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combine an array of object by id and and reduce its nested object



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Data science time! April 2019 and salary with experience
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1















I have an array of objects (old_array) that needs to be merged to become (new_array)



old_array = [
id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 ,
id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 ,
id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 ,
id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 ,
id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12
];

new_array = [
id: 'ffff55', name: 'f5', unique_cards: 2, request: device: 32, bus: 32, ship: 53 ,
id: 'vvvv44', name: 'v4', unique_cards: 1, request: device: 16, bus: 21, ship: 14 ,
id: 'cccc33', name: 'c3', unique_cards: 3, request: device: 25, bus: 23, ship: 26
];


  • merge the objects with same id and name to a single object

  • merge the nested request object (Sum of its values)

  • map card to the number of unique cards (by id)

I've been trying for 4 days straight but this array manipulation was hard



My best attempt was trying to group the array of objects by id but it become more complex with many redundant values



groupByArray(xs, key) 
return xs.reduce(function(rv, x)
let v = key instanceof Function ? key(x) : x[key];
let el = rv.find((r) => r && r.key === v);
if (el)
el.values.push(x);
else
rv.push( key: v, values: [ x ] );

return rv;
, []);


groupByArray(old_array , 'id')









share|improve this question



















  • 1





    show your best attempt in js

    – zerkms
    Mar 9 at 0:10











  • @zerkms i tried to group the arrays by id then mapping and reducing but it become more complex (more nested objects )

    – Amine Da.
    Mar 9 at 0:14











  • You have enumerated 3 different task. Try solve only the first one and publish your best attempt. That really is an EXTREMELY FREQUENT request here: at least make a research.

    – zerkms
    Mar 9 at 0:16












  • @zerkms i only ask if i'm very desperate . most of what i found about merging is either a simple array or using underscore.js loddah and other stuff. 14h straight today of searching

    – Amine Da.
    Mar 9 at 0:26











  • "is either a simple array" --- you have a simple array of simple objects. The problem here is that now you have an answer, yet you still are not able to solve such sort of tasks. You indeed can copy-paste it, but tomorrow you'll be stuck again when the conditions slightly change 🤷

    – zerkms
    Mar 9 at 0:27


















1















I have an array of objects (old_array) that needs to be merged to become (new_array)



old_array = [
id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 ,
id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 ,
id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 ,
id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 ,
id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12
];

new_array = [
id: 'ffff55', name: 'f5', unique_cards: 2, request: device: 32, bus: 32, ship: 53 ,
id: 'vvvv44', name: 'v4', unique_cards: 1, request: device: 16, bus: 21, ship: 14 ,
id: 'cccc33', name: 'c3', unique_cards: 3, request: device: 25, bus: 23, ship: 26
];


  • merge the objects with same id and name to a single object

  • merge the nested request object (Sum of its values)

  • map card to the number of unique cards (by id)

I've been trying for 4 days straight but this array manipulation was hard



My best attempt was trying to group the array of objects by id but it become more complex with many redundant values



groupByArray(xs, key) 
return xs.reduce(function(rv, x)
let v = key instanceof Function ? key(x) : x[key];
let el = rv.find((r) => r && r.key === v);
if (el)
el.values.push(x);
else
rv.push( key: v, values: [ x ] );

return rv;
, []);


groupByArray(old_array , 'id')









share|improve this question



















  • 1





    show your best attempt in js

    – zerkms
    Mar 9 at 0:10











  • @zerkms i tried to group the arrays by id then mapping and reducing but it become more complex (more nested objects )

    – Amine Da.
    Mar 9 at 0:14











  • You have enumerated 3 different task. Try solve only the first one and publish your best attempt. That really is an EXTREMELY FREQUENT request here: at least make a research.

