Complex selection and table

Question

I have an array named listSelected which feeds after each selection in another grid.

const listSelected =    
 [{    
    "_id": "10",    
    "age": 35,    
    "name": "Paige Zamora",    
    "gender": "female",    
    "company": "AUTOMON",    
    "**reference_id": "12**"    
  },    
  {    
    "_id": "11",    
    "age": 40,    
    "name": "Jennifer Carr",    
    "gender": "female",    
    "company": "SYNKGEN",    
    "**reference_id": "11**"    
  }];

I would like to check that for each value reference_id one checks in another table named data if this value exists, we get the value id in a new array, even if the value reference_id is present several times.

const data = [{    
    "_id": "**10**",    
    "age": 35,    
    "name": "Paige Zamora",    
    "gender": "female",    
    "company": "AUTOMON",    
    "**reference_id": "12**"    
  },    
  {    
    "_id": "**11**",    
    "age": 40,    
    "name": "Jennifer Carr",    
    "gender": "female",    
    "company": "SYNKGEN",    
    "**reference_id": "11**"    
  },    
  {    
    "_id": "**12**",    
    "age": 38,    
    "name": "Weaver Rosales",    
    "gender": "male",    
    "company": "ETERNIS",    
    "**reference_id": "12**"    
  },    
  {
    "_id": "13",    
    "age": 31,    
    "name": "Myers Pickett",    
    "gender": "male",    
    "company": "ETERNIS",    
    "reference_id": "13"    
  },    
  {    
    "_id": "14",    
    "age": 36,    
    "name": "Dona Nicholson",    
    "gender": "female",    
    "company": "ETERNIS",    
    "reference_id": "14"    
  },    
  {
    "_id": "15",    
    "age": 21,    
    "name": "Niki Blur",    
    "gender": "female",    
    "company": "AUTOMON",    
    "reference_id": "15"    
  },    
  {    
    "_id": "16",    
    "age": 37,    
    "name": "Bod Dennart",    
    "gender": "male",    
    "company": "SYNKGEN",    
    "reference_id": "16"    
  },    
  {    
    "_id": "17",    
    "age": 26,    
    "name": "Richard Nelson",    
    "gender": "male",    
    "company": "ETERNIS",    
    "reference_id": "17"    
  },    
  {    
    "_id": "**12**",    
    "age": 45,    
    "name": "Pedro Kluivert",    
    "gender": "female",    
    "company": "SYNKGEN",    
    "**reference_id": "18**"    
  }    
];

In the end the new grid will give the following data:

const newGrid = [11,12]

I was thinking of using two nested forEach functions, like these :

const newGrid = [];

listSelected.forEach((elt) => {
  data.forEach((item) => {
    if (item.reference_id === elt.reference_id) {
       newGrid.push(item.id);
       newGrid = Array.from(new Set(newGrid));
     }
   });
});

Is there not a simpler way to simplify this function to avoid using nested forEach ?

Your question is totally unclear. Please at least provide the end result, so that we know what to aim for. — , Feb 04 '19 at 12:39
The final result is indicated in the post. **newGrid = [11,12]** — Greg-A, Feb 04 '19 at 12:40
It is the data array ? Because in your nested loops, you seem to use it ... — , Feb 04 '19 at 12:41
No, as stated in several answers, you can use other operators on your arrays. I also have posted a solution implying a single loop in the code, to avoid nested loops (be warned that `indexOf` still makes two loops, it's just your code that looks "cleaner") — , Feb 04 '19 at 13:17

score 3 · Accepted Answer · answered Feb 04 '19 at 12:55

You could make a "reverse-unique" array :

const listSelected = [
  { "_id": "10", "age": 35, "name": "Paige Zamora", "gender": "female", "company": "AUTOMON", "reference_id": "12" }, 
  { "_id": "11", "age": 40, "name": "Jennifer Carr", "gender": "female", "company": "SYNKGEN", "reference_id": "11" }
];

