Is it possible to make these regex shorter? [closed] - ruby

Question

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.
Closed 10 years ago.
As the topic suggest, is it possible to make these regex shorter? I am are using Ruby 1.9.3
/\n\s+(\w{0,3})[\s&&[^\n]\S]+?([\d\.]+)[\S\s&&[^\n]]+?([\d\.]+)/
and this
/\s+(\w+)\s+(\d+)\s+(\d+)\s+(\d+)\s+(\d+)\s+(\d+)\s+-*\s+(\d+)\s+(\d+)\s+/
Thanks!

Answer 1

/\n\s+(\w{0,3})[\s&&[^\n]\S]+?([\d\.]+)[\S\s&&[^\n]]+?([\d\.]+)/
If I understand ruby regular expressions correctly, [\s&&[^\n]\S] means that a character should be a whitespace character AND either a non-whitespace character or not a newline. As a character cannot be both a whitespace and non-whitespace character, you could probably shorten it to [\s&&[^\n]].
You could also remove the parentheses, (\w{0,3}) becomes \w{0,3}, but if you are trying to use the characters in those groups later on in your code, then you shouldn't.
/\s+(\w+)\s+(\d+)\s+(\d+)\s+(\d+)\s+(\d+)\s+(\d+)\s+-*\s+(\d+)\s+(\d+)\s+/
You could combine some of your statements, \s+\w+(\s+\d+){5}\s+-*(\s+\d+){2}\s+, but again this would cause headaches if your code actually uses those groups to extract information.

Answer 2

Are you essentially aiming to split a fixed-width-column web page?
Regexp is one way. You may be interested in a fixed-width-column approach:
uri = URI.parse 'http://www.ida.liu.se/~TDP007/material/seminarie2/weather.txt'
page = uri.read
rows = page.split(/\n/)[9..-3]
rows.each{|r|
day, max, mnt = r[0..3].strip, r[4..11].strip, r[12..17].strip
}

Answer 3

The following might not be shorter (if you count the number of characters needed to type it), but it is a lot more readable:
arr = ['(\w+)'] # Match a word
arr += ['(\d+)']*5 # Match five numbers
arr += ['-*'] # ignore dashes
arr += ['(\d+)']*2 # Match two numbers
# All of the above separated with space, plus space before and after.
my_regexp = Regexp.new(([''] + arr + ['']).join('\s+'))

Answer 4

If that is the only file that you need to process then you can remove unnecessary data by hand, then read the file line by line, split by space characters \s+ and pick out the columns.
Even without removing unnecessary data by hand, you can also read the original file line by line, split by \s+, and test whether the first few entries are numbers. This is exactly what you are doing with regex also (test format and extract data matching the format).
Note that [\s&&[^\n]\S] means intersecting \s and [^\n]\S, which results in the set: all space characters but new line. So we can rewrite it to [\s&&[^\n]]. However, [\S\s&&[^\n]] means intersecting \S\s and [^\n], which results in the set: all characters but new line. The equivalent rewrite is . or [^\n], but I doubt this is what you mean. The result will still be correct for the current input due to the lazy quantifier, but it might not for bad input.
Another thing is . will mean literal . inside character class, so [\d.] is equivalent to [\d\.].