I am working on reports for a website and I am currently thinking of what would be the best way to handle BigDecimal -0.0's.
The database I'm working with has a lot of them. When these -0.0's are put through number_to_currency(), I get "$-0.00". My format for negative numbers is actually "-$x.xx", so note that number_to_currency is not formatting it as a negative number (otherwise there would also be a negative sign in front of the dollar sign), but for some reason the negative sign is being translated along with the 0.
Right now my solution is to do this every time I get an amount from the database:
amount *= -1 if amount == 0 && amount.sign == -1
This changes the -0.0 to a 0.0. It's simple enough, but I can't help but wonder if there is a better solution, or something on BigDecimals or number_to_currency to handle this situation that I'm just not finding.
That is so because the number is converted into a string to be displayed. And:
# to_d converts to BigDecimal, just FYI
"-0".to_d.to_s #=> "-0.0"
Therefore you will have to make it a 0 yourself. But the sign-checks are redundant - a simple comparison with 0 will do the trick:
bdn = "-0".to_d # or BigDecimal.new("-0")
value = bdn.zero? ? 0 : bdn
number_to_currency(value, other_options)
However, you wouldn't want to manually add this check everywhere you're calling number_to_currency. It would be more convenient to create your own modified_number_to_currency method, in your ApplicationHelper, like so:
def modified_number_to_currency( number, options )
value = number.zero? ? 0 : number
number_to_currency(value, options)
end
And then use modified_number_to_currency instead of number_to_currency.
Alternatively, you could overwrite number_to_currency and have it call super in the end. That might also work but I'm not 100% certain.
Coming to your check specifically:
amount *= -1 if amount == 0 && amount.sign == -1
It should simply be:
amount = 0.to_d if amount.zero? # the to_d might or might not be required
Related
I'm looking for a way to simplify the code for the following logic:
Take a value that is either a nil or an empty string
Convert that value to an integer
Set zero values to the maximum value (empty string/nil are converted to 0 when cast as an int)
.clamp the value between a minimum and a maximum
Here's the long form that works:
minimum = 1
maximum = 10_000
value = value.to_i
value = maximum if value.zero?
value = value.clamp(minimum, maximum)
So for example, if value is "", I should get 10,000. If value is "15", I should get 15. If value is "45000", I should get 10000.
Is there a way to shorten this logic, assuming that minimum and maximum are defined and that the default value is the maximum?
The biggest problem I've had in shortening it is that null-coalescing doesn't work on the zero, since Ruby considers zero a truthy value. Otherwise, it could be a one-liner.
you could still do a one-liner with your current logic
minimum, maximum = 1, 10_000
value = ( value.to_i.zero? ? maximum: value.to_i ).clamp(minimum, maximum)
but not sure if your issue is that if you enter '0' you want 1 and not 10_000 if so then try this
minimum, maximum = 1, 10_000
value = (value.to_i if Float(value) rescue maximum).clamp(minimum, maximum)
Consider Fixing the Input Object or Method
If you're messing with String objects when you expect an Integer, you're probably dealing with user input. If that's the case, the problem should really be solved through input validation and/or looping over an input prompt elsewhere in your program rather than trying to perform input transformations inline.
Duck-typing is great, but I suspect you have a broken contract between methods or objects. As a general rule, it's better to fix the source of the mismatch unless you're deliberately wrapping some piece of code that shouldn't be modified. There are a number of possible refactorings and patterns if that's the case.
One such solution is to use a collaborator object or method for information hiding. This enables you to perform your input transformations without complicating your inline logic, and allowing you to access the transformed value as a simple method call such as user_input.value.
Turning a Value into a Collaborator Object
If you are just trying to tighten up your current method you can aim for shorter code, but I'd personally recommend aiming for maintainability instead. Pragmatically, that means sending your value to the constructor of a specialized object, and then asking that object for a result. As a bonus, this allows you to use a default variable assignment to handle nil. Consider the following:
class MaximizeUnsetInputValue
MIN = 1
MAX = 10_000
def initialize value=MAX
#value = value
set_empty_to_max
end
def set_empty_to_max
#value = MAX if #value.to_i.zero?