    – zerkms
    Mar 9 at 0:16












  • @zerkms i only ask if i'm very desperate . most of what i found about merging is either a simple array or using underscore.js loddah and other stuff. 14h straight today of searching

    – Amine Da.
    Mar 9 at 0:26











  • "is either a simple array" --- you have a simple array of simple objects. The problem here is that now you have an answer, yet you still are not able to solve such sort of tasks. You indeed can copy-paste it, but tomorrow you'll be stuck again when the conditions slightly change 🤷

    – zerkms
    Mar 9 at 0:27














1












1








1








I have an array of objects (old_array) that needs to be merged to become (new_array)



old_array = [
id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 ,
id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 ,
id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 ,
id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 ,
id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12
];

new_array = [
id: 'ffff55', name: 'f5', unique_cards: 2, request: device: 32, bus: 32, ship: 53 ,
id: 'vvvv44', name: 'v4', unique_cards: 1, request: device: 16, bus: 21, ship: 14 ,
id: 'cccc33', name: 'c3', unique_cards: 3, request: device: 25, bus: 23, ship: 26
];


  • merge the objects with same id and name to a single object

  • merge the nested request object (Sum of its values)

  • map card to the number of unique cards (by id)

I've been trying for 4 days straight but this array manipulation was hard



My best attempt was trying to group the array of objects by id but it become more complex with many redundant values



groupByArray(xs, key) 
return xs.reduce(function(rv, x)
let v = key instanceof Function ? key(x) : x[key];
let el = rv.find((r) => r && r.key === v);
if (el)
el.values.push(x);
else
rv.push( key: v, values: [ x ] );

return rv;
, []);


groupByArray(old_array , 'id')









share|improve this question
















I have an array of objects (old_array) that needs to be merged to become (new_array)



old_array = [
id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 ,
id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 ,
id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 ,
id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 ,
id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 ,
id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12
];

new_array = [
id: 'ffff55', name: 'f5', unique_cards: 2, request: device: 32, bus: 32, ship: 53 ,
id: 'vvvv44', name: 'v4', unique_cards: 1, request: device: 16, bus: 21, ship: 14 ,
id: 'cccc33', name: 'c3', unique_cards: 3, request: device: 25, bus: 23, ship: 26
];


  • merge the objects with same id and name to a single object

  • merge the nested request object (Sum of its values)

  • map card to the number of unique cards (by id)

I've been trying for 4 days straight but this array manipulation was hard



My best attempt was trying to group the array of objects by id but it become more complex with many redundant values



groupByArray(xs, key) 
return xs.reduce(function(rv, x)
let v = key instanceof Function ? key(x) : x[key];
let el = rv.find((r) => r && r.key === v);
if (el)
el.values.push(x);
else
rv.push( key: v, values: [ x ] );

return rv;
, []);


groupByArray(old_array , 'id')






javascript typescript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 9 at 2:27









Dan

7,99253066




7,99253066










asked Mar 9 at 0:08









Amine Da.Amine Da.

7483917




7483917







  • 1





    show your best attempt in js

    – zerkms
    Mar 9 at 0:10











  • @zerkms i tried to group the arrays by id then mapping and reducing but it become more complex (more nested objects )

    – Amine Da.
    Mar 9 at 0:14











  • You have enumerated 3 different task. Try solve only the first one and publish your best attempt. That really is an EXTREMELY FREQUENT request here: at least make a research.

    – zerkms
    Mar 9 at 0:16












  • @zerkms i only ask if i'm very desperate . most of what i found about merging is either a simple array or using underscore.js loddah and other stuff. 14h straight today of searching

    – Amine Da.
    Mar 9 at 0:26











  • "is either a simple array" --- you have a simple array of simple objects. The problem here is that now you have an answer, yet you still are not able to solve such sort of tasks. You indeed can copy-paste it, but tomorrow you'll be stuck again when the conditions slightly change 🤷

    – zerkms
    Mar 9 at 0:27













  • 1





    show your best attempt in js

    – zerkms
    Mar 9 at 0:10











  • @zerkms i tried to group the arrays by id then mapping and reducing but it become more complex (more nested objects )

    – Amine Da.
    Mar 9 at 0:14











  • You have enumerated 3 different task. Try solve only the first one and publish your best attempt. That really is an EXTREMELY FREQUENT request here: at least make a research.

