2

Here is the input,

colA = [1,1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,3];
colB = [1.1 ,1.1 ,1.2 ,1.3, 1.3, 'ab', 2.1, 2.1, 2.2, 2.2, 'ab', 3.1, 3.2, 3.3, 3.3, 3.3, 'ab'];
  • Both the arrays have the same length.
  • if there are total 6 '1's in colA then there will be only 6 of the 1s item(1._) in colB.

I want the output in the console to look like:

1, 1.1, count of 1.1 = 2
1, 1.2, count of 1.2 = 3
1, 1.3, count of 1.3 = 2
1, AB,  count of AB  = 1
2, 2.1, count of 2.1 = 2
2, 2.2, count of 2.2 = 2
2, AB,  count of AB  = 2
.
.
.
//or something like this where I must get these three values.

Code:

for(i=0; i < colA.length; i++)
{
 for(j=0; j < colB.length; j++) 
 {
  // what to do here to compare the values.
 }
}

Should I go with two for loops because I want the code to be really optimized as there is going to be around 10k rows of data. Open to any suggestions. Thanks for the help.

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5
  • 1
    please add the data in text form. Commented Apr 15, 2020 at 14:09
  • I have edited the question. I hope this is what you wanted. @NinaScholz Commented Apr 15, 2020 at 14:20
  • come on how should s/b copy the data to try some code? btw, are the array connected by the same index? Commented Apr 15, 2020 at 14:24
  • I didn't get you earlier , my bad. will this help? yes, they are connected. thanks.@NinaScholz Commented Apr 15, 2020 at 14:35
  • do you want to count 'ab' different? Commented Apr 15, 2020 at 15:10

1 Answer 1

Reset to default
2

You could iterate with a single loop and check the value with the successor and increment count.

If unequal make the output and reset count to one.

var colA = [1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3],
    colB = [1.1, 1.1, 1.2, 1.3, 1.3, 'ab', 2.1, 2.1, 2.2, 2.2, 'ab', 3.1, 3.2, 3.3, 3.3, 3.3, 'ab'],
    count = 1,
    i,
    length = colA.length;
   
for (i = 0; i < length; i++) {
    if (colB[i] === colB[i + 1]) {
        count++;
        continue;
    }
    console.log(colA[i], colB[i], count);
    count = 1;
}

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2 Comments

Nice. But OP states 2, AB, count of AB = 2. Of course OP might not need that.
Perfect. Thanks for your valuable time. @NinaScholz

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