end
def value
#value.clamp MIN, MAX
end
end
You can easily validate that this handles your various use cases while hiding the implementation details inside the collaborator object's methods. For example:
inputs_and_expected_outputs = [
[0, 10000],
[1, 1],
[10, 10],
[10001, 10000],
[nil, 10000],
['', 10000]
]
inputs_and_expected_outputs.map do |input, expected|
MaximizeUnsetInputValue.new(input).value == expected
end
#=> [true, true, true, true, true, true]
There are certainly other approaches, but this is the one I'd recommend based on your posted code. It isn't shorter, but I think it's readable, maintainable, adaptable, and reusable. Your mileage may vary.
def digits(n):
total=0
for i in range(0,n):
if n/(10**(i))<1 and n/(10**(i-1))=>1:
total+=i
else:
total+=0
return total
I want to find the number of digits in 13 so I do the below
print digits(13)
it gives me $\0$ for every number I input into the function.
there's nothing wrong with what I've written as far as I can see:
if a number has say 4 digits say 1234 then dividing by 10^4 will make it less than 1: 0.1234 and dividing by 10^3 will make it 1.234
and by 10^3 will make it 1.234>1. when i satisfies BOTH conditions you know you have the correct number of digits.
what's failing here? Please can you advise me on the specific method I've tried
and not a different one?
Remember for every n there can only be one i which satisfies that condition.
so when you add i to the total there will only be i added so total returning total will give you i
your loop makes no sense at all. It goes from 0 to exact number - not what you want.
It looks like python, so grab a solution that uses string:
def digits(n):
return len(str(int(n))) # make sure that it's integer, than conver to string and return number of characters == number of digits
EDIT:
If you REALLY want to use a loop to count number of digits, you can do this this way:
def digits(n):
i = 0
while (n > 1):
n = n / 10
++i
return i
EDIT2:
since you really want to make your solution work, here is your problem. Provided, that you call your function like digits(5), 5 is of type integer, so your division is integer-based. That means, that 6/100 = 0, not 0.06.
def digits(n):
for i in range(0,n):
if n/float(10**(i))<1 and n/float(10**(i-1))=>1:
return i # we don't need to check anything else, this is the solution
return null # we don't the answer. This should not happen, but still, nice to put it here. Throwing an exception would be even better
I fixed it. Thanks for your input though :)
def digits(n):
for i in range(0,n):
if n/(10**(i))<1 and n/(10**(i-1))>=1:
return i
I found this code on another thread.
def is_number? string
true if Float(string) rescue false
end
Instead of using a method to return true or false, I'd like to do this "is_numeric" test in one line, in an if statement. Can someone explain if this is possible? I'm getting errors at the moment both when the string variable is null and when it contains non-numeric characters.
if Float(string)
* do something
else
* do something else
end
if Float() is pointless code, since Float() will either return a truthy value or raise an error (based on my limited look at the source code - as of writing, you can follow the code path from line #2942). I'd suggest you're asking the wrong question/looking at the problem wrong (it'd be helpful to know what you're actually trying to achieve).
To do something with Float() on one line and avoid breaking code, use rescue as a statement modifier, as has been done in the is_number? method posted.
Float(string) rescue 0.0 # trying to emulate String#to_f!
Ensuring the phone number is 10 digits, numbers only, is quite simple.
PHONE_NUM_LENGTH = 10
string.length == PHONE_NUM_LENGTH && string.count('0-9') == PHONE_NUM_LENGTH
will return the true/false value representing this check. This is more efficient than a Regex.
The first part,
string.length == PHONE_NUM_LENGTH
checks whether the string is 10 characters long. The second,
string.count('0-9') == PHONE_NUM_LENGTH
checks whether it has exactly 10 numeric characters.
I have such code in controller's method for rounding (only higher) and display ceil part of number:
#constr_num.each do |cn|
non_original_temp_var2 = get_non_tecdoc_analogs(cn.ARL_SEARCH_NUMBER, #article.supplier.SUP_BRAND, false)
non_original << non_original_temp_var2
end
#non_original = non_original.flatten!
#non_original.each do |n_original|
n_original.price = my_round2(n_original.price * markup_for_user)
end
def my_round2 a
res = (a / 1.0).ceil * 1
res
end
But for some reasons i see with every price comma with 0 after it, for example: 5142.0 but it must be 5142
Main strange part is that, if i try to write:
n_original.price = 123
in view i see 123.0
What happend?