    – zerkms
    Mar 9 at 0:16












  • @zerkms i only ask if i'm very desperate . most of what i found about merging is either a simple array or using underscore.js loddah and other stuff. 14h straight today of searching

    – Amine Da.
    Mar 9 at 0:26











  • "is either a simple array" --- you have a simple array of simple objects. The problem here is that now you have an answer, yet you still are not able to solve such sort of tasks. You indeed can copy-paste it, but tomorrow you'll be stuck again when the conditions slightly change 🤷

    – zerkms
    Mar 9 at 0:27








1




1





show your best attempt in js

– zerkms
Mar 9 at 0:10





show your best attempt in js

– zerkms
Mar 9 at 0:10













@zerkms i tried to group the arrays by id then mapping and reducing but it become more complex (more nested objects )

– Amine Da.
Mar 9 at 0:14





@zerkms i tried to group the arrays by id then mapping and reducing but it become more complex (more nested objects )

– Amine Da.
Mar 9 at 0:14













You have enumerated 3 different task. Try solve only the first one and publish your best attempt. That really is an EXTREMELY FREQUENT request here: at least make a research.

– zerkms
Mar 9 at 0:16






You have enumerated 3 different task. Try solve only the first one and publish your best attempt. That really is an EXTREMELY FREQUENT request here: at least make a research.

– zerkms
Mar 9 at 0:16














@zerkms i only ask if i'm very desperate . most of what i found about merging is either a simple array or using underscore.js loddah and other stuff. 14h straight today of searching

– Amine Da.
Mar 9 at 0:26





@zerkms i only ask if i'm very desperate . most of what i found about merging is either a simple array or using underscore.js loddah and other stuff. 14h straight today of searching

– Amine Da.
Mar 9 at 0:26













"is either a simple array" --- you have a simple array of simple objects. The problem here is that now you have an answer, yet you still are not able to solve such sort of tasks. You indeed can copy-paste it, but tomorrow you'll be stuck again when the conditions slightly change 🤷

– zerkms
Mar 9 at 0:27






"is either a simple array" --- you have a simple array of simple objects. The problem here is that now you have an answer, yet you still are not able to solve such sort of tasks. You indeed can copy-paste it, but tomorrow you'll be stuck again when the conditions slightly change 🤷

– zerkms
Mar 9 at 0:27













1 Answer
1






active

oldest

votes


















1














You can use the function reduce to group and the function Object.values to extract the grouped objects.



Inside of the reduce's handler, you need to loop over the request's keys in order to sum up each value.






let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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








share|improve this answer




















  • 1





    @AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

    – Ele
    Mar 9 at 0:42







  • 1





    @AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

    – Ele
    Mar 9 at 1:02











  • @AmineDa. you're welcome!

    – Ele
    Mar 9 at 1:05











  • @.Ele one more thing. I don't get the point of a[id].card = card inside if

    – Amine Da.
    Mar 10 at 0:05







  • 1





    @AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

    – Ele
    Mar 10 at 0:06












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1 Answer
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active

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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You can use the function reduce to group and the function Object.values to extract the grouped objects.



Inside of the reduce's handler, you need to loop over the request's keys in order to sum up each value.






let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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








share|improve this answer




















  • 1





    @AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

    – Ele
    Mar 9 at 0:42







  • 1





    @AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

    – Ele
    Mar 9 at 1:02











  • @AmineDa. you're welcome!

    – Ele
    Mar 9 at 1:05











  • @.Ele one more thing. I don't get the point of a[id].card = card inside if

    – Amine Da.
    Mar 10 at 0:05







  • 1





    @AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

    – Ele
    Mar 10 at 0:06
















1














You can use the function reduce to group and the function Object.values to extract the grouped objects.



