How to get the minimum and maximum date and grouped it by id

-2

I have this array of data

[{"id":1, "start":"2018-10-10", "end":"2018-11-10"},

 {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

 {"id":2, "start":"2018-11-22", "end":"2018-11-30"}]

I wanted to get the minimum in the start and the maximum in the end.

My desired output would be

{"id":1, "start":"2018-10-10", "end":"2018-12-10"},

{"id":2, "start":"2018-11-22", "end":"2018-11-30"}

I tried doing like this:

data.sort((a,b) => a.start.toString().localeCompare(b.start))

edited Nov 20 '18 at 8:46

HMR

13.9k113899

asked Nov 20 '18 at 8:27

sad saddest

338

There is zero jQuery in your question. Also, none of the objects have a start_date property, and you don't have to call toString() on something that's already a string.

– CertainPerformance
Nov 20 '18 at 8:28

I thought sort is jquery sorry

– sad saddest
Nov 20 '18 at 8:28

why haven't you used any of the grouping code you got in the question you asked 2 hours ago? Stackoverflow is not a free code writing service. People will happily help but we aren't here to do all the work for you

– charlietfl
Nov 20 '18 at 8:37

add a comment |

-2

I have this array of data

[{"id":1, "start":"2018-10-10", "end":"2018-11-10"},

 {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

 {"id":2, "start":"2018-11-22", "end":"2018-11-30"}]

I wanted to get the minimum in the start and the maximum in the end.

My desired output would be

{"id":1, "start":"2018-10-10", "end":"2018-12-10"},

{"id":2, "start":"2018-11-22", "end":"2018-11-30"}

I tried doing like this:

data.sort((a,b) => a.start.toString().localeCompare(b.start))

edited Nov 20 '18 at 8:46

HMR

13.9k113899

asked Nov 20 '18 at 8:27

sad saddest

338

There is zero jQuery in your question. Also, none of the objects have a start_date property, and you don't have to call toString() on something that's already a string.

– CertainPerformance
Nov 20 '18 at 8:28

I thought sort is jquery sorry

– sad saddest
Nov 20 '18 at 8:28

why haven't you used any of the grouping code you got in the question you asked 2 hours ago? Stackoverflow is not a free code writing service. People will happily help but we aren't here to do all the work for you

– charlietfl
Nov 20 '18 at 8:37

add a comment |

-2

I have this array of data

[{"id":1, "start":"2018-10-10", "end":"2018-11-10"},

 {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

 {"id":2, "start":"2018-11-22", "end":"2018-11-30"}]

I wanted to get the minimum in the start and the maximum in the end.

My desired output would be

{"id":1, "start":"2018-10-10", "end":"2018-12-10"},

{"id":2, "start":"2018-11-22", "end":"2018-11-30"}

I tried doing like this:

data.sort((a,b) => a.start.toString().localeCompare(b.start))

edited Nov 20 '18 at 8:46

HMR

13.9k113899

asked Nov 20 '18 at 8:27

sad saddest

338

I have this array of data

[{"id":1, "start":"2018-10-10", "end":"2018-11-10"},

 {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

 {"id":2, "start":"2018-11-22", "end":"2018-11-30"}]

I wanted to get the minimum in the start and the maximum in the end.

My desired output would be

{"id":1, "start":"2018-10-10", "end":"2018-12-10"},

{"id":2, "start":"2018-11-22", "end":"2018-11-30"}

I tried doing like this:

data.sort((a,b) => a.start.toString().localeCompare(b.start))

javascript

edited Nov 20 '18 at 8:46

HMR

13.9k113899

asked Nov 20 '18 at 8:27

sad saddest

338

edited Nov 20 '18 at 8:46

HMR

13.9k113899

asked Nov 20 '18 at 8:27

sad saddest

338

edited Nov 20 '18 at 8:46

HMR

13.9k113899

edited Nov 20 '18 at 8:46

HMR

13.9k113899

edited Nov 20 '18 at 8:46

HMR

13.9k113899

asked Nov 20 '18 at 8:27

sad saddest

338

asked Nov 20 '18 at 8:27

sad saddest

338

asked Nov 20 '18 at 8:27

sad saddest

338

There is zero jQuery in your question. Also, none of the objects have a start_date property, and you don't have to call toString() on something that's already a string.

– CertainPerformance
Nov 20 '18 at 8:28

I thought sort is jquery sorry

– sad saddest
Nov 20 '18 at 8:28

why haven't you used any of the grouping code you got in the question you asked 2 hours ago? Stackoverflow is not a free code writing service. People will happily help but we aren't here to do all the work for you

– charlietfl
Nov 20 '18 at 8:37

add a comment |

There is zero jQuery in your question. Also, none of the objects have a start_date property, and you don't have to call toString() on something that's already a string.

