ruby parametrized regular expression - ruby

I have a string like "{some|words|are|here}" or "{another|set|of|words}"
So in general the string consists of an opening curly bracket,words delimited by a pipe and a closing curly bracket.
What is the most efficient way to get the selected word of that string ?
I would like do something like this:
#my_string = "{this|is|a|test|case}"
#my_string.get_column(0) # => "this"
#my_string.get_column(2) # => "is"
#my_string.get_column(4) # => "case"
What should the method get_column contain ?

So this is the solution I like right now:
class String
def get_column(n)
self =~ /\A\{(?:\w*\|){#{n}}(\w*)(?:\|\w*)*\}\Z/ && $1
end
end
We use a regular expression to make sure that the string is of the correct format, while simultaneously grabbing the correct column.
Explanation of regex:
\A is the beginnning of the string and \Z is the end, so this regex matches the enitre string.
Since curly braces have a special meaning we escape them as \{ and \} to match the curly braces at the beginning and end of the string.
next, we want to skip the first n columns - we don't care about them.
A previous column is some number of letters followed by a vertical bar, so we use the standard \w to match a word-like character (includes numbers and underscore, but why not) and * to match any number of them. Vertical bar has a special meaning, so we have to escape it as \|. Since we want to group this, we enclose it all inside non-capturing parens (?:\w*\|) (the ?: makes it non-capturing).
Now we have n of the previous columns, so we tell the regex to match the column pattern n times using the count regex - just put a number in curly braces after a pattern. We use standard string substition, so we just put in {#{n}} to mean "match the previous pattern exactly n times.
the first non skipped column after that is the one we care about, so we put that in capturing parens: (\w*)
then we skip the rest of the columns, if any exist: (?:\|\w*)*.
Capturing the column puts it into $1, so we return that value if the regex matched. If not, we return nil, since this String has no nth column.
In general, if you wanted to have more than just words in your columns (like "{a phrase or two|don't forget about punctuation!|maybe some longer strings that have\na newline or two?}"), then just replace all the \w in the regex with [^|{}] so you can have each column contain anything except a curly-brace or a vertical bar.
Here's my previous solution
class String
def get_column(n)
raise "not a column string" unless self =~ /\A\{\w*(?:\|\w*)*\}\Z/
self[1 .. -2].split('|')[n]
end
end
We use a similar regex to make sure the String contains a set of columns or raise an error. Then we strip the curly braces from the front and back (using self[1 .. -2] to limit to the substring starting at the first character and ending at the next to last), split the columns using the pipe character (using .split('|') to create an array of columns), and then find the n'th column (using standard Array lookup with [n]).
I just figured as long as I was using the regex to verify the string, I might as well use it to capture the column.

Related

Splitting the content of brackets without separating the brackets ruby

I am currently working on a ruby program to calculate terms. It works perfectly fine except for one thing: brackets. I need to filter the content or at least, to put the content into an array, but I have tried for an hour to come up with a solution. Here is my code:
splitted = term.split(/\(+|\)+/)
I need an array instead of the brackets, for example:
"1-(2+3)" #=>["1", "-", ["2", "+", "3"]]
I already tried this:
/(\((?<=.*)\))/
but it returned:
Invalid pattern in look-behind.
Can someone help me with this?
UPDATE
I forgot to mention, that my program will split the term, I only need the content of the brackets to be an array.
If you need to keep track of the hierarchy of parentheses with arrays, you won't manage it just with regular expressions. You'll need to parse the string word by word, and keep a stack of expressions.
Pseudocode:
Expressions = new stack
Add new array on stack
while word in string:
if word is "(": Add new array on stack
Else if word is ")": Remove the last array from the stack and add it to the (next) last array of the stack
Else: Add the word to the last array of the stack
When exiting the loop, there should be only one array in the stack (if not, you have inconsistent opening/closing parentheses).
Note: If your ultimate goal is to evaluate the expression, you could save time and parse the string in Postfix aka Reverse-Polish Notation.
Also consider using off-the-shelf libraries.
A solution depends on the pattern you expect between the parentheses, which you have not specified. (For example, for "(st12uv)" you might want ["st", "12", "uv"], ["st12", "uv"], ["st1", "2uv"] and so on). If, as in your example, it is a natural number followed by a +, followed by another natural number, you could do this:
str = "1-( 2+ 3)"
r = /
\(\s* # match a left parenthesis followed by >= 0 whitespace chars
(\d+) # match one or more digits in a capture group
\s* # match >= 0 whitespace chars
(\+) # match a plus sign in a capture group
\s* # match >= 0 whitespace chars
(\d+) # match one or more digits in a capture group
\s* # match >= 0 whitespace chars
\) # match a right parenthesis
/x
str.scan(r0).first
=> ["2", "+", "3"]
Suppose instead + could be +, -, * or /. Then you could change:
(\+)
to:
([-+*\/])
Note that, in a character class, + needn't be escaped and - needn't be escaped if it is the first or last character of the class (as in those cases it would not signify a range).
Incidentally, you received the error message, "Invalid pattern in look-behind" because Ruby's lookarounds cannot contain variable-length matches (i.e., .*). With positive lookbehinds you can get around that by using \K instead. For example,
r = /
\d+ # match one or more digits
\K # forget everything previously matched
[a-z]+ # match one or more lowercase letters
/x
"123abc"[r] #=> "abc"

