convert phone number into words - ruby

Question

So I figured out that I can count through the hash, the problem is that for 7 and 9 I have four values. I have tried several other things with no luck. Can someone help understand what else I could do to get the values I want out of the hash. I realize that I can match the numbers with the key, but I am confused how to get the values to permute.
letters = {"1" => ["1", "1", "1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
phone_number = gets.chomp.to_s
words = []
word = []
numbers = phone_number.chomp.chars
count0 = 0
while count0 < 3
count1 = 0
while count1 < 3
count2 = 0
while count2 < 3
count3 = 0
while count3 < 3
count4 = 0
while count4 < 3
count5 = 0
while count5 < 3
count6 = 0
while count6 < 3
word[0] = letters[numbers[0]][count0]
word[1] = letters[numbers[1]][count1]
word[2] = letters[numbers[2]][count2]
word[3] = letters[numbers[3]][count3]
word[4] = letters[numbers[4]][count4]
word[5] = letters[numbers[5]][count5]
word[6] = letters[numbers[6]][count6]
words << word.join
count6 += 1
end
count5 += 1
end
count4 += 1
end
count3 += 1
end
count2 += 1
end
count1 += 1
end
count0 += 1
end
puts words
Edit:
I want to a seven digit number and print out all possible letter combinations. I am a beginner so I want to understand with the things I know now. I want to try and do this with if statements please.
numbers = phone_number.chomp.chars
if letters.key?(numbers[0])
if letters.key?(numbers[1])
if letters.key?(numbers[2])
if letters.key?(numbers[3])
if letters.key?(numbers[4])
if letters.key?(numbers[5])
if letters.key?(numbers[6])
end
end
end
end
end
end
end
I understand how to grab a value from a matching key, but don't get how I can hold the first value while going through the rest, if that makes any sense.

Answer 1

product is the function you are looking for, the following works with any number of digits:
digits = '27'
keys = digits.chars.map{|digit|letters[digit]}
p keys.shift.product(*keys).map(&:join) #=> ["ap", "aq", "ar", "as", "bp", "bq", "br", "bs", "cp", "cq", "cr", "cs"]

Answer 2

This prints all possible words for a variable-sized phone number:
letters = {"1" => ["1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
digits = gets.chomp.split ''
# Total number of combinations
n = digits.inject(1) { |a,b| a * letters[b].size }
words = []
0.upto n-1 do |q|
word = []
digits.reverse.each do |digit|
q, r = q.divmod letters[digit].size
word.unshift letters[digit][r]
end
words << word.join
end
puts words
For example, if the input is 67, then there are 12 combinations:
mp mq mr ms np nq nr ns op oq or os
Edit: I don't see a way to make use of the 7 if statements as you have written, but perhaps this is closer to the kind of answer you are looking for:
words = []
letters[digits[0]].each do |c0|
letters[digits[1]].each do |c1|
letters[digits[2]].each do |c2|
letters[digits[3]].each do |c3|
letters[digits[4]].each do |c4|
letters[digits[5]].each do |c5|
letters[digits[6]].each do |c6|
words << [c0,c1,c2,c3,c4,c5,c6].join
end
end
end
end
end
end
end
puts words
A good exercise would be to re-write this in a way that can work for phone numbers of any length, not just 7. Again, this is only for instructional purposes. In practice, one would use Array's product method as in hirolau's answer.

Answer 3

LETTERS = {"1" => ["1", "1", "1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
def convert_to_phone_number(string)
string.each_char.with_object([]) { |x, arr| LETTERS.each { |k,v| (arr.push k; break) if v.include?(x) }}.join
end
convert_to_phone_number "foobar"
#=> "366227"

Answer 4

i think so it is issue of cache memory
u need to change like below
LETTERS = {"1" => ["1", "1", "1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
def convert_to_phone_number(string)
string.each_char.with_object([]) { |x, arr| LETTERS.each { |k,v| (arr.push k; break) if v.include?(x) }}.join
end
convert_to_phone_number "foobar"