variable, which can lead to unintentional consequences if that name is used later. It is advisable to use const (or let if there was a need to re-assign it).



In that same vein, dedupedArrWithCount could be declared with the const keyword, since it is never re-assigned, just mutated using the push method.






share|improve this answer























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    1 Answer
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    1 Answer
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    active

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    active

    oldest

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    up vote
    0
    down vote













    Responding to your questions




    “Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?”




    While I like the benefits of functional approaches, they are often slower because of the extra function calls.



    Steps 1 and 2 could be simplified by using two for...of loops (since ecmascript-6 features like const are also used). Also, instead of constructing a new object to return in the nested map callback, you can just return subCategory.key2 (though perhaps you simplified the data and the original data contains more properties that aren't needed in the end).



    const arrOfObj = ;
    for (let category of arr) {
    for (let subCategory of category.key1) {
    arrOfObj.push(subCategory.key2);
    }
    }


    This generally works faster, at least for the small dataset supplied - see this jsPerf test for comparison




    “I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?”




    I was thinking of using JSON.stringify() to make an array of counts in step 1 and then use JSON.parse() in step 3 for lookups in constant time, but apparently that was slower, possibly because of the original data set which only has 1 collision. Maybe for a larger dataset that would be faster. See this jsPerf for a comparison.






    const arr = [
    {key1: [{key2: {id: 1, name: 'a'}}]},
    {key1: [{key2: {id: 2, name: 'b'}}]},
    {key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
    ];
    const counts = {};
    for (let category of arr) {
    for (let subCategory of category.key1) {
    const countKey = JSON.stringify(subCategory.key2)
    counts[countKey] = (counts[countKey] || 0) + 1;
    }
    }
    const dedupedArrWithCount = ;
    for (let key in counts) {
    const obj = JSON.parse(key);
    dedupedArrWithCount.push(Object.assign(obj, {
    count: counts[key]
    }));
    }
    console.log(dedupedArrWithCount);







    Other feedback



    The variable l is declared without any keyword:




    let dedupedArrWithCount = ;
    l = arrOfObj.length;



    Unless this code is wrapped in an IIFE, l becomes a global variable, which can lead to unintentional consequences if that name is used later. It is advisable to use const (or let if there was a need to re-assign it).



    In that same vein, dedupedArrWithCount could be declared with the const keyword, since it is never re-assigned, just mutated using the push method.






    share|improve this answer



























      up vote
      0
      down vote













      Responding to your questions




      “Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?”




      While I like the benefits of functional approaches, they are often slower because of the extra function calls.



      Steps 1 and 2 could be simplified by using two for...of loops (since ecmascript-6 features like const are also used). Also, instead of constructing a new object to return in the nested map callback, you can just return subCategory.key2 (though perhaps you simplified the data and the original data contains more properties that aren't needed in the end).



      const arrOfObj = ;
      for (let category of arr) {
      for (let subCategory of category.key1) {
      arrOfObj.push(subCategory.key2);
      }
      }


      This generally works faster, at least for the small dataset supplied - see this jsPerf test for comparison




      “I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?”




      I was thinking of using JSON.stringify() to make an array of counts in step 1 and then use JSON.parse() in step 3 for lookups in constant time, but apparently that was slower, possibly because of the original data set which only has 1 collision. Maybe for a larger dataset that would be faster. See this jsPerf for a comparison.






      const arr = [
      {key1: [{key2: {id: 1, name: 'a'}}]},
      {key1: [{key2: {id: 2, name: 'b'}}]},
      {key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
      ];
      const counts = {};
      for (let category of arr) {
      for (let subCategory of category.key1) {
      const countKey = JSON.stringify(subCategory.key2)
      counts[countKey] = (counts[countKey] || 0) + 1;
      }
      }
      const dedupedArrWithCount = ;
      for (let key in counts) {
      const obj = JSON.parse(key);
      dedupedArrWithCount.push(Object.assign(obj, {
      count: counts[key]
      }));
      }
      console.log(dedupedArrWithCount);







      Other feedback



      The variable l is declared without any keyword:




      let dedupedArrWithCount = ;
      l = arrOfObj.length;



      Unless this code is wrapped in an IIFE, l becomes a global variable, which can lead to unintentional consequences if that name is used later. It is advisable to use const (or let if there was a need to re-assign it).



      In that same vein, dedupedArrWithCount could be declared with the const keyword, since it is never re-assigned, just mutated using the push method.






      share|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Responding to your questions




        “Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?”




        While I like the benefits of functional approaches, they are often slower because of the extra function calls.



        Steps 1 and 2 could be simplified by using two for...of loops (since ecmascript-6 features like const are also used). Also, instead of constructing a new object to return in the nested map callback, you can just return subCategory.key2 (though perhaps you simplified the data and the original data contains more properties that aren't needed in the end).



        const arrOfObj = ;
        for (let category of arr) {
        for (let subCategory of category.key1) {
        arrOfObj.push(subCategory.key2);
        }
        }


        This generally works faster, at least for the small dataset supplied - see this jsPerf test for comparison




        “I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?”




