How to append an object to an array based on the number of properties other than the specified properties?











up vote
2
down vote

favorite












For example , I have an array of objects like so :



[

{
"waterfallData.PopulationName":"Population 1",
"Deaths":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


I want to add two(does not have to be two,if there is a "Extra Births" then three) intermerdiate objects between each population like so



[

{
"waterfallData.PopulationName":"Population 1",
"Death":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"Deaths" : -1000,
"open" : 6453,
"close" : 5453
},
{
"Births" : 5000,
"open" : 5453,
"close : 10453
}
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"Deaths" : -2000,
"open" : 10453,
"close" : 8453
},
{
"Births" : 500,
"open" : 8453,
"close" : 8953
}
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


So as you can see I want to add objects based on the number of properties other than the waterfallData.PopulationName ,open and close properties. Then, I want to assign open and close properties on each object based on the next "Deaths" and "Births" values.



For example , Population 1 starts with 6453 then I add two objects with the 1st object taking the next "Deaths" value in Population 2 which is -1000 then I assign the open property to be from the previous close property of Population 1 and the close property to be calculated by adding the assigned open property to "Deaths"'s value. And same goes with the 2nd extra object where I assign the open property to be the close property of the previous object and the close property to be calculated by adding open with "Births"'s value.



How do I achieve this?










share|improve this question
























  • Why not create using nested JSON?
    – HymnZ
    Nov 9 at 8:07










  • @HymnZ because I am using Amcharts and it does not support nested JSON data.
    – Syed Ariff
    Nov 9 at 8:09










  • You can try with Object.keys(your_array[1]) to create extra properties.. Just as @Grégory NEUT answered
    – HymnZ
    Nov 9 at 8:14






  • 1




    NB2: object properties are unordered, so the concept of "the number of properties after..." is unreliable. All you could say is "the number of properties other than...".
    – trincot
    Nov 9 at 8:21






  • 1




    Hey, no cheating. You changed the question from using properties such as 'State [Number]' to 'deatsh' and 'biths'. Trincot's deleted answer had a comment where he states that Object.keys is not to be guaranteed to come in any particular order so how will you know what's next and previous?
    – HMR
    Nov 9 at 8:43

















up vote
2
down vote

favorite












For example , I have an array of objects like so :



[

{
"waterfallData.PopulationName":"Population 1",
"Deaths":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


I want to add two(does not have to be two,if there is a "Extra Births" then three) intermerdiate objects between each population like so



[

{
"waterfallData.PopulationName":"Population 1",
"Death":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"Deaths" : -1000,
"open" : 6453,
"close" : 5453
},
{
"Births" : 5000,
"open" : 5453,
"close : 10453
}
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"Deaths" : -2000,
"open" : 10453,
"close" : 8453
},
{
"Births" : 500,
"open" : 8453,
"close" : 8953
}
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


So as you can see I want to add objects based on the number of properties other than the waterfallData.PopulationName ,open and close properties. Then, I want to assign open and close properties on each object based on the next "Deaths" and "Births" values.



For example , Population 1 starts with 6453 then I add two objects with the 1st object taking the next "Deaths" value in Population 2 which is -1000 then I assign the open property to be from the previous close property of Population 1 and the close property to be calculated by adding the assigned open property to "Deaths"'s value. And same goes with the 2nd extra object where I assign the open property to be the close property of the previous object and the close property to be calculated by adding open with "Births"'s value.



How do I achieve this?










share|improve this question
























  • Why not create using nested JSON?
    – HymnZ
    Nov 9 at 8:07










  • @HymnZ because I am using Amcharts and it does not support nested JSON data.
    – Syed Ariff
    Nov 9 at 8:09










  • You can try with Object.keys(your_array[1]) to create extra properties.. Just as @Grégory NEUT answered
    – HymnZ
    Nov 9 at 8:14






  • 1




    NB2: object properties are unordered, so the concept of "the number of properties after..." is unreliable. All you could say is "the number of properties other than...".
    – trincot
    Nov 9 at 8:21






  • 1




    Hey, no cheating. You changed the question from using properties such as 'State [Number]' to 'deatsh' and 'biths'. Trincot's deleted answer had a comment where he states that Object.keys is not to be guaranteed to come in any particular order so how will you know what's next and previous?
    – HMR
    Nov 9 at 8:43















up vote
2
down vote

favorite









up vote
2
down vote

favorite











For example , I have an array of objects like so :



