This is my code
stopwordlist = "a|an|all"
File.open('0_9.txt').each do |line|
line.downcase!
line.gsub!( /\b#{stopwordlist}\b/,'')
File.open('0_9_2.txt', 'w') { |f| f.write(line) }
end
I wanted to remove words - a,an and all
But, instead it matches substrings also and removes them
For an example input -
Bromwell High is a cartoon comedy. It ran at the same time as some other programs about school life
I get the output -
bromwell high is cartoon comedy. it r t the same time s some other programs bout school life
As you can see, it matched the substring.
How do I make it just match the word and not substrings ?
The | operator in regex takes the widest scope possible. Your original regex matches either \ba or an or all\b.
Change the whole regex to:
/\b(?:#{stopwordlist})\b/
or change stopwordlist into a regex instead of a string.
stopwordlist = /a|an|all/
Even better, you may want to use Regexp.union.
\ba\b|\ban\b|\ball\b
try this.this will look for word boundaries.
Related
I want to add markdown to key phrases in a (gollum) wiki page that will link to the relevant wiki page in the form:
This is the key phrase.
Becomes
This is the [[key phrase|Glossary#key phrase]].
I have a list of key phrases such as:
keywords = ["golden retriever", "pomeranian", "cat"]
And a document:
Sue has 1 golden retriever. John has two cats.
Jennifer has one pomeranian. Joe has three pomeranians.
I want to iterate over every line and find every match (that isn't already a link) for each keyword. My current attempt looks like this:
File.foreach(target_file) do |line|
glosses.each do |gloss|
len = gloss.length
# Create the regex. Avoid anything that starts with [
# or (, ends with ] or ), and ignore case.
re = /(?<![\[\(])#{gloss}(?![\]\)])/i
# Find every instance of this gloss on this line.
positions = line.enum_for(:scan, re).map {Regexp.last_match.begin(0) }
positions.each do |pos|
line.insert(pos, "[[")
# +2 because we just inserted 2 ahead.
line.insert(pos+len+2, "|#{page}\##{gloss}]]")
end
end
puts line
end
However, this will run into a problem if there are two matches for the same key phrase on the same line. Because I insert things into the line, the position I found for each match isn't accurate after the first one. I know I could adjust for the size of my insertions every time but, because my insertions are a different size for each gloss, it seems like the most brute-force, hacky solution.
Is there a solution that allows me to make multiple insertions on the same line at the same time without several arbitrary adjustments each time?
After looking at #BryceDrew's online python version, I realized ruby probably also has a way to fill in the match. I now have a much more concise and faster solution.
First, I needed to make regexes of my glosses:
glosses.push(/(?<![\[\(])#{gloss}(?![\]\)])/i)
Note: The majority of that regex is look-ahead and look-behind assertions to prevent catching a phrase that's already part of a link.
Then, I needed to make a union of all of them:
re = Regexp.union(glosses)
After that, it's as simple as doing gsub on every line, and filling in my matches:
File.foreach(target_file) do |line|
line = line.gsub(re) {|match| "[[#{match}|Glossary##{match.downcase}]]"}
puts line
end
Using Ruby 1.8.7
I need to grab everything up to a certain word - and I would like to match against words in an array. Example:
match_words = ['title','author','pages']
item = "Title: Jurassic Park\n"
item += "Author: Michael Crichton\n"
if item =~ /title: (.*)#{match any word in match_words array}/i
#do something here
end
So, this would ideally return "Jurassic Park\n". I am currently matching on newlines but have found that the data I will be matching against might have newlines in strange places, like the middle of the sentence. So, I think matching to the next match_word would be a good idea.
Is this possible, or maybe can be done another way?
Try this on for size
item.scan(/(title|author|pages):\s*?(.+)/i)
What this says is find all the results that start (case-insensitive) with either title, author or pages, are then followed by a colon and option white space and then characters. Capture the label and then the characters following the whitespace. The scan method will match as many times as it can.
Just iterate over the match words and do the regex compare as you normally would.
match_words.each do |word|
if item =~ /#{word}/ # Plus case sensitivity, start/end of item, etc.
# etc.
end
end
But if you know that the things you care about are at the beginning of the lines, then split the input string on \n and just use start_with instead of bothering with the regex--that partially depends on what the real data looks like.
First, create a | separated list of keywords from match_words.
Then, use string.scan to split the string apart, giving you an array of arrays with your results. See the end of this tutorial for a reference.
Here's my best shot:
keywords = match_words.join('|')
results = item.scan(/(#{keywords}):\s*(.+?)\s*(?= (#{keywords}):)/im)
Results: [["Title", "Jurassic Park"], ["Author", "Michael Crichton"]]
Don't forget to use the /m switch to indicate that you want . to match newlines.
