Extracting grouped objects into a flattened list












0












$begingroup$


I need to optimize this javascript method because it is a bit slow, I don't want to use the flat() command, as for some reason my angular6 app does not understand the vanilla flat() command, or display some annoying warning messages.



So check the below:





  1. my original object

  2. my result object desired

  3. my slow solution (works, but is slow)




1. my original object:



  const data = [
{
"id": 1,
"name": "Application",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 17,
"icon": "access_time",
"title": "Daily Order Cut-Off Time",
"description": "Daily Order Cut-Off Time",
"code": "daily-order-cut-off-time",
"value": "09:35",
"valueType": "Time",
"configurationTypeId": 11,
"definition": {
"step": "none"
},
"isDefault": false
}
]
}
]
},
{
"id": 3,
"name": "Processing",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 1078,
"icon": "flash_on",
"title": "Auto Process",
"description": "Will process all orders that are in the same batch",
"code": "processing-auto-process",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
},
{
"id": 1074,
"icon": "subdirectory_arrow_right",
"title": "Allow Under Picking",
"description": "Allow under pick when processing order?",
"code": "processing-allow-under-picks",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
}
]
}
]
}
];


2. object desired:



 [
{
"id": 17,
"code": "daily-order-cut-off-time",
"value": "09:35"
},
{
"id": 1078,
"code": "processing-auto-process",
"value": "0"
},
{
"id": 1074,
"code": "processing-allow-under-picks",
"value": "0"
}
]


My slow solution:



const result = data.map(module => module.groups.map(configurations => configurations.configurations.map(config => ({ id: config.id, code: config.code, value: config.value })))).reduce((l,n) => l.concat(n), ).reduce((l2,n2) => l2.concat(n2),));









share|improve this question











$endgroup$








  • 2




    $begingroup$
    Is the ES6 spread operator available to you? result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value}))))))
    $endgroup$
    – Oh My Goodness
    5 hours ago












  • $begingroup$
    yes it is available.
    $endgroup$
    – Roger Oliveira
    5 hours ago










  • $begingroup$
    AMAZING! YOUR CODE IS BEAUTIFUL!, please answer my question I will mark as the correct answer!
    $endgroup$
    – Roger Oliveira
    5 hours ago
















0












$begingroup$


I need to optimize this javascript method because it is a bit slow, I don't want to use the flat() command, as for some reason my angular6 app does not understand the vanilla flat() command, or display some annoying warning messages.



So check the below:





  1. my original object

  2. my result object desired

  3. my slow solution (works, but is slow)




1. my original object:



  const data = [
{
"id": 1,
"name": "Application",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 17,
"icon": "access_time",
"title": "Daily Order Cut-Off Time",
"description": "Daily Order Cut-Off Time",
"code": "daily-order-cut-off-time",
"value": "09:35",
"valueType": "Time",
"configurationTypeId": 11,
"definition": {
"step": "none"
},
"isDefault": false
}
]
}
]
},
{
"id": 3,
"name": "Processing",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 1078,
"icon": "flash_on",
"title": "Auto Process",
"description": "Will process all orders that are in the same batch",
"code": "processing-auto-process",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
},
{
"id": 1074,
"icon": "subdirectory_arrow_right",
"title": "Allow Under Picking",
"description": "Allow under pick when processing order?",
"code": "processing-allow-under-picks",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
}
]
}
]
}
];


2. object desired:



 [
{
"id": 17,
"code": "daily-order-cut-off-time",
"value": "09:35"
},
{
"id": 1078,
"code": "processing-auto-process",
"value": "0"
},
{
"id": 1074,
"code": "processing-allow-under-picks",
"value": "0"
}
]


My slow solution:



const result = data.map(module => module.groups.map(configurations => configurations.configurations.map(config => ({ id: config.id, code: config.code, value: config.value })))).reduce((l,n) => l.concat(n), ).reduce((l2,n2) => l2.concat(n2),));









share|improve this question











$endgroup$








  • 2




    $begingroup$
    Is the ES6 spread operator available to you? result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value}))))))
    $endgroup$
    – Oh My Goodness
    5 hours ago












  • $begingroup$
    yes it is available.
    $endgroup$
    – Roger Oliveira
    5 hours ago










  • $begingroup$
    AMAZING! YOUR CODE IS BEAUTIFUL!, please answer my question I will mark as the correct answer!
    $endgroup$
    – Roger Oliveira
    5 hours ago














0












0








0


1



$begingroup$


I need to optimize this javascript method because it is a bit slow, I don't want to use the flat() command, as for some reason my angular6 app does not understand the vanilla flat() command, or display some annoying warning messages.



