Javascript nested filters in Array of arrays

Question

I have an array of objects in this format:

var full_list = [
        {
            "pid": 1,
            "items":[
                {"item_id": '9'},
                {"item_id": '10'},
                {"item_id": '12'}
            ]
        },
        {
            "pid": 2,
            "items":[
                {"item_id": '33'},
                {"item_id": '22'},
                {"item_id": '65'}
            ]
        }...
    ];

I have a tmp array which consists of objects from the full array:

 var tmp_list =  [
        {
            "pid": 2,
            "items":[
                {"item_id": '33'},
                {"item_id": '22'},
                {"item_id": '65'}
            ]
        }, {....}

I would like to filter out objects from the full list where at least one of selectedIDs values appears in the object's item's id array

var selectedIDs = {'1', '9', '45', ....};

and then add them to the tmp list.

I tried using filters but I failed to figure it out completely.

Thank you.

selectedIDs.forEach(function(id) {
                var tmp = full_list.filter(function (obj) {
                            obj.items.forEach(function (item) {
                                if (item.id === id) {
                                    console.log('found');
                                }
                            });
                        });
                        tmp_list.push(tmp);
                 });

Do you have the option to use libraries like lodash etc? Issues like these there are solved quite nicely. — Akrion
– Akrion, Commented Jun 26, 2016 at 14:10
Is it possible some already exist in temp list? Current answers would create duplicates if so — charlietfl
– charlietfl, Commented Jun 26, 2016 at 14:18

Aditya Singh · Accepted Answer · 2016-06-27 10:02:28Z

5

First of all, this line in your question is wrong

var selectedIDs = {'1', '9', '45', ....};

You cannot declare arrays using {}. Instead use []

For your problem you can use a pure functional approach using Array#filter and Array#some methods to get your desired result as below:

var full_list = [
  {
    "pid": 1,
    "items":[
      {"item_id": '9'},
      {"item_id": '10'},
      {"item_id": '12'}
    ]
  },
  {
    "pid": 2,
    "items":[
      {"item_id": '33'},
      {"item_id": '22'},
      {"item_id": '67'}
    ]
  },
  {
    "pid": 9,
    "items":[
      {"item_id": '33'},
      {"item_id": '22'},
      {"item_id": '65'}
    ]
  },
  {
    "pid": 7,
    "items":[
      {"item_id": '7'},
      {"item_id": '22'},
      {"item_id": '65'}
    ]
  }
];

var tmp_list = [
  {
    "pid": 2,
    "items":[
      {"item_id": '7'},
      {"item_id": '22'},
      {"item_id": '65'}
    ]
  }
];


function filterResult (selectedItems) {
  return full_list.filter(function (process) {
    return process.items.some(function(item){
      return selectedItems.indexOf(item.item_id) > -1;
    });
  });
}

var selectedItems = ['9', '7', '22', '10'];

tmp_list = tmp_list.concat(filterResult(selectedItems))

console.log(tmp_list);


function flattenResults(list, selections) {
  return list.reduce(function (accumulator, current) {
    var res = current.items.filter(function(item){
      return (selections.indexOf(item.item_id) > -1 
              && checkIfAlreadyExist());
              
      function checkIfAlreadyExist () {
        return accumulator.every(function (k) {
          return k.item_id !== item.item_id;
        });
      }        
    });   

    return accumulator.concat(res);
  }, []);
}

console.log(flattenResults(full_list, selectedItems));

edited Jun 27, 2016 at 10:02

answered Jun 26, 2016 at 14:27

Aditya Singh

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7 Comments

badigard Over a year ago

honestly, I don't understand those nested 'return's, but it worked flawlessly. thanks.

Aditya Singh Over a year ago

For understanding them refer the MDN documentation for Array.filter and Array.some here: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Aditya Singh Over a year ago

@badigard: Also if the solution works for you please accept it so that others don't have to explore further for the solution of same/ similar problem

badigard Over a year ago

done. another small Q: how do I modify this code so it will return the {item_id: 'x'} object it found?

Aditya Singh Over a year ago

Do you mean only list of the found {item_id: 'x'} objects but not the whole object containing pid, items?

|

Redu · Accepted Answer · 2016-06-26 14:17:54Z

4

You may do like this;

var full_list = [
        {
            "pid": 1,
            "items":[
                {"item_id": '9'},
                {"item_id": '10'},
                {"item_id": '12'}
            ]
        },
        {
            "pid": 2,
            "items":[
                {"item_id": '33'},
                {"item_id": '22'},
                {"item_id": '65'}
            ]
        }
],
  selectedIDs = ['1', '9', '45'],
     tempList = [];
tempList.push(full_list.filter(f => f.items.some(o => selectedIDs.includes(o.item_id))));
console.log(tempList);

answered Jun 26, 2016 at 14:17

Redu

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Comments

Nina Scholz · Accepted Answer · 2016-06-26 14:31:09Z

4

You could use a hash table for the selectedID and use it for fast filtering.