To explain the pattern: we look for a keyword, then use a "look ahead" (?= ) to find the next keyword without capturing it. We capture all characters in between using a "lazy" expression .+?, so that we don't capture other keywords.
At the moment I have a regular expression that looks like this:
^(cat|dog|bird){1}(cat|dog|bird)?(cat|dog|bird)?$
It matches at least 1, and at most 3 instances of a long list of words and makes the matching words for each group available via the corresponding variable.
Is there a way to revise this so that I can return the result for each word in the string without specifying the number of groups beforehand?
^(cat|dog|bird)+$
works but only returns the last match separately , because there is only one group.
OK, so I found a solution to this.
It doesn't look like it is possible to create an unknown number of groups, so I went digging for another way of achieving the desired outcome: To be able to tell if a string was made up of words in a given list; and to match the longest words possible in each position.
I have been reading Mastering Regular Expressions by Jeffrey E. F. Friedl and it shed some light on things for me. It turns out that NFA based Regexp engines (like the one used in Ruby) are sequential as well as lazy/greedy. This means that you can dictate how a pattern is matched using the order in which you give it choices. This explains why scan was returning variable results, it was looking for the first word in the list that matched the criteria and then moved on to the next match. By design it was not looking for the longest match, but the first one. So in order to rectify this all I needed to do was reorder the array of words used to generate the regular expression from alphabetical order, to length order (longest to shortest).
array = %w[ as ascarid car id ]
list = array.sort_by {|word| -word.length }
regexp = Regexp.union(list)
Now the first match found by scan will be the longest word available. It is also pretty simple to tell if a string contains only words in the list using scan:
if "ascarid".scan(regexp).join.length == word.length
return true
else
return false
end
Thanks to everyone that posted in response to this question, I hope that this will help others in the future.
You could do it in two steps:
Use /^(cat|dog|bird)+$/ (or better /\A(cat|dog|bird)+\z/) to make sure it matches.
Then string.scan(/cat|dog|bird/) to get the pieces.
You could also use split and a Set to do both at once. Suppose you have your words in the array a and your string in s, then:
words = Set.new(a)
re = /(#{a.map{|w| Regexp.quote(w)}.join('|')})/
parts = s.split(re).reject(&:empty?)
if(parts.any? {|w| !words.include?(w) })
# 's' didn't match what you expected so throw a
# hissy fit, format the hard drive, set fire to
# the backups, or whatever is appropriate.
else
# Everything you were looking for is in 'parts'
# so you can check the length (if you care about
# how many matches there were) or something useful
# and productive.
end
When you use split with a pattern that contains groups then
the respective matches will be returned in the array as well.
In this case, the split will hand us something like ["", "cat", "", "dog"] and the empty strings will only occur between the separators that we're looking for and so we can reject them and pretend they don't exist. This may be an unexpected use of split since we're more interested in the delimiters more than what is being delimited (except to make sure that nothing is being delimited) but it gets the job done.
Based on your comments, it looks like you want an ordered alternation so that (ascarid|car|as|id) would try to match from left to right. I can't find anything in the Ruby Oniguruma (the Ruby 1.9 regex engine) docs that says that | is ordered or unordered; Perl's alternation appears to be specified (or at least strongly implied) to be ordered and Ruby's certainly behaves as though it is ordered:
>> 'pancakes' =~ /(pan|pancakes)/; puts $1
pan
So you could sort your words from longest to shortest when building your regex:
re = /(#{a.sort_by{|w| -w.length}.map{|w| Regexp.quote(w)}.join('|')})/
and hope that Oniguruma really will match alternations from left to right. AFAIK, Ruby's regexes will be eager because they support backreferences and lazy/non-greedy matching so this approach should be safe.
Or you could be properly paranoid and parse it in steps; first you'd make sure your string looks like what you want:
if(s !~ /\A(#{a.map{|w| Regexp.quote(w)}.join('|')})+\z/)
# Bail out and complain that 's' doesn't look right
end
The group your words by length:
by_length = a.group_by(&:length)
and scan for the groups from the longest words to the shortest words:
# This loses the order of the substrings within 's'...
matches = [ ]
by_length.keys.sort_by { |k| -k }.each do |group|
re = /(#{a.map{|w| Regexp.quote(w)}.join('|')})/
s.gsub!(re) { |w| matches.push(w); '' }
end
# 's' should now be empty and the matched substrings will be
# in 'matches'
There is still room for possible overlaps in these approaches but at least you'd be extracting the longest matches.
