How to create two seperate arrays from one input? - ruby

Question

DESCRIPTION:
The purpose of my code is to take in input of a sequence of R's and C's and to simply store each number that comes after the character in its proper array.
For Example: "The input format is as follows: R1C4R2C5
Column Array: [ 4, 5 ] Row Array: [1,2]
My problem is I am getting the output like this:
[" ", 1]
[" ", 4]
[" ", 2]
[" ", 5]
**How do i get all the Row integers following R in one array, and all the Column integers following C in another seperate array. I do not want to create multiple arrays, Rather just two.
Help!
CODE:
puts 'Please input: '
input = gets.chomp
word2 = input.scan(/.{1,2}/)
col = []
row = []
word2.each {|a| col.push(a.split(/C/)) if a.include? 'C' }
word2.each {|a| row.push(a.split(/R/)) if a.include? 'R' }
col.each do |num|
puts num.inspect
end
row.each do |num|
puts num.inspect
end

Answer 1

x = "R1C4R2C5"
col = []
row = []
x.chars.each_slice(2) { |u| u[0] == "R" ? row << u[1] : col << u[1] }
p col
p row

Answer 2

The main problem with your code is that you replicate operations for rows and columns. You want to write "DRY" code, which stands for "don't repeat yourself".
Starting with your code as the model, you can DRY it out by writing a method like this to extract the information you want from the input string, and invoke it once for rows and once for columns:
def doit(s, c)
...
end
Here s is the input string and c is the string "R" or "C". Within the method you want
to extract substrings that begin with the value of c and are followed by digits. Your decision to use String#scan was a good one, but you need a different regex:
def doit(s, c)
s.scan(/#{c}\d+/)
end
I'll explain the regex, but let's first try the method. Suppose the string is:
s = "R1C4R2C5"
Then
rows = doit(s, "R") #=> ["R1", "R2"]
cols = doit(s, "C") #=> ["C4", "C5"]
This is not quite what you want, but easily fixed. First, though, the regex. The regex first looks for a character #{c}. #{c} transforms the value of the variable c to a literal character, which in this case will be "R" or "C". \d+ means the character #{c} must be followed by one or more digits 0-9, as many as are present before the next non-digit (here a "R" or "C") or the end of the string.
Now let's fix the method:
def doit(s, c)
a = s.scan(/#{c}\d+/)
b = a.map {|str| str[1..-1]}
b.map(&:to_i)
end
rows = doit(s, "R") #=> [1, 2]
cols = doit(s, "C") #=> [4, 5]
Success! As before, a => ["R1", "R2"] if c => "R" and a =>["C4", "C5"] if c => "C". a.map {|str| str[1..-1]} maps each element of a into a string comprised of all characters but the first (e.g., "R12"[1..-1] => "12"), so we have b => ["1", "2"] or b =>["4", "5"]. We then apply map once again to convert those strings to their Fixnum equivalents. The expression b.map(&:to_i) is shorthand for
b.map {|str| str.to_i}
The last computed quantity is returned by the method, so if it is what you want, as it is here, there is no need for a return statement at the end.
This can be simplified, however, in a couple of ways. Firstly, we can combine the last two statements by dropping the last one and changing the one above to:
a.map {|str| str[1..-1].to_i}
which also gets rid of the local variable b. The second improvement is to "chain" the two remaining statements, which also rids us of the other temporary variable:
def doit(s, c)
s.scan(/#{c}\d+/).map { |str| str[1..-1].to_i }
end
This is typical Ruby code.
Notice that by doing it this way, there is no requirement for row and column references in the string to alternate, and the numeric values can have arbitrary numbers of digits.
Here's another way to do the same thing, that some may see as being more Ruby-like:
s.scan(/[RC]\d+/).each_with_object([[],[]]) {|n,(r,c)|
(n[0]=='R' ? r : c) << n[1..-1].to_i}
Here's what's happening. Suppose:
s = "R1C4R2C5R32R4C7R18C6C12"
Then
a = s.scan(/[RC]\d+/)
#=> ["R1", "C4", "R2", "C5", "R32", "R4", "C7", "R18", "C6", "C12"]
scan uses the regex /([RC]\d+)/ to extract substrings that begin with 'R' or 'C' followed by one or more digits up to the next letter or end of the string.
b = a.each_with_object([[],[]]) {|n,(r,c)|(n[0]=='R' ? r : c) << n[1..-1].to_i}
#=> [[1, 2, 32, 4, 18], [4, 5, 7, 6, 12]]
The row values are given by [1, 2, 32, 4, 18]; the column values by [4, 5, 7, 6, 12].
Enumerable#each_with_object (v1.9+) creates an array comprised of two empty arrays, [[],[]]. The first subarray will contain the row values, the second, the column values. These two subarrays are represented by the block variables r and c, respectively.
The first element of a is "R1". This is represented in the block by the variable n. Since
"R1"[0] #=> "R"
"R1"[1..-1] #=> "1"
we execute
r << "1".to_i #=> [1]
so now
[r,c] #=> [[1],[]]
The next element of a is "C4", so we will execute:
c << "4".to_i #=> [4]
so now
[r,c] #=> [[1],[4]]
and so on.

Answer 3

rows, cols = "R1C4R2C5".scan(/R(\d+)C(\d+)/).flatten.partition.with_index {|_, index| index.even? }
> rows
=> ["1", "2"]
> cols
=> ["4", "5"]
Or
rows = "R1C4R2C5".scan(/R(\d+)/).flatten
=> ["1", "2"]
cols = "R1C4R2C5".scan(/C(\d+)/).flatten
=> ["4", "5"]
And to fix your code use:
word2.each {|a| col.push(a.delete('C')) if a.include? 'C' }
word2.each {|a| row.push(a.delete('R')) if a.include? 'R' }