flatten array of object with map (or loop)

Question

I have an array of objects like this one:

[
    {
        "dow": 0,
        "min_price": 2000,
        "max_price": 4000,
        "is_enable": true
    },
    {
        "dow": 1,
        "min_price": 500,
        "max_price": 3000,
        "is_enable": true
    },
    {
        "dow": 2,
        "min_price": 500,
        "max_price": 3000,
        "is_enable": true
    },
    {
        "dow": 3,
        "min_price": 1000,
        "max_price": 3000,
        "is_enable": true
    }
]

How can I turn it into something like this one? The order doesn't matter.

{
    advancePrice_0_max: 4000,
    advancePrice_0_min: 2000,
    advancePrice_1_max: 3000,
    advancePrice_1_min: 500,
    advancePrice_2_max: 3000,
    advancePrice_2_min: 500,
    advancePrice_3_max: 3000,
    advancePrice_3_min: 500
    ..
    ..
}

I've tried to begin with this, but even below code has snytax error.

let temp = {};
if (!isEmpty.custom_pricing) {
  temp = custom_pricing.map(obj => {
    `advancePrice_${obj.dow}_min`: obj.min_price,
    `advancePrice_${obj.dow}_max`: obj.max_price
  })
  console.log(temp);
}

But I'm stuck and don't know how to continue..

score 1 · Answer 1 · answered May 14 '17 at 17:00

1

You want to reduce the array to a single object, so:

var result = data.reduce(function (acc,el,i) {
  acc['advancedprice_' + i + '_max'] = el.max_price;
  acc['advancedprice_' + i + '_min'] = el.min_price;
  return acc;
}, {});

answered May 14 '17 at 17:00

James

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clairestreb · Answer 2 · 2017-07-23T19:12:06.523

You have two choices:

Write your own solution
Use someone else's solution

I advise you to use someone else's solution, because you don't have to reinvent the wheel. This answer is exhaustive with examples: Fastest way to flatten / un-flatten nested JSON objects, and there is even an online example here: http://fiddle.jshell.net/blowsie/S2hsS/show/light/

See the code snippets of:

JSON.flatten = function (data) { ...

and:

JSON.unflatten = function (data) {

Mayank Shukla · Accepted Answer · 2017-05-14T17:10:46.187

Use forEach because you want the result as an object not in array format, if you use map then temp will be an array with returned data.

use this:

let temp = {};
if (!isEmpty.custom_pricing) {
   custom_pricing.forEach(obj => {
      temp[`advancePrice_${obj.dow}_min`] = obj.min_price;
      temp[`advancePrice_${obj.dow}_max`] = obj.max_price;
  })
  console.log(temp);
}

Check the working snippet:

let custom_pricing = [{
        "dow": 0,
        "min_price": 2000,
        "max_price": 4000,
        "is_enable": true
    },
    {
        "dow": 1,
        "min_price": 500,
        "max_price": 3000,
        "is_enable": true
    },
    {
        "dow": 2,
        "min_price": 500,
        "max_price": 3000,
        "is_enable": true
    },
    {
        "dow": 3,
        "min_price": 1000,
        "max_price": 3000,
        "is_enable": true
    }
]

let temp = {};
custom_pricing.forEach(obj => {
   temp[`advancePrice_${obj.dow}_min`] = obj.min_price;
   temp[`advancePrice_${obj.dow}_max`] = obj.max_price;
})
console.log(temp);

why downvote, what is wrong with this solution, please provide the reason also :( — Mayank Shukla, May 14 '17 at 17:15
I agree. There is a guy who put `-1` to each answer (including mine). This is unfair, so I give you an upvote to compensate. — Badacadabra, May 14 '17 at 17:20

score 0 · Answer 4 · answered May 14 '17 at 17:07

You could use the good old for loop...

let arr = [
  {
    "dow": 0,
    "min_price": 2000,
    "max_price": 4000,
    "is_enable": true
  },
  {
    "dow": 1,
    "min_price": 500,
    "max_price": 3000,
    "is_enable": true
  },
  {
    "dow": 2,
    "min_price": 500,
    "max_price": 3000,
    "is_enable": true
  },
  {
    "dow": 3,
    "min_price": 1000,
    "max_price": 3000,
    "is_enable": true
  }
];

let res = {};

for (let i = 0; i < arr.length; i++) {
  res[`advancePrice_${i}_max`] = arr[i].max_price;
  res[`advancePrice_${i}_min`] = arr[i].min_price;
}

console.log(res);

flatten array of object with map (or loop)

4 Answers4