[

{
"waterfallData.PopulationName":"Population 1",
"Deaths":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


I want to add two(does not have to be two,if there is a "Extra Births" then three) intermerdiate objects between each population like so



[

{
"waterfallData.PopulationName":"Population 1",
"Death":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"Deaths" : -1000,
"open" : 6453,
"close" : 5453
},
{
"Births" : 5000,
"open" : 5453,
"close : 10453
}
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"Deaths" : -2000,
"open" : 10453,
"close" : 8453
},
{
"Births" : 500,
"open" : 8453,
"close" : 8953
}
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


So as you can see I want to add objects based on the number of properties other than the waterfallData.PopulationName ,open and close properties. Then, I want to assign open and close properties on each object based on the next "Deaths" and "Births" values.



For example , Population 1 starts with 6453 then I add two objects with the 1st object taking the next "Deaths" value in Population 2 which is -1000 then I assign the open property to be from the previous close property of Population 1 and the close property to be calculated by adding the assigned open property to "Deaths"'s value. And same goes with the 2nd extra object where I assign the open property to be the close property of the previous object and the close property to be calculated by adding open with "Births"'s value.



How do I achieve this?










share|improve this question















For example , I have an array of objects like so :



[

{
"waterfallData.PopulationName":"Population 1",
"Deaths":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


I want to add two(does not have to be two,if there is a "Extra Births" then three) intermerdiate objects between each population like so



[

{
"waterfallData.PopulationName":"Population 1",
"Death":-2333,
"Births":8786,
"open":0,
"close":6453
},
{
"Deaths" : -1000,
"open" : 6453,
"close" : 5453
},
{
"Births" : 5000,
"open" : 5453,
"close : 10453
}
{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},
{
"Deaths" : -2000,
"open" : 10453,
"close" : 8453
},
{
"Births" : 500,
"open" : 8453,
"close" : 8953
}
{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}
]


So as you can see I want to add objects based on the number of properties other than the waterfallData.PopulationName ,open and close properties. Then, I want to assign open and close properties on each object based on the next "Deaths" and "Births" values.



For example , Population 1 starts with 6453 then I add two objects with the 1st object taking the next "Deaths" value in Population 2 which is -1000 then I assign the open property to be from the previous close property of Population 1 and the close property to be calculated by adding the assigned open property to "Deaths"'s value. And same goes with the 2nd extra object where I assign the open property to be the close property of the previous object and the close property to be calculated by adding open with "Births"'s value.



How do I achieve this?







javascript arrays json






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 16 at 7:01

























asked Nov 9 at 7:57









Syed Ariff

3651317




3651317












  • Why not create using nested JSON?
    – HymnZ
    Nov 9 at 8:07










  • @HymnZ because I am using Amcharts and it does not support nested JSON data.
    – Syed Ariff
    Nov 9 at 8:09










  • You can try with Object.keys(your_array[1]) to create extra properties.. Just as @Grégory NEUT answered
    – HymnZ
    Nov 9 at 8:14






  • 1




    NB2: object properties are unordered, so the concept of "the number of properties after..." is unreliable. All you could say is "the number of properties other than...".
    – trincot
    Nov 9 at 8:21






  • 1




    Hey, no cheating. You changed the question from using properties such as 'State [Number]' to 'deatsh' and 'biths'. Trincot's deleted answer had a comment where he states that Object.keys is not to be guaranteed to come in any particular order so how will you know what's next and previous?
    – HMR
    Nov 9 at 8:43




















  • Why not create using nested JSON?
    – HymnZ
    Nov 9 at 8:07










  • @HymnZ because I am using Amcharts and it does not support nested JSON data.
    – Syed Ariff
    Nov 9 at 8:09










  • You can try with Object.keys(your_array[1]) to create extra properties.. Just as @Grégory NEUT answered
    – HymnZ
    Nov 9 at 8:14






  • 1




    NB2: object properties are unordered, so the concept of "the number of properties after..." is unreliable. All you could say is "the number of properties other than...".
    – trincot
    Nov 9 at 8:21






