How to convert numbers to words in ruby?
I know there is a gem somewhere. Trying to implement it without a gem. I just need the numbers to words in English for integers. Found this but it is very messy. If you have any idea on how to implement a cleaner easier to read solution please share.
http://raveendran.wordpress.com/2009/05/29/ruby-convert-number-to-english-word/
Here is what I have been working on. But having some problem implementing the scales. The code is still a mess. I hope to make it more readable when it functions properly.
class Numberswords
def in_words(n)
words_hash = {0=>"zero",1=>"one",2=>"two",3=>"three",4=>"four",5=>"five",6=>"six",7=>"seven",8=>"eight",9=>"nine",
10=>"ten",11=>"eleven",12=>"twelve",13=>"thirteen",14=>"fourteen",15=>"fifteen",16=>"sixteen",
17=>"seventeen", 18=>"eighteen",19=>"nineteen",
20=>"twenty",30=>"thirty",40=>"forty",50=>"fifty",60=>"sixty",70=>"seventy",80=>"eighty",90=>"ninety"}
scale = [000=>"",1000=>"thousand",1000000=>" million",1000000000=>" billion",1000000000000=>" trillion", 1000000000000000=>" quadrillion"]
if words_hash.has_key?(n)
words_hash[n]
#still working on this middle part. Anything above 999 will not work
elsif n>= 1000
print n.to_s.scan(/.{1,3}/) do |number|
print number
end
#print value = n.to_s.reverse.scan(/.{1,3}/).inject([]) { |first_part,second_part| first_part << (second_part == "000" ? "" : second_part.reverse.to_i.in_words) }
#(value.each_with_index.map { |first_part,second_part| first_part == "" ? "" : first_part + scale[second_part] }-[""]).reverse.join(" ")
elsif n <= 99
return [words_hash[n - n%10],words_hash[n%10]].join(" ")
else
words_hash.merge!({ 100=>"hundred" })
([(n%100 < 20 ? n%100 : n.to_s[2].to_i), n.to_s[1].to_i*10, 100, n.to_s[0].to_i]-[0]-[10])
.reverse.map { |num| words_hash[num] }.join(" ")
end
end
end
#test code
test = Numberswords.new
print test.in_words(200)
My take on this
def in_words(int)
numbers_to_name = {
1000000 => "million",
1000 => "thousand",
100 => "hundred",
90 => "ninety",
80 => "eighty",
70 => "seventy",
60 => "sixty",
50 => "fifty",
40 => "forty",
30 => "thirty",
20 => "twenty",
19=>"nineteen",
18=>"eighteen",
17=>"seventeen",
16=>"sixteen",
15=>"fifteen",
14=>"fourteen",
13=>"thirteen",
12=>"twelve",
11 => "eleven",
10 => "ten",
9 => "nine",
8 => "eight",
7 => "seven",
6 => "six",
5 => "five",
4 => "four",
3 => "three",
2 => "two",
1 => "one"
}
str = ""
numbers_to_name.each do |num, name|
if int == 0
return str
elsif int.to_s.length == 1 && int/num > 0
return str + "#{name}"
elsif int < 100 && int/num > 0
return str + "#{name}" if int%num == 0
return str + "#{name} " + in_words(int%num)
elsif int/num > 0
return str + in_words(int/num) + " #{name} " + in_words(int%num)
end
end
end
puts in_words(4) == "four"
puts in_words(27) == "twenty seven"
puts in_words(102) == "one hundred two"
puts in_words(38_079) == "thirty eight thousand seventy nine"
puts in_words(82102713) == "eighty two million one hundred two thousand seven hundred thirteen"
Have you considered humanize ?
https://github.com/radar/humanize
Simple answer use humanize gem and you will get desired output
Install it directly
gem install humanize
Or add it to your Gemfile
gem 'humanize'
And you can use it
require 'humanize'
1.humanize #=> 'one'
345.humanize #=> 'three hundred and forty-five'
1723323.humanize #=> 'one million, seven hundred and twenty-three thousand, three hundred and twenty-three'
If you are using this in rails you can directly use this
NOTE: As mentioned by sren in the comments below. The humanize method provided by ActiveSupport is different than the gem humanize
You can also use the to_words gem.
