get fraction part of a decimal number

Question

I am trying to get a fraction part of a decimal number in rails. For example I have a number, that "1.23" and I want to get "23" It is may be too easy but, does anyone have any idea about how can I do?

score 65 · Answer 1 · answered Sep 13 '12 at 12:21

65

Try to use modulo method:

1.23.modulo(1) => 0.23

Read more here: http://www.ruby-doc.org/core-1.9.3/Numeric.html#method-i-modulo

Or you can convert float to integer and substract it from original float value.

1.23 - 1.23.to_i => 0.23

answered Sep 13 '12 at 12:21

Eugene Dorian

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I get `(1.1).modulo(1) # 0.10000000000000009` which is less than acceptable! – Augustin Riedinger Mar 14 '14 at 23:07
And I get 1.234 - 1.234.to_i # 0.2399999999999999 which is less than acceptable. This is typical of floating point arithemetic. So is EfratBlaier's .to_s way the best then? – labyrinth Apr 02 '14 at 15:48
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@labyrinth I am using `n.modulo(1).round(2)` – sixty4bit Sep 09 '15 at 15:33
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@labyrinth I'd suggest using the BigDecimal type if floating point errors aren't acceptable. The accepted String-based answer is really just a dirty hack. – bloudermilk Nov 13 '15 at 01:08
This answer did not handle negative numbers the way I expected. See my answer below – slindsey3000 Feb 25 '16 at 03:48

score 11 · Accepted Answer · answered Sep 13 '12 at 12:21

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I am not sure if it is the easiest way to do it - but you can simply split the number using "." character - like this:

number = 1.23
parts = number.to_s.split(".")
result = parts.count > 1 ? parts[1].to_s : 0

answered Sep 13 '12 at 12:21

EfratBlaier

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Future readers, please don't use this approach. It may work in a simplistic case, but will trip in any locale-aware scenario. Although not very error-prone in Ruby which hardcodes `.` as the floating point delimiter, it would promote the bad habit to be used with other, less forgiving languages. Please use math functions for math operations, see answers below. – Nic Nilov May 23 '16 at 18:16

score 10 · Answer 3 · edited Sep 26 '18 at 17:24

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n = 1.23

n.modulo(1)
=> 0.22999999999999998

n - n.to_i
=> 0.22999999999999998

Recommended read http://floating-point-gui.de/

edited Sep 26 '18 at 17:24

Joshua Pinter

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answered Sep 13 '12 at 12:21

Jonas Elfström

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Fixed the typo of `to.i` -> `to_i`. – Joshua Pinter Sep 26 '18 at 17:24

score 6 · Answer 4 · answered Apr 30 '15 at 16:35

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if you know your desired precision, this might be a tad faster to to_s solution.

1.1234.modulo(1).round(4) #0.1234

answered Apr 30 '15 at 16:35

bioffe

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score 6 · Answer 5 · edited May 23 '17 at 12:34

Deciding on the proper solution requires understanding the type that you're working with. If your values are Float (the standard type for non-whole numbers in Ruby) then logically correct answers like mod(1) may produce unexpected results due to floating point errors. For any case where Float is the proper data type to be using in the first place, this is likely acceptable.

If floating point errors are not acceptable, don't use Float! Ruby comes with the great BigDecimal class which is much more accurate at the cost of performance and a slightly more verbose syntax.

decimal = BigDecimal.new("1.23")
decimal.frac #=> 0.23

score 2 · Answer 6 · answered Sep 12 '18 at 13:36

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..."1.23" and I want to get "23".

Just remove the integer part and multiply per 100.

((number - number.to_i) * 100).to_i

answered Sep 12 '18 at 13:36

Lucas Castro

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This introduces imprecision, so I would not recommend this approach. Example: (37319.70 - 37319.70.to_i) * 100 = 69.99999999970896 – zephos2014 May 14 '19 at 16:46

dsk · Answer 7 · 2015-07-02T04:57:44.397

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 a=1.23 #value of a will be 1.23
 a=a-a.to_i #value of a will be 0.23

edited Jul 02 '15 at 04:57

answered Dec 29 '14 at 06:15

dsk

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Again, as zephos2014 says, it brings in imprecision; it produces 0.22999999999999998 – Pablo Jul 01 '20 at 12:18

score 1 · Answer 8 · answered Apr 26 '20 at 06:28

There are various ways we can achieve the output modulo is one of them

1.23.modulo(1) => 0.22999999999999998

you can round the result as per your convenience

but to achieve what you have asked in your question

I will follow a very simple approach

value = 1.23.to_s.split('.') -> ["1", "23"]
value.last.to_i -> "23"

score 0 · Answer 9 · answered Feb 25 '16 at 03:46

0

After trying everything... I feel the best answer is (num - num.to_i).abs because it also works for negative numbers.

e.g.

(1.23 - 1.23.to_i).abs = 0.23

(-1.23 - -1.23.to_i).abs = 0.23

answered Feb 25 '16 at 03:46

slindsey3000

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This will give you imprecision. Bad approach. Example: (37319.70-37319.70.to_i).abs => 0.6999999999970896 – zephos2014 May 14 '19 at 16:47

get fraction part of a decimal number

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