I'm a bit confused about answer to the following question from RubyMonk.
Let's find the second character in the string 'RubyMonk Is Pretty
Brilliant' preceded by a space, which should be 'P'
'RubyMonk Is Pretty Brilliant'.match(/ ./, 9)
Why would I place a '9' in the argument?
I would really appreciate an explanation.
The Monk clearly says
When the second parameter is present, it specifies the position
in the string to begin the search.
As mentioned in the comment, the code is 'cheating' by matching a space followed by a character starting after the 9th character. It is—frankly—a terrible example of how to do what it claims to be doing, since you cannot do this generically.
If you really wanted to find the second character that is preceded by a space, and you didn't cheat and look for yourself where it might be, you could do one of the following:
str = 'RubyMonk Is Pretty Brilliant'
Find a space, followed by a non-space, followed by one or more non-space, followed by a space, followed by a character. Capture that character:
str[/ [^ ]+ (.)/,1]
#=> "P"
Find all the characters that are preceded by a space, and then find the second one:
str.scan(/(?<= )./)[1]
#=> "P"
Split on spaces (only keeping the first three chunks, for efficiency), and then find the third match, then find the first character:
str.split(' ',3)[2][0]
#=> "P"
Related
I happened to search around everywhere and did not managed to find a solution to count number of sentence in a String using Ruby. Does anyone how to do it?
Example
string = "The best things in an artist’s work are so much a matter of intuition, that there is much to be said for the point of view that would altogether discourage intellectual inquiry into artistic phenomena on the part of the artist. Intuitions are shy things and apt to disappear if looked into too closely. And there is undoubtedly a danger that too much knowledge and training may supplant the natural intuitive feeling of a student, leaving only a cold knowledge of the means of expression in its place. For the artist, if he has the right stuff in him ... "
This string should return number 4.
You can split the text into sentences and count them. Here:
string.scan(/[^\.!?]+[\.!?]/).map(&:strip).count # scan has regex to split string and strip will remove trailing spaces.
# => 4
Explaining regex:
[^\.!?]
Caret inside of a character class [^ ] is the negation operator. Which means we are looking for characters which are not present in list: ., ! and ?.
+
is a greedy operator that returns matches between 1 and unlimited times. (capturing our sentences here and ignoring repetitions like ...)
[\.!?]
matching characters ., ! or ?.
In a nutshell, we are capturing all characters that are not ., ! or ? till we get characters that are ., ! or ?. Which basically can be treated as a sentence (in broad senses).
I think it makes sense to consider a word char followed by a ?! or . the delimiter of a sentence:
string.strip.split(/\w[?!.]/).length
#=> 4
So I'm not considering the ... a delimiter when it hangs on it's own like that:
"I waited a while ... and then I went home"
But then again, maybe I should...
It also occurs to me that maybe a better delimiter is a punctuation followed by some space and a capital letter:
string.split(/[?!.]\s+[A-Z]/).length
#=> 4
Sentences end with full stops, question marks, and exclamation marks. They can also be
separated with dashes and other punctuation, but we won’t worry about these rare cases here.
The split is simple. Instead of asking Ruby to split the text on one type of character, you simply
ask it to split on any of three types of characters, like so:
txt = "The best things in an artist’s work are so much a matter of intuition, that there is much to be said for the point of view that would altogether discourage intellectual inquiry into artistic phenomena on the part of the artist. Intuitions are shy things and apt to disappear if looked into too closely. And there is undoubtedly a danger that too much knowledge and training may supplant the natural intuitive feeling of a student, leaving only a cold knowledge of the means of expression in its place. For the artist, if he has the right stuff in him ... "
sentence_count = txt.split(/\.|\?|!/).length
puts sentence_count
#=> 7
string.squeeze('.!?').count('.!?')
#=> 4
I found something odd when using a reluctant quantifier in a negative look ahead.
When creating a regex to assert a maximum of 3 uppercase characters, I devised this:
^(?!(.*?[A-Z]){4}).*$
which works on rubular, but not on regex101.
Why is that?
^, $ matches beginning/end of line in Ruby.
While in another languages, ^, $ matches the beginning/end of the string unless multiline mode (m) is specified. (Some regular expression engine requires g flag to match multiple times.)
I just learned from a book about regular expressions in the Ruby language. I did Google it, but still got confused about {x} and {x,y}.
