在 Ruby 中拆分具有多个分隔符的字符串

Question

Take for instance, I have a string like this:

options = "Cake or pie, ice cream, or pudding"

I want to be able to split the string via or , , , and , or .

The thing is, is that I have been able to do it, but only by parsing , and , or first, and then splitting each array item at or , flattening the resultant array afterwards as such:

options = options.split(/(?:\s?or\s)*([^,]+)(?:,\s*)*/).reject(&:empty?);
options.each_index {|index| options[index] = options[index].sub("?","").split(" or "); }

The resultant array is as such: ["Cake", "pie", "ice cream", "pudding"]

Is there a more efficient (or easier) way to split my string on those three delimiters?

Answer 1

What about the following:

options.gsub(/ or /i, ",").split(",").map(&:strip).reject(&:empty?)

• replaces all delimiters but the ,
• splits it at ,
• trims each characters, since stuff like ice cream with a leading space might be left
• removes all blank strings

Answer 2

First of all, your method could be simplified a bit with Array#flatten :

>> options.split(',').map{|x|x.split 'or'}.flatten.map(&:strip).reject(&:empty?)
=> ["Cake", "pie", "ice cream", "pudding"]

I would prefer using a single regex:

>> options.split /\s*, or\s+|\s*,\s*|\s+or\s+/
=> ["Cake", "pie", "ice cream", "pudding"]

You can use | in a regex to give alternatives, and putting , or first guarantees that it won't produce an empty item. Capturing the whitespace with the regex is probably best for efficiency, since you don't have to scan the array again.

As Zabba points out, you may still want to reject empty items, prompting this solution:

>> options.split(/,|\sor\s/).map(&:strip).reject(&:empty?)
=> ["Cake", "pie", "ice cream", "pudding"]

Answer 3

As "or" and "," does the same thing, the best approach is to tell the regex that multiple cases should be treated the same as a single case:

options = "Cake or pie, ice cream, or pudding"
regex = /(?:\s*(?:,|or)\s*)+/
options.split(regex)