This Gem converts integers into words.
e.g.
1.to_words # one ,
100.to_words # one hundred ,
101.to_words # one hundred and one
It also converts negative numbers.
I can see what you're looking for, and you may wish to check out this StackOverflow post: Number to English Word Conversion Rails
Here it is in summary:
No, you have to write a function yourself. The closest thing to what
you want is number_to_human, but that does not convert 1 to One.
Here are some URLs that may be helpful:
http://codesnippets.joyent.com/posts/show/447
http://raveendran.wordpress.com/2009/05/29/ruby-convert-number-to-english-word/
http://deveiate.org/projects/Linguistics/
I am not quite sure, if this works for you. Method can be called like this.
n2w(33123) {|i| puts i unless i.to_s.empty?}
Here is the method ( I have not tested it fully. I think it works upto million. Code is ugly, there is a lot of room for re-factoring. )
def n2w(n)
words_hash = {0=>"zero",1=>"one",2=>"two",3=>"three",4=>"four",5=>"five",6=>"six",7=>"seven",8=>"eight",9=>"nine",
10=>"ten",11=>"eleven",12=>"twelve",13=>"thirteen",14=>"fourteen",15=>"fifteen",16=>"sixteen",
17=>"seventeen", 18=>"eighteen",19=>"nineteen",
20=>"twenty",30=>"thirty",40=>"forty",50=>"fifty",60=>"sixty",70=>"seventy",80=>"eighty",90=>"ninety"}
scale = {3=>"hundred",4 =>"thousand",6=>"million",9=>"billion"}
if words_hash.has_key?n
yield words_hash[n]
else
ns = n.to_s.split(//)
while ns.size > 0
if ns.size == 2
yield("and")
yield words_hash[(ns.join.to_i) - (ns.join.to_i)%10]
ns.shift
end
if ns.size > 4
yield(words_hash[(ns[0,2].join.to_i) - (ns[0,2].join.to_i) % 10])
else
yield(words_hash[ns[0].to_i])
end
yield(scale[ns.size])
ns.shift
end
end
end
def subhundred number
ones = %w{zero one two three four five six seven eight nine
ten eleven twelve thirteen fourteen fifteen
sixteen seventeen eighteen nineteen}
tens = %w{zero ten twenty thirty **forty** fifty sixty seventy eighty ninety}
subhundred = number % 100
return [ones[subhundred]] if subhundred < 20
return [tens[subhundred / 10]] if subhundred % 10 == 0
return [tens[subhundred / 10], ones[subhundred % 10]]
end
def subthousand number
hundreds = (number % 1000) / 100
tens = number % 100
s = []
s = subhundred(hundreds) + ["hundred"] unless hundreds == 0
s = s + ["and"] unless hundreds == 0 or tens == 0
s = s + [subhundred(tens)] unless tens == 0
end
def decimals number
return [] unless number.to_s['.']
digits = number.to_s.split('.')[1].split('').reverse
digits = digits.drop_while {|d| d.to_i == 0} . reverse
digits = digits.map {|d| subhundred d.to_i} . flatten
digits.empty? ? [] : ["and cents"] + digits
end
def words_from_numbers number
steps = [""] + %w{thousand million billion trillion quadrillion quintillion sextillion}
result = []
n = number.to_i
steps.each do |step|
x = n % 1000
unit = (step == "") ? [] : [step]
result = subthousand(x) + unit + result unless x == 0
n = n / 1000
end
result = ["zero"] if result.empty?
result = result + decimals(number)
result.join(' ').strip
end
def words_from_numbers(number)
ApplicationHelper.words_from_numbers(number)
end
Its been quite a while since the question was asked. Rails has something inbuilt for this now.
https://api.rubyonrails.org/classes/ActionView/Helpers/NumberHelper.html
number_to_human(1234567) # => "1.23 Million"
number_to_human(1234567890) # => "1.23 Billion"
number_to_human(1234567890123) # => "1.23 Trillion"
number_to_human(1234567890123456) # => "1.23 Quadrillion"
number_to_human(1234567890123456789) # => "1230 Quadrillion"
Related
I have a long if else statement:
rnd = rand(1..1000)
if rnd >= 600
0
elsif rnd < 600 && rnd >= 350
1
elsif rnd < 350 && rnd >= 270
2
elsif rnd < 270 && rnd >= 200
3
elsif rnd < 200 && rnd >= 150
4
elsif rnd < 150 && rnd >= 100
5
elsif rnd < 100 && rnd >= 80
6
elsif rnd < 80 && rnd >= 50
7
elsif rnd < 50 && rnd >= 30
8
else
9
end
I would like to shorten it. Is it possible?