– CertainPerformance
Nov 20 '18 at 8:28

I thought sort is jquery sorry

– sad saddest
Nov 20 '18 at 8:28

why haven't you used any of the grouping code you got in the question you asked 2 hours ago? Stackoverflow is not a free code writing service. People will happily help but we aren't here to do all the work for you

– charlietfl
Nov 20 '18 at 8:37

There is zero jQuery in your question. Also, none of the objects have a start_date property, and you don't have to call toString() on something that's already a string.

– CertainPerformance
Nov 20 '18 at 8:28

I thought sort is jquery sorry

– sad saddest
Nov 20 '18 at 8:28

why haven't you used any of the grouping code you got in the question you asked 2 hours ago? Stackoverflow is not a free code writing service. People will happily help but we aren't here to do all the work for you

– charlietfl
Nov 20 '18 at 8:37

add a comment |

4 Answers
4

active

oldest

votes

What you are trying to do will require a custom solution where you merge two entries. In the below function I have iterated your array in chunks of same id first I have sorted it based on id and kept a min and max of the index where your logic says on date

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

add a comment |

you need to sort array using date comparison and also reduce it by id.

var data = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];





let result = data.sort((a, b) => new Date(a.start) > new Date(b.start)).reduce(function(r, a) {

  if (!r[a.id] || r[a.id][0].id !== a.id) {

    r[a.id] = r[a.id] || ;

    r[a.id].push(a);

  }



  return r;

}, Object.create(null));



console.log(result)

answered Nov 20 '18 at 8:49

dganenco

2217

add a comment |

You seem to be having multiple things that need to be grouped and reduces (as shown by comment referring to other question). So I will provide some general methods in this answer.

MDN has excelent documentation on all methods used here like map, reduce and Object.value, I would advice you have a look there to understand the code better.

ES6 syntax is explained very well here

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

answered Nov 20 '18 at 9:04

HMR

13.9k113899

add a comment |

You can use reduce function and check the start and end date and if start date

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

answered Nov 20 '18 at 9:09

brk

27.4k32141

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

add a comment |

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

add a comment |

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

function groupById(arr){

  arr.sort((a,b)=>a.id-b.id);

  let arrNew = ;

  let min=0, max=0, currentid = arr[0].id;

  for(i=1; i<arr.length+1;i++){

    if(!arr[i] || arr[i].id!=currentid){

      arrNew.push({id:currentid, start:arr[min].start, end: arr[max].end});

      min = i;max=i;currentid=(arr[i]||{}).id;

    }

    if(!arr[i]){

      break;

    }

    if(arr[i].start<arr[min].start){

      min = i;

    }

    if(arr[i].end>arr[max].end){

      max = i;

    }

  }

  return arrNew;

}



var result = groupById([

  {"id":1, "start":"2018-10-10", "end":"2018-11-10"},

  {"id":1, "start":"2018-11-10", "end":"2018-12-10"},

  {"id":2, "start":"2018-11-22", "end":"2018-11-30"}

]);



console.log(result);

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

answered Nov 20 '18 at 8:48

joyBlanks

3,61611036

add a comment |

you need to sort array using date comparison and also reduce it by id.

var data = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];





let result = data.sort((a, b) => new Date(a.start) > new Date(b.start)).reduce(function(r, a) {

  if (!r[a.id] || r[a.id][0].id !== a.id) {

    r[a.id] = r[a.id] || ;

    r[a.id].push(a);

  }



  return r;

}, Object.create(null));



console.log(result)

answered Nov 20 '18 at 8:49

dganenco

2217

add a comment |

you need to sort array using date comparison and also reduce it by id.

var data = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];





let result = data.sort((a, b) => new Date(a.start) > new Date(b.start)).reduce(function(r, a) {

  if (!r[a.id] || r[a.id][0].id !== a.id) {

    r[a.id] = r[a.id] || ;

    r[a.id].push(a);

  }



  return r;

}, Object.create(null));



console.log(result)

answered Nov 20 '18 at 8:49

dganenco

2217

add a comment |

you need to sort array using date comparison and also reduce it by id.

var data = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];





let result = data.sort((a, b) => new Date(a.start) > new Date(b.start)).reduce(function(r, a) {

  if (!r[a.id] || r[a.id][0].id !== a.id) {

    r[a.id] = r[a.id] || ;

    r[a.id].push(a);

  }



  return r;

}, Object.create(null));



console.log(result)

answered Nov 20 '18 at 8:49

dganenco

2217

you need to sort array using date comparison and also reduce it by id.

var data = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];





let result = data.sort((a, b) => new Date(a.start) > new Date(b.start)).reduce(function(r, a) {

  if (!r[a.id] || r[a.id][0].id !== a.id) {

    r[a.id] = r[a.id] || ;

    r[a.id].push(a);

  }



  return r;

}, Object.create(null));



console.log(result)

answered Nov 20 '18 at 8:49

dganenco

2217

answered Nov 20 '18 at 8:49

dganenco

2217

answered Nov 20 '18 at 8:49

dganenco

2217

answered Nov 20 '18 at 8:49

dganenco

2217

add a comment |

You seem to be having multiple things that need to be grouped and reduces (as shown by comment referring to other question). So I will provide some general methods in this answer.