  • 1




    Hey, no cheating. You changed the question from using properties such as 'State [Number]' to 'deatsh' and 'biths'. Trincot's deleted answer had a comment where he states that Object.keys is not to be guaranteed to come in any particular order so how will you know what's next and previous?
    – HMR
    Nov 9 at 8:43


















Why not create using nested JSON?
– HymnZ
Nov 9 at 8:07




Why not create using nested JSON?
– HymnZ
Nov 9 at 8:07












@HymnZ because I am using Amcharts and it does not support nested JSON data.
– Syed Ariff
Nov 9 at 8:09




@HymnZ because I am using Amcharts and it does not support nested JSON data.
– Syed Ariff
Nov 9 at 8:09












You can try with Object.keys(your_array[1]) to create extra properties.. Just as @Grégory NEUT answered
– HymnZ
Nov 9 at 8:14




You can try with Object.keys(your_array[1]) to create extra properties.. Just as @Grégory NEUT answered
– HymnZ
Nov 9 at 8:14




1




1




NB2: object properties are unordered, so the concept of "the number of properties after..." is unreliable. All you could say is "the number of properties other than...".
– trincot
Nov 9 at 8:21




NB2: object properties are unordered, so the concept of "the number of properties after..." is unreliable. All you could say is "the number of properties other than...".
– trincot
Nov 9 at 8:21




1




1




Hey, no cheating. You changed the question from using properties such as 'State [Number]' to 'deatsh' and 'biths'. Trincot's deleted answer had a comment where he states that Object.keys is not to be guaranteed to come in any particular order so how will you know what's next and previous?
– HMR
Nov 9 at 8:43






Hey, no cheating. You changed the question from using properties such as 'State [Number]' to 'deatsh' and 'biths'. Trincot's deleted answer had a comment where he states that Object.keys is not to be guaranteed to come in any particular order so how will you know what's next and previous?
– HMR
Nov 9 at 8:43














2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Crude.. but works



var newArr = ;
$x = [{
"waterfallData.PopulationName":"Population 1",
"Deaths":-2333,
"Births":8786,
"open":0,
"close":6453
},{
"waterfallData.PopulationName":"Population 2",
"Deaths":-1000,
"Births":5000,
"open":0,
"close":10453
},{
"waterfallData.PopulationName":"Population 3",
"Deaths":-2000,
"Births":500,
"open":0,
"close":8953
}];
$x.forEach((p,i)=>{
var current = $x[i]
newArr.push(current)
try {
var next = $x[i+1];
var start = current.open;
var end = current.close;
var states = Object.keys(current).sort().filter((k)=>{return (['waterfallData.PopulationName','open','close'].indexOf(k) < 0)})
for (var i=0;i<states.length;i++){
var state = states[i]
var tempObj = {}
tempObj[states[i]] = next[states[i]]
tempObj['open'] = end;
end += next[states[i]];
tempObj['close'] = end;
newArr.push(tempObj)
}
} catch (e) {
return false;
}
})


The code will look for all properties other than waterfallData.Popu..,open,close and treats them as states. If you have 10 properties, there will be 7 states excluding the above 3. Then the open and close values of these states are then calculated from the next element and pushed to new array newArr.






share|improve this answer























  • what is $x? It says ReferenceError $x is not defined.
    – Syed Ariff
    Nov 9 at 9:05










  • @SyedAriff .. a line got missed.. edited and updated.
    – HymnZ
    Nov 9 at 9:13










  • Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
    – HMR
    Nov 9 at 9:21










  • The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
    – HMR
    Nov 9 at 9:33










  • @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
    – HymnZ
    Nov 9 at 10:07


















up vote
0
down vote













I'll give it a try, you can specify what keys you need to be extra data but the original data must have values for those keys. It'll create the extra items in the order that the keys are provided:






const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

const zip = (arr1, arr2) =>
[...new Array(Math.max(arr1.length, arr2.length))]
.map((_, i) => i)
.map((i) => [arr1[i], arr2[i]]);
const extras = (keys) => ([current, next]) =>
keys.map((key) => [key, next[key]]).reduce(
([result, lastClose], [key, nextValue]) => [
result.concat({
[key]: nextValue,
open: lastClose,
close: nextValue + lastClose,
}),
nextValue + lastClose,
],
[, current.close],
)[0];
const withExtras = (keys) => ([current, next]) =>
!next
? [current]
: [current].concat(extras(keys)([current, next]));
console.log(
zip(data, data.slice(1)) //having [[current,next],[current,next]...]
.map(withExtras(['Deaths', 'Births']))
.flatten(),
);

