Without jQuery, how can I round a float number to 2 non-zero decimals (but only when needed - 1.5 instead of 1.50)?

Just like this:

2.50000000004 -> 2.5
2.652 -> 2.65
2.655 -> 2.66
0.00000204 -> 0.000002
0.00000205 -> 0.0000021

I tried this code:

var r = n.toFixed(1-Math.floor(Math.log10(n)));

but n=0.00000205 implies r=0.0000020, which is in conflict with conditions above. But n=0.0000020501 implies r=0.0000021, which is OK, so the error is only for 5 as a last decimal, which should be rounded up.

Without jQuery, how can I round a float number to 2 non-zero decimals (but only when needed - 1.5 instead of 1.50)?

Just like this:

2.50000000004 -> 2.5
2.652 -> 2.65
2.655 -> 2.66
0.00000204 -> 0.000002
0.00000205 -> 0.0000021

I tried this code:

var r = n.toFixed(1-Math.floor(Math.log10(n)));

but n=0.00000205 implies r=0.0000020, which is in conflict with conditions above. But n=0.0000020501 implies r=0.0000021, which is OK, so the error is only for 5 as a last decimal, which should be rounded up.

• javascript
• math
• rounding

• 5 Possible duplicate of Round to at most 2 decimal places in JavaScript – Tot Zam Commented Aug 8, 2016 at 20:06
• 1 @PrasadShinde This will round the last 2 examples to 0, which is not what the OP wants. – Arnauld Commented Aug 8, 2016 at 20:10
• It is not a duplicate. I don't want to round to 2 decimal places, I want to round to 2 NON-ZERO decimal places. – core4096 Commented Aug 8, 2016 at 20:42
• 1 The term you're looking for is "rounding to precision". See developer.mozilla/en-US/docs/Web/JavaScript/Reference/… – JJJ Commented Aug 8, 2016 at 20:49
• 2 The reason you get 0.000020 for that is related to the imprecision of the floating point representation. 0.0000205 cannot be represented as exactly that value in floating point. For numbers that can, the result is as you would expect. NB: Ad-hoc coded functions that try to do this will probably all suffer from the same problem. – trincot Commented Aug 8, 2016 at 21:40

2 Answers 2

This should do what you want:

function twoDecimals(n) {
  var log10 = n ? Math.floor(Math.log10(n)) : 0,
      div = log10 < 0 ? Math.pow(10, 1 - log10) : 100;

  return Math.round(n * div) / div;
}

var test = [
  2.50000000004,
  2.652,
  2.655,
  0.00000204,
  0.00000205,
  0.00000605
];

test.forEach(function(n) {
  console.log(n, '->', twoDecimals(n));
});

Thank's for the @Arnauld answer. Just added decimals parameter.

function roundToDecimals(n, decimals) {
  var log10 = n ? Math.floor(Math.log10(n)) : 0,
      div = log10 < 0 ? Math.pow(10, decimals - log10 - 1) : Math.pow(10, decimals);

  return Math.round(n * div) / div;
}

var numDecimals = 2
var test = [
  2.50000000004,
  2.652,
  2.655,
  0.00000204,
  0.00000205,
  0.00000605
];

test.forEach(function(n) {
  console.log(n, '->', roundToDecimals(n, numDecimals));
});