In the code below 2/5 results to 0.0 instead of 0.4 which should be correct answer
require 'rubygems'
require 'dentaku'
expression = '2/5'
calculator = Dentaku::Calculator.new
result = calculator.evaluate(expression).to_f
puts "#{result}"
I am using dentaku to evaluate the the math expression the documentation for this gem can be found here: https://github.com/rubysolo/dentaku/blob/master/README.md
Your first stop when searching for an answer to this should have been the official documentation for /:
fix / numeric → numeric_result
Performs division: the class of the resulting object depends on the class of numeric and on the magnitude of the result. It may return a Bignum.
For example:
3/2 # => 1
(3/2).class # => Fixnum
3/2.0 # => 1.5
(3/2.0).class # => Float
Ruby will return a Float if either value is a Float:
3.0/2 # => 1.5
(3.0/2).class # => Float
As far as I can tell, Dentaku treats division of two integers as integer division. Try 2/5.0.
Please try expression = '2.0/5.0' instead, maybe your numbers are considered as integers (you want floats) because of the missing ".0" and so you get an integer result of zero
Related
How to convert 9999999999999999999999.001 to "9999999999999999999999.001" in ruby
I have tried
>> 9999999999999999999999.001.to_s
=> "1.0e+22"
>> "%f" % 9999999999999999999999.001
=> "10000000000000000000000.000000"
Truth is you can't. The number you write is 22 digits and floats in ruby have only 15 digits precision. So when you use this variable already part of its value is kind of "lost" as it is of the class Float.
Use BigDecimal from the standard library.
1.8.7 :005 > require 'bigdecimal'
=> true
1.8.7 :006 > BigDecimal('9999999999999999999999.001')
=> #<BigDecimal:7fe0cbcead70,'0.9999999999 9999999999 99001E22',36(36)>
1.8.7 :007 > BigDecimal('9999999999999999999999.001').to_s
=> "0.9999999999999999999999001E22"
Of course, this example only shows that BigDecimal can handle numbers that big. Wherever you're initially getting your 9999999999999999999999.001 number from needs to get it into BigDecimal as soon as it's calculated / inputted.
You can not do that. The reason is simple: from the very beginning the value of the number is not going to be exactly 9999999999999999999999.001. Floats will have just 15 digits of precision.
However, you can use other type to achieve what you want:
require 'bigdecimal'
a = BigDecimal("9999999999999999999999.001")
a.to_s("F")
>> "9999999999999999999999.001"
For BigDecimal the precision is extended with the requests of bigger real numbers - no restrictions are applied.
Float is faster in calculations, because its meant to use the FPU of the processor directly, but because of that comes the restriction in the precision.
EDIT Especially for #izomorphius and his argument, just a very short code sample:
a = "34.101"
b = BigDecimal(a.to_s)
c = b ** 15
c.to_s("F")
>>> 98063348952510709441484.183684987951811295085234607613193907150561501
Now tell me how otherwise you get the last string?
It would be nice to have an equivalent of R's signif function in Ruby.
For example:
>> (11.11).signif(1)
10
>> (22.22).signif(2)
22
>> (3.333).signif(2)
3.3
>> (4.4).signif(3)
4.4 # It's usually 4.40 but that's OK. R does not print the trailing 0's
# because it returns the float data type. For Ruby we want the same.
>> (5.55).signif(2)
5.6
There is probably better way, but this seems to work fine:
class Float
def signif(signs)
Float("%.#{signs}g" % self)
end
end
(1.123).signif(2) # => 1.1
(11.23).signif(2) # => 11.0
(11.23).signif(1) # => 10.0
Here's an implementation that doesn't use strings or other libraries.
class Float
def signif(digits)
return 0 if self.zero?
self.round(-(Math.log10(self).ceil - digits))
end
end
I don't see anything like that in Float. Float is mostly a wrapper for the native double type and given the usual binary/decimal issues, I'm not that surprised that Float doesn't allow you to manipulate the significant digits.
However, BigDecimal in the standard library does understand significant digits but again, I don't see anything that allows you to directly alter the significant digits in a BigDecimal: you can ask for it but you can't change it. But, you can kludge around that by using a no-op version of the mult or add methods:
require 'bigdecimal'
a = BigDecimal.new('11.2384')
a.mult(1, 2) # the result is 0.11E2 (i.e. 11)
a.add(0, 4) # the result is 0.1124E2 (i.e. 11.24)
The second argument to these methods:
If specified and less than the number of significant digits of the result, the result is rounded to that number of digits, according to BigDecimal.mode.