MDN has excelent documentation on all methods used here like map, reduce and Object.value, I would advice you have a look there to understand the code better.

ES6 syntax is explained very well here

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

answered Nov 20 '18 at 9:04

HMR

13.9k113899

add a comment |

You seem to be having multiple things that need to be grouped and reduces (as shown by comment referring to other question). So I will provide some general methods in this answer.

MDN has excelent documentation on all methods used here like map, reduce and Object.value, I would advice you have a look there to understand the code better.

ES6 syntax is explained very well here

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

answered Nov 20 '18 at 9:04

HMR

13.9k113899

add a comment |

You seem to be having multiple things that need to be grouped and reduces (as shown by comment referring to other question). So I will provide some general methods in this answer.

MDN has excelent documentation on all methods used here like map, reduce and Object.value, I would advice you have a look there to understand the code better.

ES6 syntax is explained very well here

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

answered Nov 20 '18 at 9:04

HMR

13.9k113899

You seem to be having multiple things that need to be grouped and reduces (as shown by comment referring to other question). So I will provide some general methods in this answer.

MDN has excelent documentation on all methods used here like map, reduce and Object.value, I would advice you have a look there to understand the code better.

ES6 syntax is explained very well here

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

const data = [

  { id: 1, start: '2018-10-10', end: '2018-11-10' },

  { id: 1, start: '2018-11-10', end: '2018-12-10' },

  { id: 2, start: '2018-11-22', end: '2018-11-30' },

];



const groupBy = (arr, key) =>

  arr.reduce(

    (result, item) => (

      result[item[key]].push(item), result

    ),

    arr.reduce(

      (result, item) => ((result[item[key]] = ), result),

      {},

    ),

  );

const returnLowHigh = (comp) => (a, b) =>

  a.localeCompare(b) === comp ? a : b;

const lowest = returnLowHigh(-1);

const highest = returnLowHigh(1);

console.log(

  Object.values(groupBy(data, 'id')).map((items) =>

    items.reduce((result, { id, start, end }) => ({

      id,

      start: lowest(result.start, start),

      end: highest(result.end, end),

    })),

  ),

);

answered Nov 20 '18 at 9:04

HMR

13.9k113899

answered Nov 20 '18 at 9:04

HMR

13.9k113899

answered Nov 20 '18 at 9:04

HMR

13.9k113899

answered Nov 20 '18 at 9:04

HMR

13.9k113899

add a comment |

You can use reduce function and check the start and end date and if start date

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

answered Nov 20 '18 at 9:09

brk

27.4k32141

add a comment |

You can use reduce function and check the start and end date and if start date

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

answered Nov 20 '18 at 9:09

brk

27.4k32141

add a comment |

You can use reduce function and check the start and end date and if start date

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

answered Nov 20 '18 at 9:09

brk

27.4k32141

You can use reduce function and check the start and end date and if start date

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

let old = [{

    "id": 1,

    "start": "2018-10-10",

    "end": "2018-11-10"

  },

  {

    "id": 1,

    "start": "2018-11-10",

    "end": "2018-12-10"

  },

  {

    "id": 2,

    "start": "2018-11-22",

    "end": "2018-11-30"

  }

];



let k = old.reduce(function(acc, curr) {

  let findId = acc.findIndex((item) => {

    return item.id === curr.id

  });



  if (findId === -1) {

    acc.push(curr)

  } else {

    let oldStartDate = createDate(acc[findId].start);

    let newStartDate = createDate(curr.start);

    let oldEndDate = createDate(acc[findId].end);

    let newEndDate = createDate(curr.end);





    if (newStartDate < oldStartDate) {



      acc[findId].start = curr.start

    }

    if (newEndDate > oldEndDate) {



      acc[findId].end = curr.end

    }



  }



  return acc;



}, );



console.log(k);



function createDate(dte) {

  let dt = new Date(dte);

  return `${dt.getYear()}-${dt.getMonth()}-${dt.getDate()}`



}

answered Nov 20 '18 at 9:09

brk

27.4k32141

answered Nov 20 '18 at 9:09

brk

27.4k32141

answered Nov 20 '18 at 9:09

brk

27.4k32141

answered Nov 20 '18 at 9:09

brk

27.4k32141

add a comment |

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