console.log(
'diffenrent order different result',
zip(data, data.slice(1))
.map(withExtras(['Births', 'Deaths']))
.flatten(),
);








share|improve this answer























    Your Answer






    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: "1"
    };
    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',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    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%2fstackoverflow.com%2fquestions%2f53221806%2fhow-to-append-an-object-to-an-array-based-on-the-number-of-properties-other-than%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Crude.. but works



    var newArr = ;
    $x = [{
    "waterfallData.PopulationName":"Population 1",
    "Deaths":-2333,
    "Births":8786,
    "open":0,
    "close":6453
    },{
    "waterfallData.PopulationName":"Population 2",
    "Deaths":-1000,
    "Births":5000,
    "open":0,
    "close":10453
    },{
    "waterfallData.PopulationName":"Population 3",
    "Deaths":-2000,
    "Births":500,
    "open":0,
    "close":8953
    }];
    $x.forEach((p,i)=>{
    var current = $x[i]
    newArr.push(current)
    try {
    var next = $x[i+1];
    var start = current.open;
    var end = current.close;
    var states = Object.keys(current).sort().filter((k)=>{return (['waterfallData.PopulationName','open','close'].indexOf(k) < 0)})
    for (var i=0;i<states.length;i++){
    var state = states[i]
    var tempObj = {}
    tempObj[states[i]] = next[states[i]]
    tempObj['open'] = end;
    end += next[states[i]];
    tempObj['close'] = end;
    newArr.push(tempObj)
    }
    } catch (e) {
    return false;
    }
    })


    The code will look for all properties other than waterfallData.Popu..,open,close and treats them as states. If you have 10 properties, there will be 7 states excluding the above 3. Then the open and close values of these states are then calculated from the next element and pushed to new array newArr.






    share|improve this answer























    • what is $x? It says ReferenceError $x is not defined.
      – Syed Ariff
      Nov 9 at 9:05










    • @SyedAriff .. a line got missed.. edited and updated.
      – HymnZ
      Nov 9 at 9:13










    • Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
      – HMR
      Nov 9 at 9:21










    • The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
      – HMR
      Nov 9 at 9:33










    • @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
      – HymnZ
      Nov 9 at 10:07















    up vote
    1
    down vote



    accepted










    Crude.. but works



    var newArr = ;
    $x = [{
    "waterfallData.PopulationName":"Population 1",
    "Deaths":-2333,
    "Births":8786,
    "open":0,
    "close":6453
    },{
    "waterfallData.PopulationName":"Population 2",
    "Deaths":-1000,
    "Births":5000,
    "open":0,
    "close":10453
    },{
    "waterfallData.PopulationName":"Population 3",
    "Deaths":-2000,
    "Births":500,
    "open":0,
    "close":8953
    }];
    $x.forEach((p,i)=>{
    var current = $x[i]
    newArr.push(current)
    try {
    var next = $x[i+1];
    var start = current.open;
    var end = current.close;
    var states = Object.keys(current).sort().filter((k)=>{return (['waterfallData.PopulationName','open','close'].indexOf(k) < 0)})
    for (var i=0;i<states.length;i++){
    var state = states[i]
    var tempObj = {}
    tempObj[states[i]] = next[states[i]]
    tempObj['open'] = end;
    end += next[states[i]];
    tempObj['close'] = end;
    newArr.push(tempObj)
    }
    } catch (e) {
    return false;
    }
    })


    The code will look for all properties other than waterfallData.Popu..,open,close and treats them as states. If you have 10 properties, there will be 7 states excluding the above 3. Then the open and close values of these states are then calculated from the next element and pushed to new array newArr.






    share|improve this answer























    • what is $x? It says ReferenceError $x is not defined.
      – Syed Ariff
      Nov 9 at 9:05










    • @SyedAriff .. a line got missed.. edited and updated.
      – HymnZ
      Nov 9 at 9:13










    • Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
      – HMR
      Nov 9 at 9:21










    • The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
      – HMR
      Nov 9 at 9:33