Using BigDecimal will be slower but it might be your only choice if you need fine grained control or if you need to avoid the usual floating point problems.
Some of the previous answers and comments have alluded to this solution but this is what worked for me:
# takes in a float value and returns another float value rounded to
# given significant figures.
def round_to_sig_figs(val, sig_figs)
BigDecimal.new(val, sig_figs).to_f
end
You are probably looking for Ruby's Decimal.
You could then write:
require 'decimal/shortcut'
num = 1.23541764
D.context.precision = 2
num_with_2_significant_digits = +D(num.to_s) # => Decimal('1.2')
num_with_2_significant_digits.to_f # => 1.2000000000000002
Or if you prefer to use the same syntax add this as a function to class Float like this:
class Float
def signif num_digits
require 'decimal/shortcut'
D.context.precision = num_digits
(+D(self.to_s)).to_f
end
end
Usage would then be the same, i.e.
(1.23333).signif 3
# => 1.23
To use it, install the gem
gem install ruby-decimal
#Blou91's answer is nearly there, but it returns a string, instead of a float. This below works for me:
(sprintf "%.2f", 1.23456).to_f
So as a function,
def round(val, sig_figs)
(sprintf "%.#{sig_figs}f", val).to_f
end
Use sprintf if you want to print trailing zeros
2.0.0-p353 :001 > sprintf "%.3f", 500
=> "500.000"
2.0.0-p353 :002 > sprintf "%.4f", 500
=> "500.0000"
2.0.0-p353 :003 >
Is there any worthy Ruby method to count the number of digits in a float? Also, how do I specify the precise when to_s float numbers?
# Number of digits
12345.23.to_s.split("").size -1 #=> 7
# The precious part
("." + 12345.23.to_s.split(".")[1]).to_f #=> .023
# I would rather used
# 12345.23 - 12345.23.to_i
# but this gives 0.22999999999563
to specify precision of a float in Ruby. you can use the round method.
number.round(2)
2 is the precision.
53.819.round(2) -> 53.82
I think you should check out the number_with_precision helper.
number_with_precision(13, :precision => 5) # => 13.00000
In ruby, I want to convert a float to an int if it's a whole number. For example
a = 1.0
b = 2.5
a.to_int_if_whole # => 1
b.to_int_if_whole # => 2.5
Basically I'm trying to avoid displaying ".0" on any number that doesn't have a decimal. I'm looking for an elegant (or built-in) way to do
def to_int_if_whole(float)
(float % 1 == 0) ? float.to_i : float
end
One simple way to it would be:
class Float
def prettify
to_i == self ? to_i : self
end
end
That's because:
irb> 1.0 == 1
=> true
irb> 1 == 1.0
=> true
Then you could do:
irb> 1.0.prettify
=> 1
irb> 1.5.prettify
=> 1.5
A one liner sprintf...
sprintf("%g", 5.0)
=> "5"
sprintf("%g", 5.5)
=> "5.5"
This is the solution that ended up working the way I want it to:
class Float
alias_method(:original_to_s, :to_s) unless method_defined?(:original_to_s)
def is_whole?
self % 1 == 0
end
def to_s
self.is_whole? ? self.to_i.to_s : self.original_to_s
end
end
This way I can update the is_whole? logic (I seems like tadman's is the most sophisticated) if needed, and it ensures that anywhere a Float outputs to a string (eg, in a form) it appears the way I want it to (ie, no zeros on the end).
Thanks to everybody for your ideas - they really helped.
I'm don't know much about Ruby.
But this is a display issue. I would be extremely surprised if the libraries you are using don't have a way to format a number when you convert it to a string.
There might not be a catch-all formatting option that does exactly what you want but you could set up a method that returns true if the float is the float representation of a whole number and false otherwise. Inside a formatting routine that you create (so you only have to do this in once place) just change the formatting based on if this is true or false.
This discusses how to control the number of digits that appear after the decimal when displaying a number.
Watch out for the intricacies of floating point representations. Math might say the answer is 3 but you may get 3.000000000000000000001. I'd suggest using a delta to see if the number is almost an integer number.