Only when i write in view (when displaying price):
price.ceil
i see normal numbers, without comma
What i di wrong? How to ceil my numbers with rounding (but only high, for example 2.24 is 3 3.51 is 4 and 2.0 is 2)? Becouse now for some reasons i see comma and nul after my number, even if i try to "hardcode" number in controller.
How about using the next or succ function of the Integer class? Try something like the following:
def my_round2 a
(a.is_a? Integer) ? a : a.to_i.next
end
If a is an Integer then return a otherwise cast it to Integer using the to_i method and call next or succ method on it.
Reference: http://www.ruby-doc.org/core-2.0/Integer.html
I guess I missed the second part of your question. To avoid the decimal places I guess you would have to use the a.to_i like Philip Hallstrom has suggested.
My guess is that your price field is a Float. Floats will be printed with a decimal spot by default. You need to either cast it to an Integer earlier on (say in my_round2 method) or in your view task a .to_i onto the output.
I'm trying to recalculate percentages in an after_update callback of my model.
def update_percentages
if self.likes_changed? or self.dislikes_changed?
total = self.likes + self.dislikes
self.likes_percent = (self.likes / total) * 100
self.dislikes_percent = (self.dislikes / total) * 100
self.save
end
end
This doesn't work. The percentage always comes out as a 100 or 0, which completely wrecks everything.
Where am I slipping up? I guarantee that self.likes and self.dislikes are being incremented correctly.
The Problem
When you divide an integer by an integer (aka integer division), most programming languages, including Ruby, assume you want your result to be an Integer. This is mostly due to History, because with lower level representations of numbers, an integer is very different than a number with a decimal point, and division with integers is much faster. So your percentage, a number between 0 and 1, has its decimal truncated, and so becomes either 0 or 1. When multiplied by 100, becomes either 0 or 100.
A General Solution
If any of the numbers in the division are not integers, then integer division will not be performed. The alternative is a number with a decimal point. There are several types of numbers like this, but typically they are referred to as floating point numbers, and in Ruby, the most typical floating point number is of the class Float.
1.0.class.ancestors
# => [Float, Precision, Numeric, Comparable, Object, Kernel]
1.class.ancestors
# => [Fixnum, Integer, Precision, Numeric, Comparable, Object, Kernel]
In Rails' models, floats are represented with the Ruby Float class, and decimal with the Ruby BigDecimal class. The difference is that BigDecimals are much more accurate (ie can be used for money).
Typically, you can "typecaste" your number to a float, which means that you will not be doing integer division any more. Then, you can convert it back to an integer after your calculations if necessary.
x = 20 # => 20
y = 30 # => 30
y.to_f # => 30.0
x.class # => Fixnum
y.class # => Fixnum
y.to_f.class # => Float
20 / 30 # => 0
20 / 30.0 # => 0.666666666666667
x / y # => 0
x / y.to_f # => 0.666666666666667
(x / y.to_f).round # => 1
A Solution For You
In your case, assuming you are wanting integer results (ie 42 for 42%) I think the easiest way to do this would be to multiply by 100 before you divide. That pushes your decimal point as far out to the right as it will ever go, before the division, which means that your number is as accurate as it will ever get.
before_save :update_percentages
def update_percentages
total = likes + dislikes
self.likes_percent = 100 * likes / total
self.dislikes_percent = 100 * dislikes / total
end
Notes:
I removed implicit self you only need them on assignment to disambiguate from creating a local variable, and when you have a local variable to disambiguate that you wish to invoke the method rather than reference the variable
As suggested by egarcia, I moved it to a callback that happens before the save (I selected before_save because I don't know why you would need to calculate this percentage on an update but not a create, and I feel like it should happen after you validate that the numbers are correct -- ie within range, and integers or decimal or whatever)
Because it is done before saving, we remove the call to save in the code, that is already going to happen
Because we are not explicitly saving in the callback, we do not risk an infinite loop, and thus do not need to check if the numbers have been updated. We just calculate the percentages every time we save.
Because likes/dislikes is an integer value and integer/integer = integer.
so you can do one of two things, convert to Float or change your order of operations.
self.likes_percent = (self.likes.to_f/total.to_f) * 100
Or, to keep everything integers
self.likes_percent = (self.likes * 100)/total
I'm not sure that this is the only problem that you have, but after_update gets called after the object is saved.
Try changing the update_percentages before - on a before_update or a before_validate instead. Also, remove the self.save line - it will be called automatically later on if you use one of those callbacks.