    • @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
      – HymnZ
      Nov 9 at 10:07













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Crude.. but works



    var newArr = ;
    $x = [{
    "waterfallData.PopulationName":"Population 1",
    "Deaths":-2333,
    "Births":8786,
    "open":0,
    "close":6453
    },{
    "waterfallData.PopulationName":"Population 2",
    "Deaths":-1000,
    "Births":5000,
    "open":0,
    "close":10453
    },{
    "waterfallData.PopulationName":"Population 3",
    "Deaths":-2000,
    "Births":500,
    "open":0,
    "close":8953
    }];
    $x.forEach((p,i)=>{
    var current = $x[i]
    newArr.push(current)
    try {
    var next = $x[i+1];
    var start = current.open;
    var end = current.close;
    var states = Object.keys(current).sort().filter((k)=>{return (['waterfallData.PopulationName','open','close'].indexOf(k) < 0)})
    for (var i=0;i<states.length;i++){
    var state = states[i]
    var tempObj = {}
    tempObj[states[i]] = next[states[i]]
    tempObj['open'] = end;
    end += next[states[i]];
    tempObj['close'] = end;
    newArr.push(tempObj)
    }
    } catch (e) {
    return false;
    }
    })


    The code will look for all properties other than waterfallData.Popu..,open,close and treats them as states. If you have 10 properties, there will be 7 states excluding the above 3. Then the open and close values of these states are then calculated from the next element and pushed to new array newArr.






    share|improve this answer














    Crude.. but works



    var newArr = ;
    $x = [{
    "waterfallData.PopulationName":"Population 1",
    "Deaths":-2333,
    "Births":8786,
    "open":0,
    "close":6453
    },{
    "waterfallData.PopulationName":"Population 2",
    "Deaths":-1000,
    "Births":5000,
    "open":0,
    "close":10453
    },{
    "waterfallData.PopulationName":"Population 3",
    "Deaths":-2000,
    "Births":500,
    "open":0,
    "close":8953
    }];
    $x.forEach((p,i)=>{
    var current = $x[i]
    newArr.push(current)
    try {
    var next = $x[i+1];
    var start = current.open;
    var end = current.close;
    var states = Object.keys(current).sort().filter((k)=>{return (['waterfallData.PopulationName','open','close'].indexOf(k) < 0)})
    for (var i=0;i<states.length;i++){
    var state = states[i]
    var tempObj = {}
    tempObj[states[i]] = next[states[i]]
    tempObj['open'] = end;
    end += next[states[i]];
    tempObj['close'] = end;
    newArr.push(tempObj)
    }
    } catch (e) {
    return false;
    }
    })


    The code will look for all properties other than waterfallData.Popu..,open,close and treats them as states. If you have 10 properties, there will be 7 states excluding the above 3. Then the open and close values of these states are then calculated from the next element and pushed to new array newArr.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 9 at 10:08

























    answered Nov 9 at 8:53









    HymnZ

    1,3691918




    1,3691918












    • what is $x? It says ReferenceError $x is not defined.
      – Syed Ariff
      Nov 9 at 9:05










    • @SyedAriff .. a line got missed.. edited and updated.
      – HymnZ
      Nov 9 at 9:13










    • Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
      – HMR
      Nov 9 at 9:21










    • The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
      – HMR
      Nov 9 at 9:33










    • @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
      – HymnZ
      Nov 9 at 10:07


















    • what is $x? It says ReferenceError $x is not defined.
      – Syed Ariff
      Nov 9 at 9:05










    • @SyedAriff .. a line got missed.. edited and updated.
      – HymnZ
      Nov 9 at 9:13










    • Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
      – HMR
      Nov 9 at 9:21










    • The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
      – HMR
      Nov 9 at 9:33










    • @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
      – HymnZ
      Nov 9 at 10:07
















    what is $x? It says ReferenceError $x is not defined.
    – Syed Ariff
    Nov 9 at 9:05




    what is $x? It says ReferenceError $x is not defined.
    – Syed Ariff
    Nov 9 at 9:05












    @SyedAriff .. a line got missed.. edited and updated.
    – HymnZ
    Nov 9 at 9:13




    @SyedAriff .. a line got missed.. edited and updated.
    – HymnZ
    Nov 9 at 9:13












    Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
    – HMR
    Nov 9 at 9:21




    Now try to run your code where you switch the order of Deaths and Births in the $x data: $x = [{... "Births":8786, "Deaths":-2333 The output will differ
    – HMR
    Nov 9 at 9:21












    The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
    – HMR
    Nov 9 at 9:33




    The point is that the output of your code relies on Object.keys (look at the first line) to return the keys in a specific order but this is not the case. The specification says that there is no guaranteed order of the keys.
    – HMR
    Nov 9 at 9:33












    @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
    – HymnZ
    Nov 9 at 10:07




    @HMR.. obviously, because by switching order of births & deaths, your representation itself changes.. to make it standardized.. I've added a sort to ensure consistency in the order. On a side note, where can I learn to code minimally like you. :) It's glorious.. :D
    – HymnZ
    Nov 9 at 10:07












    up vote
    0
    down vote













    I'll give it a try, you can specify what keys you need to be extra data but the original data must have values for those keys. It'll create the extra items in the order that the keys are provided:






    const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

    const zip = (arr1, arr2) =>
    [...new Array(Math.max(arr1.length, arr2.length))]
    .map((_, i) => i)
    .map((i) => [arr1[i], arr2[i]]);
    const extras = (keys) => ([current, next]) =>
    keys.map((key) => [key, next[key]]).reduce(
    ([result, lastClose], [key, nextValue]) => [
    result.concat({
    [key]: nextValue,
    open: lastClose,
    close: nextValue + lastClose,
    }),
    nextValue + lastClose,
    ],
    [, current.close],
    )[0];
    const withExtras = (keys) => ([current, next]) =>
    !next
    ? [current]
    : [current].concat(extras(keys)([current, next]));
    console.log(
    zip(data, data.slice(1)) //having [[current,next],[current,next]...]
    .map(withExtras(['Deaths', 'Births']))
    .flatten(),
    );

    console.log(
    'diffenrent order different result',
    zip(data, data.slice(1))
    .map(withExtras(['Births', 'Deaths']))
    .flatten(),
    );








    share|improve this answer



























      up vote
      0
      down vote













      I'll give it a try, you can specify what keys you need to be extra data but the original data must have values for those keys. It'll create the extra items in the order that the keys are provided:






      const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

      const zip = (arr1, arr2) =>
      [...new Array(Math.max(arr1.length, arr2.length))]
      .map((_, i) => i)
      .map((i) => [arr1[i], arr2[i]]);
      const extras = (keys) => ([current, next]) =>
      keys.map((key) => [key, next[key]]).reduce(
      ([result, lastClose], [key, nextValue]) => [
      result.concat({
      [key]: nextValue,
      open: lastClose,
      close: nextValue + lastClose,
      }),
      nextValue + lastClose,
      ],
      [, current.close],
      )[0];
      const withExtras = (keys) => ([current, next]) =>
      !next
      ? [current]
      : [current].concat(extras(keys)([current, next]));
      console.log(
      zip(data, data.slice(1)) //having [[current,next],[current,next]...]
      .map(withExtras(['Deaths', 'Births']))
      .flatten(),
      );

      console.log(
      'diffenrent order different result',
      zip(data, data.slice(1))
      .map(withExtras(['Births', 'Deaths']))
      .flatten(),
      );








      share|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        I'll give it a try, you can specify what keys you need to be extra data but the original data must have values for those keys. It'll create the extra items in the order that the keys are provided:






        const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

        const zip = (arr1, arr2) =>
        [...new Array(Math.max(arr1.length, arr2.length))]
        .map((_, i) => i)
        .map((i) => [arr1[i], arr2[i]]);
        const extras = (keys) => ([current, next]) =>
        keys.map((key) => [key, next[key]]).reduce(
        ([result, lastClose], [key, nextValue]) => [
        result.concat({
        [key]: nextValue,
        open: lastClose,
        close: nextValue + lastClose,
        }),
        nextValue + lastClose,
        ],
        [, current.close],
        )[0];
        const withExtras = (keys) => ([current, next]) =>
        !next
        ? [current]
        : [current].concat(extras(keys)([current, next]));
        console.log(
        zip(data, data.slice(1)) //having [[current,next],[current,next]...]
        .map(withExtras(['Deaths', 'Births']))
        .flatten(),
        );