If you are using rails you can use helper number_to_rounded with option strip_insignificant_zeros, for example:
ActiveSupport::NumberHelper.number_to_rounded(42.0, strip_insignificant_zeros: true)
or like this:
42.0.to_s(:rounded, strip_insignificant_zeros: true)
Although I'd tend to agree with the above post, if you must do this:
(float == float.floor) ? float.to_i : float
Here's my horribly hacktastic implementation provided for educational purposes:
class Float
def to_int_if_whole(precision = 2)
("%.#{precision}f" % self).split(/\./).last == '0' * precision and self.to_i or self
end
end
puts 1.0.to_int_if_whole # => 1
puts 2.5.to_int_if_whole # => 2.5
puts 1.9999999999999999999923.to_int_if_whole # => 2
The reason for using the sprintf-style call is that it handles floating point approximations much more reliably than the Float#round method tends to.
I don't know much about Ruby either.
But in C++, I'd do this:
bool IsWholeNumber( float f )
{
const float delta = 0.0001;
int i = (int) f;
return (f - (float)i) < delta;
}
And then I'd format the output precision based on that.
Additionally, how can I format it as a string padded with zeros?
To generate the number call rand with the result of the expression "10 to the power of 10"
rand(10 ** 10)
To pad the number with zeros you can use the string format operator
'%010d' % rand(10 ** 10)
or the rjust method of string
rand(10 ** 10).to_s.rjust(10,'0')
I would like to contribute probably a simplest solution I know, which is a quite a good trick.
rand.to_s[2..11]
=> "5950281724"
This is a fast way to generate a 10-sized string of digits:
10.times.map{rand(10)}.join # => "3401487670"
The most straightforward answer would probably be
rand(1e9...1e10).to_i
The to_i part is needed because 1e9 and 1e10 are actually floats:
irb(main)> 1e9.class
=> Float
DON'T USE rand.to_s[2..11].to_i
Why? Because here's what you can get:
rand.to_s[2..9] #=> "04890612"
and then:
"04890612".to_i #=> 4890612
Note that:
4890612.to_s.length #=> 7
Which is not what you've expected!
To check that error in your own code, instead of .to_i you may wrap it like this:
Integer(rand.to_s[2..9])
and very soon it will turn out that:
ArgumentError: invalid value for Integer(): "02939053"
So it's always better to stick to .center, but keep in mind that:
rand(9)
sometimes may give you 0.
To prevent that:
rand(1..9)
which will always return something withing 1..9 range.
I'm glad that I had good tests and I hope you will avoid breaking your system.
Random number generation
Use Kernel#rand method:
rand(1_000_000_000..9_999_999_999) # => random 10-digits number
Random string generation
Use times + map + join combination:
10.times.map { rand(0..9) }.join # => random 10-digit string (may start with 0!)
Number to string conversion with padding
Use String#% method:
"%010d" % 123348 # => "0000123348"
Password generation
Use KeePass password generator library, it supports different patterns for generating random password:
KeePass::Password.generate("d{10}") # => random 10-digit string (may start with 0!)
A documentation for KeePass patterns can be found here.
Just because it wasn't mentioned, the Kernel#sprintf method (or it's alias Kernel#format in the Powerpack Library) is generally preferred over the String#% method, as mentioned in the Ruby Community Style Guide.
Of course this is highly debatable, but to provide insight:
The syntax of #quackingduck's answer would be
# considered bad
'%010d' % rand(10**10)
# considered good
sprintf('%010d', rand(10**10))
The nature of this preference is primarily due to the cryptic nature of %. It's not very semantic by itself and without any additional context it can be confused with the % modulo operator.
Examples from the Style Guide:
# bad
'%d %d' % [20, 10]
# => '20 10'
# good
sprintf('%d %d', 20, 10)
# => '20 10'
# good
sprintf('%{first} %{second}', first: 20, second: 10)
# => '20 10'
format('%d %d', 20, 10)
# => '20 10'
# good
format('%{first} %{second}', first: 20, second: 10)
# => '20 10'
To make justice for String#%, I personally really like using operator-like syntaxes instead of commands, the same way you would do your_array << 'foo' over your_array.push('123').