My rubocop swears at this long method.
I would start with something like this:
RANGES = {
(0...30) => 9,
(30...50) => 8,
(50...80) => 7,
# ...
(350...600) => 1,
(600...1000) => 0
}
rnd = rand(1..1000)
RANGES.find { |k, _| k.cover?(rnd) }.last
Great answers already! Just chiming in since I had a suspicion that ruby could handle this with a case statement, and it appears to be able to do so:
rnd = rand(1..1000)
case rnd
when 600.. then 0
when 350...600 then 1
when 270...350 then 2
...
else 9
end
Regardless of the approach taken, you're going to have to specify the ranges somewhere, so I think using something like a case statement is appropriate here (sorry! It doesn't shorten the code more than a few lines). Using a hash would also be a great approach (and might allow you to move the hash elsewhere), as other commenters have already shown.
It's worth mentioning, with ruby ranges, .. means that the range is inclusive and includes the last value (1..10 includes the number 10), and ... means the range is exclusive where it does not include the last value.
The top case 600.. is an endless range, which means it will match anything greater than 600. (That functionality was added in ruby 2.6)
You can simplify your conditions by using only the lower bound.
And you can avoid repeting elsif because it is cumbersome
rnd = rand(1..1000)
lower_bounds = {
600 => 0,
350 => 1,
270 => 2,
200 => 3,
150 => 4,
100 => 5,
80 => 6,
50 => 7,
30 => 8,
0 => 9,
}
lower_bounds.find { |k, _| k <= rnd }.last
MX = 1000
LIMITS = [600, 350, 270, 200, 150, 100, 80, 50, 30, 0]
The required index can be computed as follows.
def doit
rnd = rand(1..MX)
LIMITS.index { |n| n <= rnd }
end
doit
#=> 5
In this example rnd #=> 117.
If this must be repeated many times, and speed is paramount, you could do the following.
LOOK_UP = (1..MX).each_with_object({}) do |m,h|
h[m] = LIMITS.index { |n| n <= m }
end
#=> {1=>9, 2=>9,..., 29=>9,
# 30=>8, 31=>8,..., 49=>8,
# ...
# 600=>0, 601=>0,..., 1000=>0}
Then simply
def doit
LOOK_UP[rand(1..MX)]
end
doit
#=> 3
In this example rand(1..MX) #=> 262.
If speed were paramount but MX were so large that the previous approach would require excessive memory, you could use a binary search.
def doit
rnd = rand(1..MX)
LIMITS.bsearch_index { |n| n <= rnd }
end
doit
#=> 5
In this example rnd #=> 174.
See Array#bsearch_index. bsearch_index returns the correct index in O(log n), n being LIMITS.size). bsearch_index requires the array on which it operates to be ordered.
Looking to work with a dataset of strings that store money amounts in these formats. For example:
$217.3M
$1.6B
$34M
€1M
€2.8B
I looked at the money gem but it doesn't look like it handles the "M, B, k"'s back to numbers. Looking for a gem that does do that so I can convert exchange rates and compare quantities. I need the opposite of the number_to_human method.
I would start with something like this:
MULTIPLIERS = { 'k' => 10**3, 'm' => 10**6, 'b' => 10**9 }
def human_to_number(human)
number = human[/(\d+\.?)+/].to_f
factor = human[/\w$/].try(:downcase)
number * MULTIPLIERS.fetch(factor, 1)
end
human_to_number('$217.3M') #=> 217300000.0
human_to_number('$1.6B') #=> 1600000000.0
human_to_number('$34M') #=> 34000000.0
human_to_number('€1M') #=> 1000000.0
human_to_number('€2.8B') #=> 2800000000.0
human_to_number('1000') #=> 1000.0
human_to_number('10.88') #=> 10.88
I decided to not be lazy and actually write my own function if anyone else wants this:
def text_to_money(text)
returnarray = []
if (text.count('k') >= 1 || text.count('K') >= 1)
multiplier = 1000
elsif (text.count('M') >= 1 || text.count('m') >= 1)
multiplier = 1000000
elsif (text.count('B') >= 1 || text.count('b') >= 1)
multiplier = 1000000000
else
multiplier = 1
end
num = text.to_s.gsub(/[$,]/,'').to_f
total = num * multiplier
returnarray << [text[0], total]
return returnarray
end
Thanks for the help!
