How to change key value based on comparison of array of objects in jQuery?

Question

I have two arrays in jQuery as below.

var arr1 = [{A: "t1"},{A: "t3"}];

var arr2 = [{A: "t1",B:""},{A: "t2",B:""},{A: "t3",B:""},{A: "t4",B:""},{A: "t5",B:""},{A: "t6",B:""}];

var arr3 = []; //Only for output

Now, the output in arr3 should be:

all the values from arr2
the 'B' key should be Yes if 'A' key exists in arr1
the 'B' key should be No if 'A' key doesn't exist in arr1

So, the output I want in arr3 should be:

arr3 = [{A: "t1",B:"Yes"},{A: "t2",B:"No"},{A: "t3",B:"Yes"},{A: "t4",B:"No"},{A: "t5",B:"No"},{A: "t6",B:"No"}];

This is what I have tried so far

var arr1 = [{A: "t1"},{A: "t3"}];
var arr2 = [{A: "t1",B:""},{A: "t2",B:""},{A: "t3",B:""},{A: "t4",B:""},{A: "t5",B:""},{A: "t6",B:""}];
var arr3 = []; //Only for output

arr1.forEach(function(x) {
  arr2.forEach(function(y) {
    if (x.A == y.A) {
      arr3.push({
        A: x.A,
        B: "Yes"
      });
    } else {
      arr3.push({
        A: x.A,
        B: "No"
      });
    }
  });
});

console.log(arr3);

//Output
0: {A: "t1", B: "Yes"}
1: {A: "t1", B: "No"}
2: {A: "t1", B: "No"}
3: {A: "t1", B: "No"}
4: {A: "t1", B: "No"}
5: {A: "t1", B: "No"}
6: {A: "t3", B: "No"}
7: {A: "t3", B: "No"}
8: {A: "t3", B: "Yes"}
9: {A: "t3", B: "No"}
10: {A: "t3", B: "No"}
11: {A: "t3", B: "No"}

Issue here is that it repeats for each item in arr1, and count gets doubled. I know it should break somewhere between if and else, so that it does not repeat.

I have tried checking below links too, but couldn't get much help.

filter array of objects based on separate object keys, values

js - How to change one key's value based on another key's value in array of objects

Object comparison based on keys

compare array of objects

Thanks in advance.

Thanks for the edit, I was unaware how to add it as a snippet. — foodiepanda
– foodiepanda, Commented Apr 8, 2020 at 6:46
How long can these arrays become? If not long you can do arr1.find(o => o.A===x.A); — mplungjan
– mplungjan, Commented Apr 8, 2020 at 6:49

mplungjan · Accepted Answer · 2020-04-08 06:58:42Z

2

Use map and find if the arrays are not huge.

Saves a reduce and push

var arr1 = [{ A: "t1" }, { A: "t3" }];
var arr2 = [{ A: "t1", B: ""}, { A: "t2", B: "" }, { A: "t3", B: "" }, { A: "t4", B: "" }, { A: "t5", B: "" }, { A: "t6", B: "" }];
var arr3 = arr2.map(
  item2 => ({
    A: item2.A,
    B: arr1.find(item1 => item1.A === item2.A) ? 'Yes' : 'No'
  }));

console.log(arr3);

answered Apr 8, 2020 at 6:58

mplungjan

180k29 gold badges183 silver badges246 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Nick · Accepted Answer · 2020-04-08 06:52:01Z

1

You can make an array of the A values in arr1 using Array.reduce(), then use Array.includes to check if each of the A values in arr2 is present in that array, using that to decide whether to set B to Yes or No:

var arr1 = [{ A: "t1" }, { A: "t3" }];
var arr2 = [{ A: "t1", B: ""}, { A: "t2", B: "" }, { A: "t3", B: "" }, { A: "t4", B: "" }, { A: "t5", B: "" }, { A: "t6", B: "" }];
var arr3 = []; //Only for output

var arr1values = arr1.reduce((c, v) => c.concat([v.A]), []);

arr2.forEach(function(y) {
  arr3.push({
    A: y.A,
    B: arr1values.includes(y.A) ? 'Yes' : 'No'
  });
});

console.log(arr3);

answered Apr 8, 2020 at 6:52

Nick

147k23 gold badges67 silver badges106 bronze badges

3 Comments

foodiepanda Over a year ago

Great, that worked like a charm. Array.reduce() and Array.includes is something new that I learned today. Much appreciated.

mplungjan Over a year ago

Would be interesting to compare our two answer in some benchmark test for 5000 items

Nick Over a year ago

@mplungjan indeed. I guess it will largely come down to whether includes is enough faster than find to make up for the reduce.

Collectives™ on Stack Overflow

How to change key value based on comparison of array of objects in jQuery?

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related