changing integers into words ruby without gems - ruby

Question

I am trying to change numbers up to 100 from integers into words, but have run into some trouble, can anyone point out what is missing with my code:
def in_words(integer)
numWords = {
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=>"fourty",
50=>"fifty",
60=>"sixty",
70=>"seventy",
80=>"eighty",
90=>"ninety",
100=>"one hundred"
}
array = integer.to_s.split('')
new_array = []
numWords.each do |k,v|
array.each do |x|
if x = k
new_array.push(v)
end
end
end
new_array.join('')
end
Right now when I do:
inwords(0)
I get the following:
=>"zeroonetwothreefourfivesixseveneightnineteneleventwelvethirteenfourteenfiftee nsixteenseventeeneighteennineteentwentythirtyfourtyfiftysixtyseventyeightyninetyone hundred"

Answer 1

Edit
I noticed your code iterates through the array a lot of times and uses the = instead of the == in your if statements.
Your code could be more efficient using the Hash's #[] method in combination with the #map method.., here's a one-line alternative:
integer.to_s.split('').map {|i| numWords[i.to_i]} .join ' '
Also, notice that the integer.to_s.split('') will split the array into one-digit strings, so having numbers up to a hundred isn't relevant for the code I proposed.
To use all the numbers in the Hash, you might want to use a Regexp to identify the numbers you have. One way is to do the following (I write it in one line, but it's easy to break it down using variable names for each step):
integer.to_s.gsub(/(\d0)|([1]?\d)/) {|v| v + " "} .split.map {|i| numWords[i.to_i]} .join ' '
# or:
integer.to_s.gsub(/(#{numWords.keys.reverse.join('|')})/) {|v| v + " "} .split.map {|i| numWords[i.to_i]} .join ' '
# out = integer.to_s
# out = out.gsub(/(#{numWords.keys.reverse.join('|')})/) {|v| v + " "}
# out = out.split
# out = out.map {|i| numWords[i.to_i]}
# out = out.join ' '
Edit 2
Since you now mention that you want the method to accept numbers up to a hundred and return the actual number (23 => twenty three), maybe a different approach should be taken... I would recommend that you update your question as well.
def in_words(integer)
numWords = {
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=>"fourty",
50=>"fifty",
60=>"sixty",
70=>"seventy",
80=>"eighty",
90=>"ninety",
100=>"one hundred"
}
raise "cannot accept such large numbers" if integer > 100
raise "cannot accept such small numbers" if integer < 0
return "one hundred" if integer == 100
if integer < 20 || integer %10 == 0
numWords[integer]
else
[numWords[integer / 10 * 10], numWords[integer % 10]].join ' '
end
end
the integer / 10 * 10 makes the number a round number (ten, twenty, etc') because integers don't have fractions (so, 23/10 == 2 and 2 * 10 == 20). The same could be achieved using integer.round(-1), which is probably better.

Answer 2

It seems like all you're trying to do is find a mapping from an implicit hash
module NumWords
INT2STR = {
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=>"fourty",
50=>"fifty",
60=>"sixty",
70=>"seventy",
80=>"eighty",
90=>"ninety",
100=>"one hundred"
}
module_function
def in_words(integer)
INT2STR[integer]
end
end
The above code separates the hash definition from the method call so that the hash doesn't get recreated every time you call in_words.
You can also use Hash#fetch instead of Hash#[] as Andrey pointed out.

Answer 3

Your test whether x = k is your first problem (in two ways).
Firstly, if x = k means assign the value of k to x and then execute the if block if that value is true (basically anything other than false or nil).
What you should actually be testing is x == k which will return true if x is equal to k.
The second problem is that you converted your number into an array of string representation so you are comparing, for example, if "0" == 0. This won't return true because they are different types.
If you convert it to if x.to_i == k then your if block will be executed and you'll get:
> in_words(0)
=> "zero"
Then you get to move onto the next problem which is that you're looking at your number digit by digit and some of the values you are testing against need two digits to be recognised:
> in_words(10)
=> "zeroone"
You might be in looking at a different question then - or maybe that is the question you wanted answered all along!

Answer 4

Here's another way you might do it:
ONES_TO_TEXT = { 0=>"zero", 1=>"one", 2=>"two", 3=>"three", 4=>"four",
5=>"five", 6=>"six", 7=>"seven", 8=>"eight", 9=>"nine" }
TEENS_TO_TEXT = { 10=>"ten", 11=>"eleven", 12=>"twelve",
13=>"thirteen", 15=>"fifteen" }
TENS_TO_TEXT = { 2=>"twenty", 3=>"thirty", 5=>"fifty", 8=>"eighty" }
def in_words(n)
raise ArgumentError, "#{n} is out-of_range" unless (0..100).cover?(n)
case n.to_s.size
when 1 then ONES_TO_TEXT[n]
when 3 then "one hundred"
else
case n
when (10..19)
TEENS_TO_TEXT.key?(n) ? TEENS_TO_TEXT[n] : ONES_TO_TEXT[n]+"teen"
else
t,o = n.divmod(10)
(TENS_TO_TEXT.key?(t) ? TENS_TO_TEXT[t] : ONES_TO_TEXT[t]+"ty") +
(o.zero? ? '' : "-#{ONES_TO_TEXT[o]}")
end
end
end
Let's try it:
in_words(5) #=> "five"
in_words(10) #=> "ten"
in_words(15) #=> "fifteen"
in_words(20) #=> "twenty"
in_words(22) #=> "twenty-two"
in_words(30) #=> "thirty"
in_words(40) #=> "fourty"
in_words(45) #=> "fourty-five"
in_words(50) #=> "fifty"
in_words(80) #=> "eighty"
in_words(99) #=> "ninety-nine"
in_words(100) #=> "one hundred"
Here the increased complexity may not be justified, but this approach may in fact simplify the calculations when the maximum permitted value of n is much greater than 100.