Inside of the reduce's handler, you need to loop over the request's keys in order to sum up each value.






let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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








share|improve this answer




















  • 1





    @AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

    – Ele
    Mar 9 at 0:42







  • 1





    @AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

    – Ele
    Mar 9 at 1:02











  • @AmineDa. you're welcome!

    – Ele
    Mar 9 at 1:05











  • @.Ele one more thing. I don't get the point of a[id].card = card inside if

    – Amine Da.
    Mar 10 at 0:05







  • 1





    @AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

    – Ele
    Mar 10 at 0:06














1












1








1







You can use the function reduce to group and the function Object.values to extract the grouped objects.



Inside of the reduce's handler, you need to loop over the request's keys in order to sum up each value.






let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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








share|improve this answer















You can use the function reduce to group and the function Object.values to extract the grouped objects.



Inside of the reduce's handler, you need to loop over the request's keys in order to sum up each value.






let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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








let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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





let array = [ id: 'ffff55', name: 'f5', card: 'a', request: device: 0, bus: 1, ship: 21 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 3, bus: 10, ship: 2 , id: 'cccc33', name: 'c3', card: 'a', request: device: 0, bus: 1, ship: 2 , id: 'ffff55', name: 'f5', card: 'b', request: device: 32, bus: 31, ship: 32 , id: 'cccc33', name: 'c3', card: 'e', request: device: 21, bus: 21, ship: 22 , id: 'cccc33', name: 'c3', card: 'd', request: device: 4, bus: 1, ship: 2 , id: 'vvvv44', name: 'v4', card: 'c', request: device: 13, bus: 11, ship: 12 ];
let result = Object.values(array.reduce((a, id, name, card, request) => 0) + request[k]);
return a;
, Object.create(null)));

result.forEach(o => delete o.card);

console.log(result);

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






share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 9 at 0:41

























answered Mar 9 at 0:20









EleEle

25.7k52252




25.7k52252







  • 1





    @AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

    – Ele
    Mar 9 at 0:42







  • 1





    @AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

    – Ele
    Mar 9 at 1:02











  • @AmineDa. you're welcome!

    – Ele
    Mar 9 at 1:05











  • @.Ele one more thing. I don't get the point of a[id].card = card inside if

    – Amine Da.
    Mar 10 at 0:05







  • 1





    @AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

    – Ele
    Mar 10 at 0:06













  • 1





    @AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

    – Ele
    Mar 9 at 0:42







  • 1





    @AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

    – Ele
    Mar 9 at 1:02











  • @AmineDa. you're welcome!

    – Ele
    Mar 9 at 1:05











  • @.Ele one more thing. I don't get the point of a[id].card = card inside if

    – Amine Da.
    Mar 10 at 0:05







  • 1





    @AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

    – Ele
    Mar 10 at 0:06








1




1





@AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

– Ele
Mar 9 at 0:42






@AmineDa. you're right, so you need to check for the last validated card's value to increase the counter. See the updated answer, unfortunately, the last step should be loop the result and delete the card property.

– Ele
Mar 9 at 0:42





1




1





@AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

– Ele
Mar 9 at 1:02





@AmineDa. basically, I'm asking for a previously captured object by id. So, If a[id] is different than undefined means that I got an object before so I need to work on it, otherwise, is a new id which I didn't get before, so I should create the desired object structure.

– Ele
Mar 9 at 1:02













@AmineDa. you're welcome!

– Ele
Mar 9 at 1:05





@AmineDa. you're welcome!

– Ele
Mar 9 at 1:05













@.Ele one more thing. I don't get the point of a[id].card = card inside if

– Amine Da.
Mar 10 at 0:05






@.Ele one more thing. I don't get the point of a[id].card = card inside if

– Amine Da.
Mar 10 at 0:05





1




1





@AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

– Ele
Mar 10 at 0:06






@AmineDa. this is like a breadcrumb, because I need to know if the previously checked card is different in order to increment the count.

– Ele
Mar 10 at 0:06




















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