JavaScript iterate loop every x number of times in same loop?

Question

I am not sure how to phrase this, so please re-title this question if it doesn't make sense. Anyways, this is what I am trying to do.

I have a variable with the length of 9. And then I have another variable with the length of 3.

How do I write a loop that iterates through all 9 but starts over every third time?

For example: I have this,

x = 3;
l = 9;

for ( var i = 0; i < l; i++)
{ 
     console.log(i + 1);
}

output = 1,2,3,4,5,6,7,8,9

The output I want to create

output = 1,2,3,1,2,3,1,2,3

I was thinking there might be away to do this with an if statement or possibly modulus, but wasn't quite sure how to implement it. What would be a good way to do this? Thanks in advance.

@LightStyle, Yeah that is what I was looking for... I must have been doing it wrong when I tried it. — khollenbeck
– khollenbeck, Commented Mar 28, 2014 at 19:19

fny · Accepted Answer · 2014-03-28 19:21:35Z

2

Embrace the modulus:

function expand(length, loop_length) {
  for (var i = 0; i < length; i++) {
    console.log(i % loop_length + 1);
  } 
}
expand(9, 3) // =>  1, 2, 3, 1, 2, 3, 1, 2, 3

edited Mar 28, 2014 at 19:21

answered Mar 28, 2014 at 19:16

fny

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Comments

Shomz · Accepted Answer · 2014-03-28 19:21:13Z

1

x = 3;
l = 9;

for ( var i = 0; i < l; i++)
{ 
     console.log(i % x + 1);
}

output = 1,2,3,1,2,3,1,2,3

See it in action here: http://jsfiddle.net/BgBGL/

edited Mar 28, 2014 at 19:21

answered Mar 28, 2014 at 19:13

Shomz

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12 Comments

khollenbeck Over a year ago

I thought you could do it with modulus, I must have been doing it wrong. This is what I was looking for. Thanks!

Shomz Over a year ago

No problem, I missed it at first as well.

Mauno Vähä Over a year ago

I'm crying, code snipped with syntax errors won. :(

Shomz Over a year ago

Don't cry, what syntax errors? Or are you referring to that you finally saw what works and used for your answer?

khollenbeck Over a year ago

Just FYI, I am using var x = somevar; or whatever in my actual code that was something I had just typed up quick to illustrate my problem.

|

p.s.w.g · Accepted Answer · 2014-03-28 19:09:49Z

0

If you want to loop from a min value to a max value a specific number of times, the easiest way is just to have 2 loops, like this:

var min = 1, max = 3, times = 3;
for (var i = 0; i < times; i++) {
    for (var j = min; j <= max; j++) {
         console.log(j);
    }
}

Or if you want to fix total length of the sequence, then yes, you can do it with a single loop and a little bit of math:

var min = 1, max = 3, totalLength = 9;
for (var i = 0; i < totalLength; i++) {
    console.log((i % (max - min + 1)) + min);
}

answered Mar 28, 2014 at 19:09

p.s.w.g

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Comments

juvian · Accepted Answer · 2014-03-28 19:13:28Z

0

For that case, this works:

x = 3;
l = 9;

for ( var i = 0; i < l; i++)
{ 
     var num=(i %(x+1)) || 1;
     console.log(num);
}

answered Mar 28, 2014 at 19:13

juvian

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Comments

Mauno Vähä · Accepted Answer · 2014-03-28 19:37:04Z

0

You could go mad with following syntax:

var i = 0;

do {
  console.log(i++ % 3 + 1);
} while (i < 9);

Alternative, you could define 3 and 9 as a variable also.

I took an advantage of the fact that calling i++ will display old variable and increases it by 1 after, so I saved some bits!

See: fiddle example

edited Mar 28, 2014 at 19:37

answered Mar 28, 2014 at 19:13

Mauno Vähä

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Comments

Alaeddine · Accepted Answer · 2014-03-30 19:25:53Z

0

x = 3;
l = 9;

for ( var i = 0; i <= l; i++)
{ 
  for ( var j = 1; j <= x; j++)
   { 
     console.log(j);
   }
}

edited Mar 30, 2014 at 19:25

answered Mar 28, 2014 at 19:20

Alaeddine

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4 Comments

Shomz Over a year ago

I suggest you test this. How are j + x going to give 1 when x is 3, and j is >= 0?

Shomz Over a year ago

Better, there are two more things - the outer loop will run 10 times instead of 9 and the inner one 3 times each, making the total number of printed numbers 30 instead of 9.

Alaeddine Over a year ago

I tested it and it gives 123 nine times. but you're right 30 loop is far answer from the main question.

Shomz Over a year ago

Yes, sorry, my bad - it's 27 instead of 9. :) The easiest fix is probably: for ( var i = 0; i < l/x; i++) for the outer loop.

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