I know there is a gem somewhere. Trying to implement it without a gem. I just need the numbers to words in English for integers. Found this but it is very messy. If you have any idea on how to implement a cleaner easier to read solution please share.
If you don't want to use any Gem then....
Try this one(this will convert upto million):
def in_words(int)
numbers_to_name = {
1000000 => "million",
1000 => "thousand",
100 => "hundred",
90 => "ninety",
80 => "eighty",
70 => "seventy",
60 => "sixty",
50 => "fifty",
40 => "forty",
30 => "thirty",
20 => "twenty",
19=>"nineteen",
18=>"eighteen",
17=>"seventeen",
16=>"sixteen",
15=>"fifteen",
14=>"fourteen",
13=>"thirteen",
12=>"twelve",
11 => "eleven",
10 => "ten",
9 => "nine",
8 => "eight",
7 => "seven",
6 => "six",
5 => "five",
4 => "four",
3 => "three",
2 => "two",
1 => "one"
}
str = ""
numbers_to_name.each do |num, name|
if int == 0
return str
elsif int.to_s.length == 1 && int/num > 0
return str + "#{name}"
elsif int < 100 && int/num > 0
return str + "#{name}" if int%num == 0
return str + "#{name} " + in_words(int%num)
elsif int/num > 0
return str + in_words(int/num) + " #{name} " + in_words(int%num)
end
end
end
puts in_words(4) == "four"
puts in_words(27) == "twenty seven"
puts in_words(102) == "one hundred two"
puts in_words(38_079) == "thirty eight thousand seventy nine"
puts in_words(82102713) == "eighty two million one hundred two thousand seven hundred thirteen"
Noob to Ruby here. Working through some exercises and have hit a wall.
Exercise: Calculate the letter grade of a series of grades
Create a method get_grade that accepts an Array of test scores. Each score in the array should be between 0 and 100, where 100 is the max score.
Compute the average score and return the letter grade as a String, i.e., 'A', 'B', 'C', 'D', 'E', or 'F'.
I keep returning the error:
avg.rb:1: syntax error, unexpected tLBRACK, expecting ')'
def get_grade([100,90,80])
^
avg.rb:1: syntax error, unexpected ')', expecting $end
Here's what I have so far. I'd like to stick with the methods below or .join as I'm trying to work with the methods we're learning in class. So sum, inject, etc won't necessarily be helpful. And I apologize in advance for the specificity of the request :) I'm sure there's a way better way that is way less code, but I'm just trying to learn it this way to start.
def get_grade([100,90,80])
get_grade = (array[0] + array[1] + array[2]).to_i / array.length.to_i
case get_grade
when 90..100
"A"
when 80..90
"B"
when 70..80
"C"
when 60..70
"D"
when 0..60
"F"
else
"Error"
end
end
puts get_grade([100,90,80])
You can't just randomly dump an array literal like [100,90,80] into the parameter list of a function definition. Judging by the function body, I think you meant to accept a single parameter array:
def get_grade(array)
grade = (array[0].to_i + array[1].to_i + array[2].to_i) / array.length
case grade
# unchanged
end
end
A terse replacement of the big case statement, for fun:
def letter_grade( score ) # assumes that score is between 0 and 100 (not 0-1)
%w[F F F F F F D C B A][ (score/10.0).floor ] || 'A' # handles grades >=100
end
Or, for more granularity:
def letter_grade( score ) # score is between 0 and 100 (not 0-1)
grades = %w[F F F F F F F F F F F F F F F F F F D- D D+ C- C C+ B- B B+ A- A A+ A+]
grades[ (3.0*score/10).floor ]
end
Thanks for the help today! Here's what I ended up doing to make it work with more than just 3 arguments. I used an Array#each method. I imagine there's a more elegant solution out there, but it worked! Worked on this since 10:00 AM, greatly appreciate the help!