        I was thinking of using JSON.stringify() to make an array of counts in step 1 and then use JSON.parse() in step 3 for lookups in constant time, but apparently that was slower, possibly because of the original data set which only has 1 collision. Maybe for a larger dataset that would be faster. See this jsPerf for a comparison.






        const arr = [
        {key1: [{key2: {id: 1, name: 'a'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
        ];
        const counts = {};
        for (let category of arr) {
        for (let subCategory of category.key1) {
        const countKey = JSON.stringify(subCategory.key2)
        counts[countKey] = (counts[countKey] || 0) + 1;
        }
        }
        const dedupedArrWithCount = ;
        for (let key in counts) {
        const obj = JSON.parse(key);
        dedupedArrWithCount.push(Object.assign(obj, {
        count: counts[key]
        }));
        }
        console.log(dedupedArrWithCount);







        Other feedback



        The variable l is declared without any keyword:




        let dedupedArrWithCount = ;
        l = arrOfObj.length;



        Unless this code is wrapped in an IIFE, l becomes a global variable, which can lead to unintentional consequences if that name is used later. It is advisable to use const (or let if there was a need to re-assign it).



        In that same vein, dedupedArrWithCount could be declared with the const keyword, since it is never re-assigned, just mutated using the push method.






        share|improve this answer














        Responding to your questions




        “Is there a simpler, more semantic, faster, or otherwise better way than to use two nested map in step 1?”




        While I like the benefits of functional approaches, they are often slower because of the extra function calls.



        Steps 1 and 2 could be simplified by using two for...of loops (since ecmascript-6 features like const are also used). Also, instead of constructing a new object to return in the nested map callback, you can just return subCategory.key2 (though perhaps you simplified the data and the original data contains more properties that aren't needed in the end).



        const arrOfObj = ;
        for (let category of arr) {
        for (let subCategory of category.key1) {
        arrOfObj.push(subCategory.key2);
        }
        }


        This generally works faster, at least for the small dataset supplied - see this jsPerf test for comparison




        “I realise that step 3 is longer than it should be and there has to be a better way to do it. How can step 3 be improved?”




        I was thinking of using JSON.stringify() to make an array of counts in step 1 and then use JSON.parse() in step 3 for lookups in constant time, but apparently that was slower, possibly because of the original data set which only has 1 collision. Maybe for a larger dataset that would be faster. See this jsPerf for a comparison.






        const arr = [
        {key1: [{key2: {id: 1, name: 'a'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
        ];
        const counts = {};
        for (let category of arr) {
        for (let subCategory of category.key1) {
        const countKey = JSON.stringify(subCategory.key2)
        counts[countKey] = (counts[countKey] || 0) + 1;
        }
        }
        const dedupedArrWithCount = ;
        for (let key in counts) {
        const obj = JSON.parse(key);
        dedupedArrWithCount.push(Object.assign(obj, {
        count: counts[key]
        }));
        }
        console.log(dedupedArrWithCount);







        Other feedback



        The variable l is declared without any keyword:




        let dedupedArrWithCount = ;
        l = arrOfObj.length;



        Unless this code is wrapped in an IIFE, l becomes a global variable, which can lead to unintentional consequences if that name is used later. It is advisable to use const (or let if there was a need to re-assign it).



        In that same vein, dedupedArrWithCount could be declared with the const keyword, since it is never re-assigned, just mutated using the push method.






        const arr = [
        {key1: [{key2: {id: 1, name: 'a'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
        ];
        const counts = {};
        for (let category of arr) {
        for (let subCategory of category.key1) {
        const countKey = JSON.stringify(subCategory.key2)
        counts[countKey] = (counts[countKey] || 0) + 1;
        }
        }
        const dedupedArrWithCount = ;
        for (let key in counts) {
        const obj = JSON.parse(key);
        dedupedArrWithCount.push(Object.assign(obj, {
        count: counts[key]
        }));
        }
        console.log(dedupedArrWithCount);





        const arr = [
        {key1: [{key2: {id: 1, name: 'a'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}]},
        {key1: [{key2: {id: 2, name: 'b'}}, {key2: {id: 3, name: 'c'}}]}
        ];
        const counts = {};
        for (let category of arr) {
        for (let subCategory of category.key1) {
        const countKey = JSON.stringify(subCategory.key2)
        counts[countKey] = (counts[countKey] || 0) + 1;
        }
        }
        const dedupedArrWithCount = ;
        for (let key in counts) {
        const obj = JSON.parse(key);
        dedupedArrWithCount.push(Object.assign(obj, {
        count: counts[key]
        }));
        }
        console.log(dedupedArrWithCount);






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 7 hours ago

























        answered Nov 1 at 16:40









        Sᴀᴍ Onᴇᴌᴀ

        7,71061748




        7,71061748






























             

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