var full_list = [{ "pid": 1, "items": [{ "item_id": '9' }, { "item_id": '10' }, { "item_id": '12' }] }, { "pid": 2, "items": [{ "item_id": '33' }, { "item_id": '22' }, { "item_id": '65' }] }],
    tmp_list,
    selectedIDs = ['1', '9', '45'],
    selected = Object.create(null);

selectedIDs.forEach(function (a) {
    selected[a] = true;
});

tmp_list = full_list.filter(function (a) {
    return !a.items.some(function (b) {
        return selected[b.item_id];
    });
});

console.log(tmp_list);

ES6

var full_list = [{ "pid": 1, "items": [{ "item_id": '9' }, { "item_id": '10' }, { "item_id": '12' }] }, { "pid": 2, "items": [{ "item_id": '33' }, { "item_id": '22' }, { "item_id": '65' }] }],
    tmp_list,
    selectedIDs = ['1', '9', '45'],
    selected = Object.create(null);

selectedIDs.forEach(a => selected[a] = true);
tmp_list = full_list.filter(a => !a.items.some(b=> selected[b.item_id]));

console.log(tmp_list);

edited Jun 26, 2016 at 14:31

answered Jun 26, 2016 at 14:25

Nina Scholz

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4 Comments

Morteza Tourani Over a year ago

You made it wrong. you must check of the item_id is in selectedIDs.

Nina Scholz Over a year ago

i use a hash table selected and the check appears here: !a.items.some(function (b) { return selected[b.item_id]; });.

Morteza Tourani Over a year ago

That would be the best answer if it returns the correct answer. but non of selected ids are in returned value

Nina Scholz Over a year ago

sorry, that is not right, it should filter out where some of item_id appears in selectedIDs.

Allex Daemon · Accepted Answer · 2019-09-06 16:02:50Z

3

array1
array2

array1.filter(el=>{
   return array2.filter(el2=>{
          return el.id == el2.id
     })
})

It will take array1 and run it against the second one ... for each value.

edited Sep 6, 2019 at 16:02

answered Sep 6, 2019 at 15:57

Allex Daemon

392 bronze badges

Comments

Morteza Tourani · Accepted Answer · 2016-06-26 14:19:19Z

0

I think this could be a solution.

var full_list = [
  { 
    "pid": 1, 
    "items": [
    { "item_id": '9' }, 
    { "item_id": '10' }, 
    { "item_id": '12' }
  ]
}, {
  "pid": 2,
  "items": [{
    "item_id": '33'
  }, {
    "item_id": '22'
  }, {
    "item_id": '65'
  }]
}];

var selectedIDs = ['33', '3'];
var tempList = [];

full_list
  .filter(item => 
    item.items
      .some(i => selectedIDs.indexOf(i.item_id) != -1)
   ).forEach(item => tempList.push(item));
console.log(tempList)

edited Jun 26, 2016 at 14:19

answered Jun 26, 2016 at 14:13

Morteza Tourani

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Comments

Rohit Shedage · Accepted Answer · 2016-06-26 14:21:35Z

-2

Looking at your job 1. There is nothing like market

obj.items.forEach(function (item) { if (item.item_id === id) { console.log('found'); return true; } return false;
});

answered Jun 26, 2016 at 14:21

Rohit Shedage

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1 Comment

charlietfl Over a year ago

where does forEach return to?

Jose Hermosilla Rodrigo · Accepted Answer · 2016-06-26 14:59:03Z

-2

Spread operator should simplify some code:

var full_list = [
  { 
    "pid": 1, 
    "items": [
    { "item_id": '9' }, 
    { "item_id": '10' }, 
    { "item_id": '12' }
  ]
}, {
  "pid": 2,
  "items": [{
    "item_id": '33'
  }, {
    "item_id": '22'
  }, {
    "item_id": '65'
  }]
}];

var selectedIDs = ['65', '6'];
var tempList = [];

tempList = [...tempList, 
            ...full_list.filter(item => item.items.some(i => selectedIDs.includes(i.item_id)))
           ];

console.log(tempList);

edited Jun 26, 2016 at 14:59

answered Jun 26, 2016 at 14:52

Jose Hermosilla Rodrigo

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5 Comments

Jose Hermosilla Rodrigo Over a year ago

@mortezaT why redundant? It's necessary for destructuring the two list into new one. I think it's equivalent to the approach with Array.concat() method but with this new operator, that makes it more readable, unless for me. =)

Morteza Tourani Over a year ago

None of the above will append to tempList.

Jose Hermosilla Rodrigo Over a year ago

I don't get you! The OP says: "I would like to filter out objects from the full list where at least one of selectedIDs values appears in the object's item's id array and then add them to the tmp list." That's exactly what I'm doing...

Morteza Tourani Over a year ago

The exact part which saying and then add them to the tmp list.

Nina Scholz Over a year ago

then the example of tmp_list makes no sense.

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