String gsub - Replace characters between two elements, but leave surrounding elements

Suppose I have the following string:
mystring = "start/abc123/end"
How can you splice out the abc123 with something else, while leaving the "/start/" and "/end" elements intact?
I had the following to match for the pattern, but it replaces the entire string. I was hoping to just have it replace the abc123 with 123abc.
mystring.gsub(/start\/(.*)\/end/,"123abc") #=> "123abc"
Edit: The characters between the start & end elements can be any combination of alphanumeric characters, I changed my example to reflect this.
You can do it using this character class : [^\/] (all that is not a slash) and lookarounds
mystring.gsub(/(?<=start\/)[^\/]+(?=\/end)/,"7")
For your example, you could perhaps use:
mystring.gsub(/\/(.*?)\//,"/7/")
This will match the two slashes between the string you're replacing and putting them back in the substitution.
Alternatively, you could capture the pieces of the string you want to keep and interpolate them around your replacement, this turns out to be much more readable than lookaheads/lookbehinds:
irb(main):010:0> mystring.gsub(/(start)\/.*\/(end)/, "\\1/7/\\2")
=> "start/7/end"
\\1 and \\2 here refer to the numbered captures inside of your regular expression.
The problem is that you're replacing the entire matched string, "start/8/end", with "7". You need to include the matched characters you want to persist:
mystring.gsub(/start\/(.*)\/end/, "start/7/end")
Alternatively, just match the digits:
mystring.gsub(/\d+/, "7")
You can do this by grouping the start and end elements in the regular expression and then referring to these groups in in the substitution string:
mystring.gsub(/(?<start>start\/).*(?<end>\/end)/, "\\<start>7\\<end>")

Ruby Regex gsub - everything after string

I have a string something like:
test:awesome my search term with spaces
And I'd like to extract the string immediately after test: into one variable and everything else into another, so I'd end up with awesome in one variable and my search term with spaces in another.
Logically, what I'd so is move everything matching test:* into another variable, and then remove everything before the first :, leaving me with what I wanted.
At the moment I'm using /test:(.*)([\s]+)/ to match the first part, but I can't seem to get the second part correctly.
The first capture in your regular expression is greedy, and matches spaces because you used .. Instead try:
matches = string.match(/test:(\S*) (.*)/)
# index 0 is the whole pattern that was matched
first = matches[1] # this is the first () group
second = matches[2] # and the second () group
Use the following:
/^test:(.*?) (.*)$/
That is, match "test:", then a series of characters (non-greedily), up to a single space, and another series of characters to the end of the line.
I am guessing you want to remove all the leading spaces before the second match too, hence I have \s+ in the expression. Otherwise, remove the \s+ from the expression, and you'll have what you want:
m = /^test:(\w+)\s+(.*)/.match("test:awesome my search term with spaces")
a = m[1]
b = m[2]
http://codepad.org/JzuNQxBN