const data = [
  { "_id": "10", "age": 35, "name": "Paige Zamora", "gender": "female", "company": "AUTOMON", "reference_id": "12" }, 
  { "_id": "11", "age": 40, "name": "Jennifer Carr", "gender": "female", "company": "SYNKGEN", "reference_id": "11" }, 
  { "_id": "12", "age": 38, "name": "Weaver Rosales", "gender": "male", "company": "ETERNIS", "reference_id": "12" }, 
  { "_id": "13", "age": 31, "name": "Myers Pickett", "gender": "male", "company": "ETERNIS", "reference_id": "13" }, 
  { "_id": "14", "age": 36, "name": "Dona Nicholson", "gender": "female", "company": "ETERNIS", "reference_id": "14" }, 
  { "_id": "15", "age": 21, "name": "Niki Blur", "gender": "female", "company": "AUTOMON", "reference_id": "15" }, 
  { "_id": "16", "age": 37, "name": "Bod Dennart", "gender": "male", "company": "SYNKGEN", "reference_id": "16" }, 
  { "_id": "17", "age": 26, "name": "Richard Nelson", "gender": "male", "company": "ETERNIS", "reference_id": "17" }, 
  { "_id": "12", "age": 45, "name": "Pedro Kluivert", "gender": "female", "company": "SYNKGEN", "reference_id": "18" }
];

const merged = [
  ...listSelected.map(item => item.reference_id),
  ...data.map(item => item.reference_id)
];

const reverseUnique = merged.filter((item, index, array) => array.indexOf(item) === index && array.lastIndexOf(item) !== index);

console.log(reverseUnique);

This only makes one loop, instead of every other method doing two nested (because yes, filter/find/some do a loop over the array)

You start by concatenating the two arrays in a single one, and using only the reference ID.

Then, you use the opposite of a unique filtering function.

so you are saying others are using two nested loops! what do you think the runtime of array.indexof() method? — Md Golam Rahman Tushar, Feb 04 '19 at 13:18
@GolamRahmanTushar maybe you should check the comments on the question. — , Feb 04 '19 at 13:19
so you think, instead of using find method inside forEach, first merging the arrays and then writing a single for loop with two nested forloops inside it, is much more clean? — Md Golam Rahman Tushar, Feb 04 '19 at 13:24
Where did I say that ? I provided a solution implying a single loop (understand, "not nested functions") instead of two nested functions. But if you ask me what is the cleanest, then `reduce` would be. Maybe you should stop talking to me, you seem to have an issue with my self and you don't actually criticize the answer, but rather my personal opinions. — , Feb 04 '19 at 13:27
I think you criticized other's answer and devoted them saying your answer is better without stating any valid reasoning. stop doing this. and this is my last comment. — Md Golam Rahman Tushar, Feb 04 '19 at 13:37
@GolamRahmanTushar never said my answer was better. You did about yours though. Sure, bye. — , Feb 04 '19 at 13:37

Md Golam Rahman Tushar · Answer 2 · 2019-02-04T13:03:55.467

-1

You can use the find method of array like following:

newGrid = [];

listSelected.forEach((elt) => {

    newGrid.push(data.find(function check(item) {

        return item.reference_id === elt.reference_id;

    }).reference_id);
});

edited Feb 04 '19 at 13:03

answered Feb 04 '19 at 12:53

Md Golam Rahman Tushar

2,175
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14
29

`find` still does a loop over the array. – Feb 04 '19 at 12:57
yes, I know. but he/she asked for a way to avoid using nested ForEach. – Md Golam Rahman Tushar Feb 04 '19 at 13:06
true that, but I hardly believe he just wants to just replace the operator. – Feb 04 '19 at 13:07
what you believe doesn't matter here. clearly he is searching for cleaner code not thinking about the runtime. – Md Golam Rahman Tushar Feb 04 '19 at 13:13
"Clean code" is a matter of beliefs too ... Why do you think nested `forEach` are worse than `forEach/find` ? That's totally biased of you. – Feb 04 '19 at 13:16
because it looks better. if not why do you think they provide these utility methods? – Md Golam Rahman Tushar Feb 04 '19 at 13:19
No it doesn't. it looks the same. Two function nested into one another. And they provide those operators to do different things. If you think `find` and `forEach` do the same, then learn about them ! – Feb 04 '19 at 13:21

Sandra Willford · Answer 3 · 2019-02-04T13:00:35.780

-1

const newGrig = listSeleced.map(listItem => {
    if (data.some(dataItem => dataItem.item.reference_id === listItem.item.reference_id) {
        return listItem.reference_id;
    }
    return;
});

Have not tested this, but it should work. You could also do something similar with reduce.

edited Feb 04 '19 at 13:00

answered Feb 04 '19 at 12:59

Sandra Willford

3,459
11
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Complex selection and table

3 Answers3