So check the below:





  1. my original object

  2. my result object desired

  3. my slow solution (works, but is slow)




1. my original object:



  const data = [
{
"id": 1,
"name": "Application",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 17,
"icon": "access_time",
"title": "Daily Order Cut-Off Time",
"description": "Daily Order Cut-Off Time",
"code": "daily-order-cut-off-time",
"value": "09:35",
"valueType": "Time",
"configurationTypeId": 11,
"definition": {
"step": "none"
},
"isDefault": false
}
]
}
]
},
{
"id": 3,
"name": "Processing",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 1078,
"icon": "flash_on",
"title": "Auto Process",
"description": "Will process all orders that are in the same batch",
"code": "processing-auto-process",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
},
{
"id": 1074,
"icon": "subdirectory_arrow_right",
"title": "Allow Under Picking",
"description": "Allow under pick when processing order?",
"code": "processing-allow-under-picks",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
}
]
}
]
}
];


2. object desired:



 [
{
"id": 17,
"code": "daily-order-cut-off-time",
"value": "09:35"
},
{
"id": 1078,
"code": "processing-auto-process",
"value": "0"
},
{
"id": 1074,
"code": "processing-allow-under-picks",
"value": "0"
}
]


My slow solution:



const result = data.map(module => module.groups.map(configurations => configurations.configurations.map(config => ({ id: config.id, code: config.code, value: config.value })))).reduce((l,n) => l.concat(n), ).reduce((l2,n2) => l2.concat(n2),));









share|improve this question











$endgroup$




I need to optimize this javascript method because it is a bit slow, I don't want to use the flat() command, as for some reason my angular6 app does not understand the vanilla flat() command, or display some annoying warning messages.



So check the below:





  1. my original object

  2. my result object desired

  3. my slow solution (works, but is slow)




1. my original object:



  const data = [
{
"id": 1,
"name": "Application",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 17,
"icon": "access_time",
"title": "Daily Order Cut-Off Time",
"description": "Daily Order Cut-Off Time",
"code": "daily-order-cut-off-time",
"value": "09:35",
"valueType": "Time",
"configurationTypeId": 11,
"definition": {
"step": "none"
},
"isDefault": false
}
]
}
]
},
{
"id": 3,
"name": "Processing",
"groups": [
{
"groupName": "",
"configurations": [
{
"id": 1078,
"icon": "flash_on",
"title": "Auto Process",
"description": "Will process all orders that are in the same batch",
"code": "processing-auto-process",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
},
{
"id": 1074,
"icon": "subdirectory_arrow_right",
"title": "Allow Under Picking",
"description": "Allow under pick when processing order?",
"code": "processing-allow-under-picks",
"value": "0",
"valueType": "Boolean",
"configurationTypeId": 6,
"definition": null,
"isDefault": false
}
]
}
]
}
];


2. object desired:



 [
{
"id": 17,
"code": "daily-order-cut-off-time",
"value": "09:35"
},
{
"id": 1078,
"code": "processing-auto-process",
"value": "0"
},
{
"id": 1074,
"code": "processing-allow-under-picks",
"value": "0"
}
]


My slow solution:



const result = data.map(module => module.groups.map(configurations => configurations.configurations.map(config => ({ id: config.id, code: config.code, value: config.value })))).reduce((l,n) => l.concat(n), ).reduce((l2,n2) => l2.concat(n2),));






javascript ecmascript-6






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 mins ago









200_success

130k17155419




130k17155419










asked 6 hours ago









Roger OliveiraRoger Oliveira

150111




150111








  • 2




    $begingroup$
    Is the ES6 spread operator available to you? result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value}))))))
    $endgroup$
    – Oh My Goodness
    5 hours ago












