JavaScript for-loop: Skip 2 Numbers Followed by a set of 3

Question

I made a JavaScript loop that skips every set of two numbers that are followed by a set of three.

Consider the following set of integers:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

This is what the array returns:

1, 2, 3, 6, 7, 8

See how it skipped 4-5 and 9-10? (every set of two numbers followed by a set of three)

Visual:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (skips bolded numbers)

Here is what I came up with:

var y = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

    for (var i = 0; i < y.length; i++) {
        if (y[i] >= 4 && y[i] <= 5) {
            continue;
        }
        if (y[i] >= 9 && y[i] <= 10) {
            continue;
        }
        document.write(y[i]);
    }

The above for-loop returns 123678 from 12345678910. The problem is that I had to manually say which ones to skip. Is there a way, possibly with % modulus, to automatically skip every set of two numbers followed by a set of three without limiting the length to ten?

Answer 1

The pattern for the processing repeats after 5 items, so yes, you can use the modulo operator:

if (i % 5 < 3)
    document.write(y[i]);

Answer 2

after seeing @ysf comment i have to explain what i did for anyone who is seeing this answer - sorry for bad English if exists -

first of all you need to analyze things, what do we want? base on this everything will be solved nice and clear

so we have an array and you want to pick numbers and skips other right? in order to achieve this and to know we will use the mod operator "%"

if we want to pick 2 numbers and skip three numbers, pick 2 skip 3 ...etc) we first sum the pick + skip values in order to make it on the right hand of the mod operation ... in our case 2 + 3 = 5

let all = pick + skip;

in order to pick from array i don't need the value of the index ... i need the index it self pick index[0] and index[1] skip index 2,3,4

because i want to pick only 2, then our expression will be limited to only 2 and to the sum of the pick and skip values

[index] % all < pick so if the index mod all is less than the limit which is 2 then take it otherwise leave (skip) it

so the results of our loop will look like this:

i % all < pick ===> 0 % 5 < 2 = true
i % all < pick ===> 1 % 5 < 2 = true
i % all < pick ===> 2 % 5 < 2 = false
i % all < pick ===> 3 % 5 < 2 = false
i % all < pick ===> 4 % 5 < 2 = false
i % all < pick ===> 5 % 5 < 2 = true
i % all < pick ===> 6 % 5 < 2 = true
i % all < pick ===> 7 % 5 < 2 = false
i % all < pick ===> 8 % 5 < 2 = false
i % all < pick ===> 9 % 5 < 2 = false
i % all < pick ===> 10 % 5 < 2 = true
i % all < pick ===> 11 % 5 < 2 = true

and the last thing we use filter because this is a JS function that returns an Array

const skip = (arr, pick, skip) => {
  let all = pick + skip;
  let filtered = arr.filter((_, i) => {
    return i % all < pick;
  });
  console.log(filtered);
};

}

skip([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],2,3);

Answer 3

Here is JSFiddle link

var seq = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
  pick = 3,
  skip = 2;

var r = skipFilter(seq, pick, skip);
console.log(r);

function skipFilter(seq, pick, skip) {
  var stride = pick + skip;

  return seq.filter(function(_, i) {
    return i % stride < pick;
  });
}

Answer 4

for (var i=0; i < y.length; i++) {

  if (i % 5 < 3) {
    document.write(y[i] + ' ');
  } else {
    i += 1;
    document.write('<br>');
  }
}

Answer 5

You could just keep a counter instead

for (var i=0, j=0; i < y.length; i++) {

    if (j < 3) {
        document.write(y[i]);
    } else if ( j > 3) {
        j = -1;
    }

    j++;
}

FIDDLE

Answer 6

Move the "i" increment to the body of "for"

    var y = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

    for (var i=0; i<y.length;) {
        console.log(y[i]);
        console.log(y[i+1]);
        console.log(y[i+2]);

        i = i + 5;
    }