I am wondering what is a convenient function in Rails to convert a string with a negative sign into a number. e.g. -1005.32
When I use the .to_f method, the number becomes 1005 with the negative sign and decimal part being ignored.
-
Are you storing the result in an int rather than a float?Sharjeel Aziz– Sharjeel Aziz2010-05-06 02:51:29 +00:00Commented May 6, 2010 at 2:51
Add a comment
|
2 Answers
.to_f is the right way.
Example:
irb(main):001:0> "-10".to_f
=> -10.0
irb(main):002:0> "-10.33".to_f
=> -10.33
Maybe your string does not include a regular "-" (dash)? Or is there a space between the dash and the first numeral?
Added:
If you know that your input string is a string version of a floating number, eg, "10.2", then .to_f is the best/simplest way to do the conversion.
If you're not sure of the string's content, then using .to_f will give 0 in the case where you don't have any numbers in the string. It will give various other values depending on your input string too. Eg
irb(main):001:0> "".to_f
=> 0.0
irb(main):002:0> "hi!".to_f
=> 0.0
irb(main):003:0> "4 you!".to_f
=> 4.0
The above .to_f behavior may be just what you want, it depends on your problem case.
Depending on what you want to do in various error cases, you can use Kernel::Float as Mark Rushakoff suggests, since it raises an error when it is not perfectly happy with converting the input string.
2 Comments
hakunin
Look at the other answer,
"abcd".to_f === 0 unfortunately
Vishwas Nahar
" '11.0' " with this can we change to 11.0 using to_fYou should be using Kernel::Float to convert the number; on invalid input, this will raise an error instead of just "trying" to convert it.
>> "10.5".to_f
=> 10.5
>> "asdf".to_f # do you *really* want a zero for this?
=> 0.0
>> Float("asdf")
ArgumentError: invalid value for Float(): "asdf"
from (irb):11:in `Float'
from (irb):11
>> Float("10.5")
=> 10.5
2 Comments
Andrew Grimm
One of the nice things about
Float() as opposed to Integer is that the former doesn't convert 010 to 8 (Integer regards something starting with 0 as octal)
Ajedi32
@AndrewGrimm Or you could just pass the correct base in as the second argument:
Integer("010", 10) #=> 10 (See Kernel#integer)