In Ruby,
x = "this is a test".match(/(\w+) (\w+)/)
puts x[0], x[1], x[2]
why is the output
this is
this
is
Nothing special is going on here. You have the pattern
(\w+) (\w+)
namely two words separated by a space. That would be "this is" in your example (since we start looking for matches from the beginning of the string). The full match goes into the zeroth element of the return value, in your case x[0].
Now parentheses capture matches. The first left parenthesis starts at the first word, namely "this" so that value goes into x[1]. The second left parenthesis starts a group that matches the word "is", which will be captured into x[2].
Again, nothing special. This is how regular expression matching and grouping work in many, many languages.
Related
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"
Is there a better way to write the following regular expression in Ruby? The first regex matches a string that begins with a (lower case) consonant, the second with a vowel.
I'm trying to figure out if there's a way to write a regular expression that matches the negative of the second expression, versus writing the first expression with several ranges.
string =~ /\A[b-df-hj-np-tv-z]/
string =~ /\A[aeiou]/
The statement
$string =~ /\A[^aeiou]/
will test whether the string starts with a non-vowel character, which includes digits, punctuation, whitespace and control characters. That is fine if you know beforehand that the string begins with a letter, but to check that it starts with a consonant you can use forward look-ahead to test that it starts with both a letter and a non-vowel, like this
$string =~ /\A(?=[^aeiou])(?=[a-z])/i
To match an arbitrary number of consonants, you can use the sub-expression (?i:(?![aeiou])[a-z]) to match a consonant. It is atomic, so you can put a repetition count like {3} right after it. For example, this program finds all the strings in a list that contain three consonants in a row
list = %w/ aab bybt xeix axei AAsE SAEE eAAs xxsa Xxsr /
puts list.select { |word| word =~ /\A(?i:(?![aeiou])[a-z]){3}/ }
output
bybt
xxsa
Xxsr
I modified the answer provided by #Alexander Cherednichenko in order to get rid of the if statements.
/^[^aeiou\W]/i.match(s) != nil
If you want to catch a string that doesn't start with vowels, but only starts with consonants you can use this code below. It returns true if a string starts with any letter other than A, E, I, O, U. s is any string we give to a function
if /^[^aeiou\W]/i.match(s) == nil
return false
else
return true
end
i added at the end to make regular expression case insensitive.
\W is used to catch any non-word character, for example if a string starts with a digit like: "1something"
[^aeiou] means a range of character except a e i o u
And we put ^ at the beginning before [ to indicate that the following range [^aeiou\W] if for the 1st character
Note that ^[^aeiou\W] pattern is not correct because it also matches a line that starts with a digit, or underscore. Borodin's solution is working well, but there is one more possible solution without lookaheads, based on character class subtraction (more here) and using the more contemporary Regexp#match?:
/\A[a-z&&[^aeiou]]/i.match?(word)
See the Rubular demo.
Details
\A - start of a string (^ in Ruby is start of any line)
[a-z&&[^aeiou]] - an a-z character range matching any ASCII letter (/i flag makes it case insensitive) except for the aeiou chars.
See the Ruby demo:
test = %w/ 1word _word ball area programming /
puts test.select { |w| /\A[a-z&&[^aeiou]]/i.match?(w) }
# => ['ball', 'programming']
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
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.
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