        console.log(
        'diffenrent order different result',
        zip(data, data.slice(1))
        .map(withExtras(['Births', 'Deaths']))
        .flatten(),
        );








        share|improve this answer














        I'll give it a try, you can specify what keys you need to be extra data but the original data must have values for those keys. It'll create the extra items in the order that the keys are provided:






        const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

        const zip = (arr1, arr2) =>
        [...new Array(Math.max(arr1.length, arr2.length))]
        .map((_, i) => i)
        .map((i) => [arr1[i], arr2[i]]);
        const extras = (keys) => ([current, next]) =>
        keys.map((key) => [key, next[key]]).reduce(
        ([result, lastClose], [key, nextValue]) => [
        result.concat({
        [key]: nextValue,
        open: lastClose,
        close: nextValue + lastClose,
        }),
        nextValue + lastClose,
        ],
        [, current.close],
        )[0];
        const withExtras = (keys) => ([current, next]) =>
        !next
        ? [current]
        : [current].concat(extras(keys)([current, next]));
        console.log(
        zip(data, data.slice(1)) //having [[current,next],[current,next]...]
        .map(withExtras(['Deaths', 'Births']))
        .flatten(),
        );

        console.log(
        'diffenrent order different result',
        zip(data, data.slice(1))
        .map(withExtras(['Births', 'Deaths']))
        .flatten(),
        );








        const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

        const zip = (arr1, arr2) =>
        [...new Array(Math.max(arr1.length, arr2.length))]
        .map((_, i) => i)
        .map((i) => [arr1[i], arr2[i]]);
        const extras = (keys) => ([current, next]) =>
        keys.map((key) => [key, next[key]]).reduce(
        ([result, lastClose], [key, nextValue]) => [
        result.concat({
        [key]: nextValue,
        open: lastClose,
        close: nextValue + lastClose,
        }),
        nextValue + lastClose,
        ],
        [, current.close],
        )[0];
        const withExtras = (keys) => ([current, next]) =>
        !next
        ? [current]
        : [current].concat(extras(keys)([current, next]));
        console.log(
        zip(data, data.slice(1)) //having [[current,next],[current,next]...]
        .map(withExtras(['Deaths', 'Births']))
        .flatten(),
        );

        console.log(
        'diffenrent order different result',
        zip(data, data.slice(1))
        .map(withExtras(['Births', 'Deaths']))
        .flatten(),
        );





        const data = [{"waterfallData.PopulationName":"Population 1","Deaths":-2333,"Births":8786,"open":0,"close":6453},{"waterfallData.PopulationName":"Population 2","Deaths":-1000,"Births":5000,"open":0,"close":10453},{"waterfallData.PopulationName":"Population 3","Deaths":-2000,"Births":500,"open":0,"close":8953}];

        const zip = (arr1, arr2) =>
        [...new Array(Math.max(arr1.length, arr2.length))]
        .map((_, i) => i)
        .map((i) => [arr1[i], arr2[i]]);
        const extras = (keys) => ([current, next]) =>
        keys.map((key) => [key, next[key]]).reduce(
        ([result, lastClose], [key, nextValue]) => [
        result.concat({
        [key]: nextValue,
        open: lastClose,
        close: nextValue + lastClose,
        }),
        nextValue + lastClose,
        ],
        [, current.close],
        )[0];
        const withExtras = (keys) => ([current, next]) =>
        !next
        ? [current]
        : [current].concat(extras(keys)([current, next]));
        console.log(
        zip(data, data.slice(1)) //having [[current,next],[current,next]...]
        .map(withExtras(['Deaths', 'Births']))
        .flatten(),
        );

        console.log(
        'diffenrent order different result',
        zip(data, data.slice(1))
        .map(withExtras(['Births', 'Deaths']))
        .flatten(),
        );






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 9 at 10:08

























        answered Nov 9 at 9:12









        HMR

        13.2k103898




        13.2k103898






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • 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.


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





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2fstackoverflow.com%2fquestions%2f53221806%2fhow-to-append-an-object-to-an-array-based-on-the-number-of-properties-other-than%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







            這個網誌中的熱門文章

            Tangent Lines Diagram Along Smooth Curve

            Yusuf al-Mu'taman ibn Hud

            Zucchini