def get_grade(array)
sum = 0
array.each do |element|
sum += element
end
average = sum / array.length
if average >= 90
grade = "A"
elsif average >= 80
grade = "B"
elsif average >= 70
grade = "C"
elsif average >= 60
grade = "D"
elsif average >= 0
grade = "F"
else
"Error"
end
end
puts get_grade([70,80,80,90,100])
puts get_grade([100,80,90,11,20])
puts get_grade([30,20,10,60,75])
Remember that the max score is 100 (and it can be assumed that the min is 0).
def get_grade(array)
sum = 0
array.each do |x|
sum += x
end
average = sum / array.length
if average > 100
print "Grades must be no more than 100!"
elsif average >= 90
grade = "A"
elsif average >= 80
grade = "B"
elsif average >= 70
grade = "C"
elsif average >= 60
grade = "D"
elsif average >=0
grade = "F"
else
print "Grades must be no less than 0!"
end
grade
end
puts get_grade([100,90,80]) == "A"
puts get_grade([98,90,80]) == "B"
puts get_grade([80,80,80]) == "B"
puts get_grade([55,45,35]) == "F"
puts get_grade([101,100,104])
puts get_grade([-2,-3,-4])
Added a proc so that even if a user enters a score over 100 it won't be calculated into the average.
Also refactored the switch statements to one line each. Let me know if this helps. Good luck.
def get_grade array
scores_under_100 = Proc.new {|score| score <= 100 && score > 0}
scores = array.select(&scores_under_100)
average = scores.inject(:+) / scores.size
case average
when 90..100 then puts "A."
when 80..89 then puts "B."
when 70..79 then puts "C."
when 60..69 then puts "D."
else puts "F."
end
end
puts get_grade([100, 100, 90, 67, 85, 200, 290, 299, 299])
Trying to create a simple regular expression that can extract numbers(between 7 - 14) after a keyword starting with g letter and some id, something like following :
(g)(\d{1,6})\s+(\d{7,14}\s*)+
Lets assume :
m = (/(g)(\d{1,6})\s+(\d{7,14}\s*)+/i.match("g12 327638474 83873478 2387327683 44 437643673476"))
I have results of :
#<MatchData "g23333 327638474 83873478 2387327683 " "g" "12" "2387327683 ">
But what I need as a final result , to include, 327638474, 83873478, 2387327683 and exclude 44.
For now I just getting the last number 2387327683 with not including the previous numbers
Any help here .
cheers
Instead of a regex, you can use something like that:
s = "g12 327638474 83873478 2387327683 44 437643673476"
s.split[1..-1].select { |x| (7..14).include?(x.size) }.map(&:to_i)
# => [327638474, 83873478, 2387327683, 437643673476]
Just as a FYI, here is a benchmark showing a bit faster way of accomplishing the selected answer:
require 'ap'
require 'benchmark'
n = 100_000
s = "g12 327638474 83873478 2387327683 44 437643673476"
ap s.split[1..-1].select { |x| (7..14).include? x.size }.map(&:to_i)
ap s.split[1..-1].select { |x| 7 <= x.size && x.size <= 14 }.map(&:to_i)
Benchmark.bm(11) do |b|
b.report('include?' ) { n.times{ s.split[1..-1].select { |x| (7..14).include? x.size }.map(&:to_i) } }
b.report('conditional') { n.times{ s.split[1..-1].select { |x| 7 <= x.size && x.size <= 14 }.map(&:to_i) } }
end
ruby ~/Desktop/test.rb
[
[0] 327638474,
[1] 83873478,
[2] 2387327683,
[3] 437643673476
]
[
[0] 327638474,
[1] 83873478,
[2] 2387327683,
[3] 437643673476
]
user system total real
include? 1.010000 0.000000 1.010000 ( 1.011725)
conditional 0.830000 0.000000 0.830000 ( 0.825746)
For speed I'll use the conditional test. It's a tiny bit more verbose, but is still easily read.