How using regex would I take a string like "ratings-small star rating-4 field_stars_rating csm_review" and using gsub have it only return "rating-4", where 4 could be any digit? Anything I use replaces only partial bits
gsub is the wrong choice here. It would make much more sense to do something like this:
"ratings-small star rating-4 field_stars_rating csm_review".match(/\brating-\d\b/).to_s
Because you're looking for a specific part of the string, it makes more sense to search directly for that.
To just get the number after the hyphen, use this:
"ratings-small star rating-4 field_stars_rating csm_review".match(/\brating-(\d)\b/)[0]
Since you are trying to keep a bit of the string, instead of thinking how you can remove anything else to leave only the interesting bit, you should think how to extract the relevant part of the string. The String#[] method with a regexp argument would be my choice:
string = "ratings-small star rating-4 field_stars_rating csm_review"
string[/\brating-\d\b/]
# => "rating-4"
Instead of trying to replace everything up to the position of the word or after the position of the digit you want matched, a better approach would be to match that subpattern throughout your string.
string.match(/\b[a-z]+-\d+\b/i)
Explanation:
A word boundary does not consume any characters. It asserts that on one side there is a word character, and on the other side there is not.
\b # the boundary between a word char (\w) and not a word char
[a-z]+ # any character of: 'a' to 'z' (1 or more times)
- # '-'
\d+ # digits (0-9) (1 or more times)
\b # the boundary between a word char (\w) and not a word char
I wouldn't go with pure regex for this as it would make it pretty hard to read:
string = "ratings-small star rating-4 field_stars_rating csm_review"
string.split.select {|s| s =~ /^rating-\d$/}.join(' ')
If you expect only one element:
string[/\brating-\d\b/]
Related
I want to select all the commas in a string that do not have any white space around. Suppose I have this string:
"He,she, They"
I want to select only the comma between he and she. I tried this in rubular and came up with this regex:
(,[^(,\s)(\s,)])
This selects the comma that I want, but also selects an s which is a character after it.
In your regex (,[^(,\s)(\s,)]) you capture a comma followed by a negated character class that matches not any of the specified characters, which could also be written as (,[^)(,\s]) which will capture for example ,s in a group,
What you could do is use a positive lookahead and a positve lookbehind to check what is on the left and what is on the right is not a \S whitespace character:
(?<=\S),(?=\S)
Regex demo
In Ruby, you may use [[:space:]] to match any (Unicode) whitespace and [^[:space:]] to match any char other than whitespace. Using these character classes inside lookarounds solves the problem:
/(?<=[^[:space:]]),(?=[^[:space:]])/
See the Rubular demo
Here,
(?<=[^[:space:]]) - a positive lookbehind that matches a location that is immediately preceded with a non-whitespace char (if the string start position should also be matched, replace with (?<![[:space:]]))
, - a comma
(?=[^[:space:]]) - a positive lookahead that matches a location that is immediately followed with a non-whitespace char (if the string end position should also be matched, replace with (?![[:space:]])).
Check the regex below and use the code hope it will help you!
re = /[^\s](,)[^\s]/m
str = 'check ,my,domain, qwe,sd'
# Print the match result
str.scan(re) do |match|
puts match.to_s
end
Check LIVE DEMO HERE
I have a string with chars inside and I would like to match only the chars around a string.
"This is a [1]test[/1] string. And [2]test[/2]"
Rubular http://rubular.com/r/f2Xwe3zPzo
Currently, the code in the link matches the text inside the special chars, how can I change it?
Update
To clarify my question. It should only match if the opening and closing has the same number.
"[2]first[/2] [1]second[/2]"
In the code above, only first should match and not second. The text inside the special chars (first), should be ignored.
Try this:
(\[[0-9]\]).+?(\[\/[0-9]\])
Permalink to the example on Rubular.
Update
Since you want to remove the 'special' characters, try this instead:
foo = "This is a [1]test[/1] string. And [2]test[/2]"
foo.gsub /\[\/?\d\]/, ""
# => "This is a test string. And test"
Update, Part II
You only want to remove the 'special' characters when the surrounding tags match, so what about this:
foo = "This is a [1]test[/1] string. And [2]test[/2], but not [3]test[/2]"
foo.gsub /(?:\[(?<number>\d)\])(?<content>.+?)(?:\[\/\k<number>\])/, '\k<content>'
# => "This is a test string. And test, but not [3]test[/2]"
\[([0-9])\].+?\[\/\1\]
([0-9]) is a capture since it is surrounded with parentheses. The \1 tells it to use the result of that capture. If you had more than one capture, you could reference them as well, \2, \3, etc.