If you need to repeat parts of a regex, one option is to store the repeated part in a variable and just reference that, for example:
r = "(cat|dog|bird)"
str.match(/#{r}#{r}?#{r}?/)
You can do it with .Net regular expressions. If I write the following in PowerShell
$pat = [regex] "^(cat|dog|bird)+$"
$m = $pat.match('birddogcatbird')
$m.groups[1].captures | %{$_.value}
then I get
bird
dog
cat
bird
when I run it. I know even less about IronRuby than I do about PowerShell, but perhaps this means you can do it in IronRuby as well.
I'm trying to add conditional logic to determine if there's one regex match for a URL in a string. Here's an example of the string:
string_to_match = "http://www.twitpic.com/23456 ran to catch the bus, http://www.twitpic.com/3456 dodged a bullet at work."
I only want to match if I determine there's one URL in the string, so the above string wouldn't be a match in the case I'm trying to solve. I thought something like this would work:
if string_to_match =~ /[http\:\/\/]?/
puts "you're matching more then once. bad man!"
end
But it doesn't! How do I determine that there's only one match in a string?
The answer from Mladen is fine (counting the return from scan), but regular expressions already include the idea of matching the same thing multiple times or a particular number of times. In your case, you want to print the warning if your text occurs 2 or more times:
/(http:\/\/.+?){2,}/
Use .+ or .*, depending on whether you want to require the URL to have some content or not. As it stands, the .+? will match 1 or more characters in a non-greedy fashion, which is what you want. A greedy quantifier would gobble up the entire string on the first try and then have to do a bunch of backtracking before ultimately finding multiple URLs.
Take a look at String#scan, you can use it this way:
if string_to_match.scan(/[http\:\/\/]/).count > 1
puts "you're matching more then once. bad man!"
end
you could do it like this:
if string_to_match =~ /((http:\/\/.*?)http:\/\/)+/
this would match only if you have 2 (or more) occurrences of http://
I'm not sure how to use regular expressions in a function so that I could grab all the words in a sentence starting with a particular letter. I know that I can do:
word =~ /^#{letter}/
to check if the word starts with the letter, but how do I go from word to word. Do I need to convert the string to an array and then iterate through each word or is there a faster way using regex? I'm using ruby so that would look like:
matching_words = Array.new
sentance.split(" ").each do |word|
matching_words.push(word) if word =~ /^#{letter}/
end
Scan may be a good tool for this:
#!/usr/bin/ruby1.8
s = "I think Paris in the spring is a beautiful place"
p s.scan(/\b[it][[:alpha:]]*/i)
# => ["I", "think", "in", "the", "is"]
\b means 'word boundary."
[:alpha:] means upper or lowercase alpha (a-z).
You can use \b. It matches word boundaries--the invisible spot just before and after a word. (You can't see them, but oh they're there!) Here's the regex:
/\b(a\w*)\b/
The \w matches a word character, like letters and digits and stuff like that.
You can see me testing it here: http://rubular.com/regexes/13347
Similar to Anon.'s answer:
/\b(a\w*)/g
and then see all the results with (usually) $n, where n is the n-th hit. Many libraries will return /g results as arrays on the $n-th set of parenthesis, so in this case $1 would return an array of all the matching words. You'll want to double-check with whatever library you're using to figure out how it returns matches like this, there's a lot of variation on global search returns, sadly.
As to the \w vs [a-zA-Z], you can sometimes get faster execution by using the built-in definitions of things like that, as it can easily have an optimized path for the preset character classes.
The /g at the end makes it a "global" search, so it'll find more than one. It's still restricted by line in some languages / libraries, though, so if you wish to check an entire file you'll sometimes need /gm, to make it multi-line
If you want to remove results, like your title (but not question) suggests, try:
/\ba\w*//g
which does a search-and-replace in most languages (/<search>/<replacement>/). Sometimes you need a "s" at the front. Depends on the language / library. In Ruby's case, use:
string.gsub(/(\b)a\w*(\b)/, "\\1\\2")
to retain the non-word characters, and optionally put any replacement text between \1 and \2. gsub for global, sub for the first result.
/\ba[a-z]*\b/i
will match any word starting with 'a'.
The \b indicates a word boundary - we want to only match starting from the beginning of a word, after all.
Then there's the character we want our word to start with.
Then we have as many as possible letter characters, followed by another word boundary.
To match all words starting with t, use:
\bt\w+
That will match test but not footest; \b means "word boundary".
Personally i think that regex is overkill for this application, simply running a select is more than capable of solving this particular problem.
"this is a test".split(' ').select{ |word| word[0,1] == 't' }
result => ["this", "test"]
or if you are determined to use regex then go with grep
"this is a test".split(' ').grep(/^t/)
result => ["this", "test"]
Hope this helps.