Use regular expression to evaluate and modify string in Ruby?

Question

I want to modify part of a string I have using Ruby.

The string is [x, y] where y is an integer that I want to change to its alphabetical letter. So say [1, 1] would become [1, A] and [1, 26] would become [1, Z].

Would a regular expression help me do this? or is there an easier way? I am not to strong with regular expressions, I am reading up on those now.

Well what I am doing is changing an index I have for excel cells. The gem I am using gives me the index as [1, 1], etc. I just want to change that to the column formats [1, A] — IZI_Shadow_IZI
– IZI_Shadow_IZI, Commented Aug 22, 2012 at 14:49
MS Excel can handle that: choose Tools->Options->General tab-> R1C1 reference style. — steenslag
– steenslag, Commented Aug 22, 2012 at 15:44

Florian Feldhaus · Accepted Answer · 2012-08-22 16:58:06Z

1

The shortest way I can think of is the following

string = "[1,1]"
array = string.chop.reverse.chop.reverse.split(',')
new_string="[#{array.first},#{(array.last.to_i+64).chr}]"

edited Aug 22, 2012 at 16:58

answered Aug 22, 2012 at 16:51

Florian Feldhaus

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Noxx · Accepted Answer · 2012-08-22 16:40:12Z

Maybe this helps:

Because we do not have an alphabet yet we can look up the position, create one. This is a range converted to an array so you don't need to specify it yourself.

alphabet = ("A".."Z").to_a

Then we try to get the integer/position out of the string:

string_to_match = "[1,5]"
/(\d+)\]$/.match(string_to_match)

Maybe the regexp can be improved, however for this example it is working. The first reference in the MatchData is holding the second integer in your "string_to_match". Or you can get it via "$1". Do not forget to convert it to an integer.

position_in_alphabet = $1.to_i

Also we need to remember that the index of arrays starts at 0 and not 1

position_in_alphabet -= 1

Finally, we can take a look which char we really get

char = alphabet[position_in_alphabet]

Example:

alphabet = ("A".."Z").to_a #=> ["A", "B", "C", ..*snip*.. "Y", "Z"]
string_to_match = "[1,5]" #=> "[1,5]"
/(\d+)\]$/.match(string_to_match) #=> #<MatchData "5]" 1:"5">
position_in_alphabet = $1.to_i #=> 5
position_in_alphabet -= 1 #=> 4
char = alphabet[position_in_alphabet] #=> "E"

Greetings~

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