Rubular
You can also use a named capture, rather than \1 to make it a little less cryptic. As in: \[(?<number>[0-9])\].+?\[\/\k<number>\]
Here's a way to do it that uses the form of String#gsub that takes a block. The idea is to pull strings such as "[1]test[/1]" into the block, and there remove the unwanted bits.
str = "This is a [1]test[/1] string. And [2]test[/2], plus [3]test[/99]"
r = /
\[ # match a left bracket
(\d+) # capture one or more digits in capture group 1
\] # match a right bracket
.+? # match one or more characters lazily
\[\/ # match a left bracket and forward slash
\1 # match the contents of capture group 1
\] # match a right bracket
/x
str.gsub(r) { |s| s[/(?<=\]).*?(?=\[)/] }
#=> "This is a test string. And test, plus [3]test[/99]"
Aside: When I first heard of named capture groups, they seemed like a great idea, but now I wonder if they really make regexes easier to read than \1, \2....
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 unfortunately wandered into a situation where I need regex using Ruby. Basically I want to match this string after the underscore and before the first parentheses. So the end result would be 'table salt'.
_____ table salt (1) [F]
As usual I tried to fight this battle on my own and with rubular.com. I got the first part
^_____ (Match the beginning of the string with underscores ).
Then I got bolder,
^_____(.*?) ( Do the first part of the match, then give me any amount of words and letters after it )
Regex had had enough and put an end to that nonsense and crapped out. So I was wondering if anyone on stackoverflow knew or would have any hints on how to say my goal to the Ruby Regex parser.
EDIT: Thanks everyone, this is the pattern I ended up using after creating it with rubular.
ingredientNameRegex = /^_+([^(]*)/;
Everything got better once I took a deep breath, and thought about what I was trying to say.
str = "_____ table salt (1) [F]"
p str[ /_{3}\s(.+?)\s+\(/, 1 ]
#=> "table salt"
That says:
Find at least three underscores
and a whitespace character (\s)
and then one or more (+) of any character (.), but as little as possible (?), up until you find
one or more whitespace characters,
and then a literal (
The parens in the middle save that bit, and the 1 pulls it out.
Try this: ^[_]+([^(]*)\(
It will match lines starting with one or more underscores followed by anything not equal to an opening bracket: http://rubular.com/r/vthpGpVr4y
Here's working regex:
str = "_____ table salt (1) [F]"
match = str.match(/_([^_]+?)\(/)
p match[1].strip # => "table salt"
You could use
^_____\s*([^(]+?)\s*\(
^_____ match the underscore from the beginning of string
\s* matches any whitespace character
( grouping start
[^(]+ matches all non ( character at least once
? matches the shortest possible string (non greedy)
) grouping end
\s* matches any whitespace character
\( find the (
"_____ table salt (1) [F]".gsub(/[_]\s(.+)\s\(/, ' >>>\1<<< ')
# => "____ >>>table salt<<< 1) [F]"
It seems to me the simplest regex to do what you want is:
/^_____ ([\w\s]+) /
That says:
leading underscores, space, then capture any combination of word chars or spaces, then another space.
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
Note that [^\d]+ means at least one non digit character.
On second thought, here's a more generic solution that doesn't need to deal with regular expressions:
str.gsub(/[^\d.-]+/, " ").split.collect{|d| d.to_f}
Example:
str = "blabla9999 some more text -1.1"
Parsed:
[9999.0, -1.1]
The parenthesis have different meanings.
[] defines a character class, that means one character is matched that is part of this class
() is defining a capturing group, the string that is matched by this part in brackets is put into a variable.
You did not define any anchors so your pattern will match your second string
blabla9999 some more text 1.1
^^^^ here ^^^ and here
Maybe this is more what you wanted
^(\s*-?\d+(?:\.\d+)?\s*)+$
See it here on Regexr
^ anchors the pattern to the start of the string and $ to the end.
it allows Whitespace \s before and after the number and an optional fraction part (?:\.\d+)? This kind of pattern will be matched at least once.
maybe /(-?\d+(.\d+)?)+/
irb(main):010:0> "910 -6.258000 6.290".scan(/(\-?\d+(\.\d+)?)+/).map{|x| x[0]}
=> ["910", "-6.258000", "6.290"]
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map(&:to_f)
# => [910.0, -6.258, 6.29]
If you don't want integers to be converted to floats, try this:
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map do |ns|
ns[/\./] ? ns.to_f : ns.to_i
end
# => [910, -6.258, 6.29]