  • $begingroup$
    yes it is available.
    $endgroup$
    – Roger Oliveira
    5 hours ago










  • $begingroup$
    AMAZING! YOUR CODE IS BEAUTIFUL!, please answer my question I will mark as the correct answer!
    $endgroup$
    – Roger Oliveira
    5 hours ago














  • 2




    $begingroup$
    Is the ES6 spread operator available to you? result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value}))))))
    $endgroup$
    – Oh My Goodness
    5 hours ago












  • $begingroup$
    yes it is available.
    $endgroup$
    – Roger Oliveira
    5 hours ago










  • $begingroup$
    AMAZING! YOUR CODE IS BEAUTIFUL!, please answer my question I will mark as the correct answer!
    $endgroup$
    – Roger Oliveira
    5 hours ago








2




2




$begingroup$
Is the ES6 spread operator available to you? result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value}))))))
$endgroup$
– Oh My Goodness
5 hours ago






$begingroup$
Is the ES6 spread operator available to you? result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value}))))))
$endgroup$
– Oh My Goodness
5 hours ago














$begingroup$
yes it is available.
$endgroup$
– Roger Oliveira
5 hours ago




$begingroup$
yes it is available.
$endgroup$
– Roger Oliveira
5 hours ago












$begingroup$
AMAZING! YOUR CODE IS BEAUTIFUL!, please answer my question I will mark as the correct answer!
$endgroup$
– Roger Oliveira
5 hours ago




$begingroup$
AMAZING! YOUR CODE IS BEAUTIFUL!, please answer my question I will mark as the correct answer!
$endgroup$
– Roger Oliveira
5 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

As commented, the ES6 spread operator flattens like you want. The syntax is a little weird for multiple invocations:



result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value})))))) 





share|improve this answer









$endgroup$













  • $begingroup$
    also, the use of the command .flat() would help somehow in this solution? thanks.
    $endgroup$
    – Roger Oliveira
    2 hours ago












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f216454%2fextracting-grouped-objects-into-a-flattened-list%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

As commented, the ES6 spread operator flattens like you want. The syntax is a little weird for multiple invocations:



result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value})))))) 





share|improve this answer









$endgroup$













  • $begingroup$
    also, the use of the command .flat() would help somehow in this solution? thanks.
    $endgroup$
    – Roger Oliveira
    2 hours ago
















2












$begingroup$

As commented, the ES6 spread operator flattens like you want. The syntax is a little weird for multiple invocations:



result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value})))))) 





share|improve this answer









$endgroup$













  • $begingroup$
    also, the use of the command .flat() would help somehow in this solution? thanks.
    $endgroup$
    – Roger Oliveira
    2 hours ago














2












2








2





$begingroup$

As commented, the ES6 spread operator flattens like you want. The syntax is a little weird for multiple invocations:



result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value})))))) 





share|improve this answer









$endgroup$



As commented, the ES6 spread operator flattens like you want. The syntax is a little weird for multiple invocations:



result = .concat(....concat(...data.map( d => d.groups.map( g => g.configurations.map( c => ({id:c.id, code:c.code, value:c.value})))))) 






share|improve this answer












share|improve this answer



share|improve this answer










answered 5 hours ago









Oh My GoodnessOh My Goodness

1,874314




1,874314












  • $begingroup$
    also, the use of the command .flat() would help somehow in this solution? thanks.
    $endgroup$
    – Roger Oliveira
    2 hours ago


















  • $begingroup$
    also, the use of the command .flat() would help somehow in this solution? thanks.
    $endgroup$
    – Roger Oliveira
    2 hours ago
















$begingroup$
also, the use of the command .flat() would help somehow in this solution? thanks.
$endgroup$
– Roger Oliveira
2 hours ago




$begingroup$
also, the use of the command .flat() would help somehow in this solution? thanks.
$endgroup$
– Roger Oliveira
2 hours ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Code Review Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f216454%2fextracting-grouped-objects-into-a-flattened-list%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Список кардиналов, возведённых папой римским Каликстом III

Deduzione

Mysql.sock missing - “Can't connect to local MySQL server through socket”