This just illustrates a tendency in the community, what's "best" is up to you.
More info in this blogpost.
I ended up with using Ruby kernel srand
srand.to_s.last(10)
Docs here: Kernel#srand
Here is an expression that will use one fewer method call than quackingduck's example.
'%011d' % rand(1e10)
One caveat, 1e10 is a Float, and Kernel#rand ends up calling to_i on it, so for some higher values you might have some inconsistencies. To be more precise with a literal, you could also do:
'%011d' % rand(10_000_000_000) # Note that underscores are ignored in integer literals
('%010d' % rand(0..9999999999)).to_s
or
"#{'%010d' % rand(0..9999999999)}"
I just want to modify first answer. rand (10**10) may generate 9 digit random no if 0 is in first place. For ensuring 10 exact digit just modify
code = rand(10**10)
while code.to_s.length != 10
code = rand(11**11)
end
Try using the SecureRandom ruby library.
It generates random numbers but the length is not specific.
Go through this link for more information: http://ruby-doc.org/stdlib-2.1.2/libdoc/securerandom/rdoc/SecureRandom.html
Simplest way to generate n digit random number -
Random.new.rand((10**(n - 1))..(10**n))
generate 10 digit number number -
Random.new.rand((10**(10 - 1))..(10**10))
This technique works for any "alphabet"
(1..10).map{"0123456789".chars.to_a.sample}.join
=> "6383411680"
Just use straightforward below.
rand(10 ** 9...10 ** 10)
Just test it on IRB with below.
(1..1000).each { puts rand(10 ** 9...10 ** 10) }
To generate a random, 10-digit string:
# This generates a 10-digit string, where the
# minimum possible value is "0000000000", and the
# maximum possible value is "9999999999"
SecureRandom.random_number(10**10).to_s.rjust(10, '0')
Here's more detail of what's happening, shown by breaking the single line into multiple lines with explaining variables:
# Calculate the upper bound for the random number generator
# upper_bound = 10,000,000,000
upper_bound = 10**10
# n will be an integer with a minimum possible value of 0,
# and a maximum possible value of 9,999,999,999
n = SecureRandom.random_number(upper_bound)
# Convert the integer n to a string
# unpadded_str will be "0" if n == 0
# unpadded_str will be "9999999999" if n == 9_999_999_999
unpadded_str = n.to_s
# Pad the string with leading zeroes if it is less than
# 10 digits long.
# "0" would be padded to "0000000000"
# "123" would be padded to "0000000123"
# "9999999999" would not be padded, and remains unchanged as "9999999999"
padded_str = unpadded_str.rjust(10, '0')
rand(9999999999).to_s.center(10, rand(9).to_s).to_i
is faster than
rand.to_s[2..11].to_i
You can use:
puts Benchmark.measure{(1..1000000).map{rand(9999999999).to_s.center(10, rand(9).to_s).to_i}}
and
puts Benchmark.measure{(1..1000000).map{rand.to_s[2..11].to_i}}
in Rails console to confirm that.
An alternative answer, using the regexp-examples ruby gem:
require 'regexp-examples'
/\d{10}/.random_example # => "0826423747"
There's no need to "pad with zeros" with this approach, since you are immediately generating a String.
This will work even on ruby 1.8.7:
rand(9999999999).to_s.center(10, rand(9).to_s).to_i
A better approach is use Array.new() instead of .times.map. Rubocop recommends it.
Example:
string_size = 9
Array.new(string_size) do
rand(10).to_s
end
Rubucop, TimesMap:
https://www.rubydoc.info/gems/rubocop/RuboCop/Cop/Performance/TimesMap
In my case number must be unique in my models, so I added checking block.
module StringUtil
refine String.singleton_class do
def generate_random_digits(size:)
proc = lambda{ rand.to_s[2...(2 + size)] }
if block_given?
loop do
generated = proc.call
break generated if yield(generated) # check generated num meets condition
end
else
proc.call
end
end
end
end
using StringUtil
String.generate_random_digits(3) => "763"
String.generate_random_digits(3) do |num|
User.find_by(code: num).nil?
end => "689"(This is unique in Users code)
I did something like this
x = 10 #Number of digit
(rand(10 ** x) + 10**x).to_s[0..x-1]
Random 10 numbers:
require 'string_pattern'
puts "10:N".gen