Regex: Substring the second last value between two slashes of a url string

I have a string like this:
http://www.example.com/value/1234/different-value
How can I extract the 1234?
Note: There may be a slash at the end:
http://www.example.com/value/1234/different-value
http://www.example.com/value/1234/different-value/
/([^/]+)(?=/[^/]+/?$)
should work. You might need to format it differently according to the language you're using. For example, in Ruby, it's
if subject =~ /\/([^\/]+)(?=\/[^\/]+\/?\Z)/
match = $~[1]
else
match = ""
end
Use Slice for Positional Extraction
If you always want to extract the 4th element (including the scheme) from a URI, and are confident that your data is regular, you can use Array#slice as follows.
'http://www.example.com/value/1234/different-value'.split('/').slice 4
#=> "1234"
'http://www.example.com/value/1234/different-value/'.split('/').slice 4
#=> "1234"
This will work reliably whether there's a trailing slash or not, whether or not you have more than 4 elements after the split, and whether or not that fourth element is always strictly numeric. It works because it's based on the element's position within the path, rather than on the contents of the element. However, you will end up with nil if you attempt to parse a URI with fewer elements such as http://www.example.com/1234/.
Use Scan/Match for Pattern Extraction
Alternatively, if you know that the element you're looking for is always the only one composed entirely of digits, you can use String#match with look-arounds to extract just the numeric portion of the string.
'http://www.example.com/value/1234/different-value'.match %r{(?<=/)\d+(?=/)}
#=> #<MatchData "1234">
$&
#=> "1234"
The look-behind and look-ahead assertions are needed to anchor the expression to a path. Without them, you'll match things like w3.example.com too. This solution is a better approach if the position of the target element may change, and if you can guarantee that your element of interest will be the only one that matches the anchored regex.
If there will be more than one match (e.g. http://www.example.com/1234/5678/) then you might want to use String#scan instead to select the first or last match. This is one of those "know your data" things; if you have irregular data, then regular expressions aren't always the best choice.
Javascript:
var myregexp = /:\/\/.*?\/.*?\/(\d+)/;
var match = myregexp.exec(subject);
if (match != null) {
result = match[1];
}
Works with your examples... But I am sure it will fail in general...
Ruby edit:
if subject =~ /:\/\/.*?\/.*?\/(.+?)\//
match = $~[1]
It does work.
I think this is a little simpler than the accepted answer, because it doesn't use any positive lookahead (?=), but rather simply makes the last slash optional via the ? character:
^.+\/(.+)\/.+\/?$
In Ruby:
STDIN.read.split("\n").each do |nextline|
if nextline =~ /^.+\/(.+)\/.+\/?$/
printf("matched %s in %s\n", $~[1], nextline);
else
puts "no match"
end
end
Live Demo
Let's break down what's happening:
^: start of the line
.+\/: match anything (greedily) up to a slash
Since we're going to later match at least 1, at most 2 more slashes, this slash will be either the second last slash (as in http://www.example.com/value/1234/different-value) or the third last slash as in (http://www.example.com/value/1234/different-value/)
Up to this point we've matched http://www.example.com/value/ (due to greediness)
(.+)\/: Our capturing group for 1234 indicated by the parenthesis. It's anything followed by another slash.
Since the previous match matched up to the second or third last slash, this will match up to the last slash or second last slash, respectively
.+: match anything. This would be after our 1234, so we're assuming there are characters after 1234/ (different-value)
\/?: optionally match another slash (the slash after different-value)
$: match the end of the line
Note that in a url, you probably won't have spaces. I used the . character because it's easily distinguished, but perhaps you might use \S instead to match non-spaces.
Also, you might use \A instead of ^ to match start of string (instead of after line break) and \Z instead of $ to match end of string (instead of at line break)

How to remove the first 4 characters from a string if it matches a pattern in Ruby

I have the following string:
"h3. My Title Goes Here"
I basically want to remove the first four characters from the string so that I just get back:
"My Title Goes Here".
The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first four characters blindly.
I checked the docs and the closest thing I could find was chomp, but that only works for the end of a string.
Right now I am doing this:
"h3. My Title Goes Here".reverse.chomp(" .3h").reverse
This gives me my desired output, but there has to be a better way. I don't want to reverse a string twice for no reason. Is there another method that will work?
To alter the original string, use sub!, e.g.:
my_strings = [ "h3. My Title Goes Here", "No h3. at the start of this line" ]
my_strings.each { |s| s.sub!(/^h3\. /, '') }
To not alter the original and only return the result, remove the exclamation point, i.e. use sub. In the general case you may have regular expressions that you can and want to match more than one instance of, in that case use gsub! and gsub—without the g only the first match is replaced (as you want here, and in any case the ^ can only match once to the start of the string).
You can use sub with a regular expression:
s = 'h3. foo'
s.sub!(/^h[0-9]+\. /, '')
puts s
Output:
foo
The regular expression should be understood as follows:
^ Match from the start of the string.
h A literal "h".
[0-9] A digit from 0-9.
+ One or more of the previous (i.e. one or more digits)
\. A literal period.
A space (yes, spaces are significant by default in regular expressions!)
You can modify the regular expression to suit your needs. See a regular expression tutorial or syntax guide, for example here.
A standard approach would be to use regular expressions:
"h3. My Title Goes Here".gsub /^h3\. /, '' #=> "My Title Goes Here"
gsub means globally substitute and it replaces a pattern by a string, in this case an empty string.
The regular expression is enclosed in / and constitutes of:
^ means beginning of the string
h3 is matched literally, so it means h3
\. - a dot normally means any character so we escape it with a backslash
is matched literally

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