The book says:
{x}→Match x occurrences of the preceding character.
{x,y}→Match at least x occurrences and at most y occurrences.
Can anyone explain this better, or provide some examples?
Sure, look at these examples:
http://rubular.com/r/sARHv0vf72
http://rubular.com/r/730Zo6rIls
/a{4}/
is the short version for:
/aaaa/
It says: Match exact 4 (consecutive) characters of 'a'.
where
/a{2,4}/
says: Match at least 2, and at most 4 characters of 'a'.
it will match
/aa/
/aaa/
/aaaa/
and it won't match
/a/
/aaaaa/
/xxx/
Limiting Repetition good online tutorial for this.
I highly recommend regexbuddy.com and very briefly, the regex below does what you refer to:
[0-9]{3}|\w{3}
The [ ] characters indicate that you must match a number between 0 and 9. It can be anything, but the [ ] is literal match. The { } with a 3 inside means match sets of 3 numbers between 0 and 9. The | is an or statement. The \w, is short hand for any word character and once again the {3} returns only sets of 3.
If you go to RegexPal.com you can enter the code above and test it. I used the following data to test the expression:
909 steve kinzey
and the expression matched the 909, the 'ste', the 'kin' and the 'zey'. It did not match the 've' because it is only 2 word characters long and a word character does not span white space so it could not carry over to the second word.
Interval Expressions
GNU awk refers to these as "interval expressions" in the Regexp Operators section of its manual. It explains the expressions as follows:
{n}
{n,}
{n,m}
One or two numbers inside braces denote an interval expression. If there is one number in the braces, the preceding regexp is repeated n times. If there are two numbers separated by a comma, the preceding regexp is repeated n to m times. If there is one number followed by a comma, then the preceding regexp is repeated at least n times:
The manual also includes these reference examples:
wh{3}y
Matches ‘whhhy’, but not ‘why’ or ‘whhhhy’.
wh{3,5}y
Matches ‘whhhy’, ‘whhhhy’, or ‘whhhhhy’, only.
wh{2,}y
Matches ‘whhy’ or ‘whhhy’, and so on.
See Also
Ruby's Regexp class.
Quantifiers section of Ruby's oniguruma engine.
I want to match single letters in a sentence. So in ...
I want to have my turkey. May I. I 20,000-t bar-b-q
I'd like to match
*I* want to have my turkey. May *I*. *I* 20,000-t bar-b-q
right now I'm using
/\b\w\b/
as my regular expression, but that is matching
*I* want to have my turkey. May *I*. *I* 20,000-*t* bar-*b*-*q*
Any suggestions on how to get past that last mile?
Use a negative lookbehind and negative lookahead to fail if the previous character is a word or a hyphen, or if the next character is a word a or a hyphen:
/(?<![\w\-])\w(?![\w\-])/
Example: http://www.rubular.com/r/9upmgfG9u4
Note that as mentioned by rtcherry, this will also match single numbers. To prevent this you may want to change the \w that is outside of the character classes to [a-zA-Z].
F.J's answer will also include numbers. This is restricted to ASCII characters, but you really need to define what characters can be side by side an still count as a single letter.
/(?<![0-9a-zA-Z\-])[a-zA-Z](?![0-9a-zA-Z\-])/
That will also avoid things like This -> 1a <- is not a single letter. Neither is -> 2 <- that.
As long as we're being picky, non-ASCII letters are easy to include:
/(?<![[:alnum:]-])[[:alpha:]](?![[:alnum:]-])/
This will avoid matching the t in 'Cómo eres tú'
Notice that it's not necessary to escape the - when it is the last character in a character class (which I'm not sure that this technically is).
You are asking far too much of a regular expression. \w matches a word character, which includes upper and lower case alphabetics, the ten digits, and underscore. So it is the same as [0-9A-Z_a-z].
\b matches the (zero-width) boundary where a word character doesn't have another word character next to it, for instance at the beginning or end of a string, or next to some punctuation or white space.
Using negative look-behinds and look-aheads, this amounts to \b\w\b being equivalent to
(?<!\w)\w(?!\w)
i.e. a word character that doesn't have another word character before or after it.
As you have found, that finds t, b and q in 20,000-t bar-b-q. So it's back in your court to define what you really mean by "single letters in a sentence".
It nearly works to say "any letter that isn't preceded or followed by a printable character, which is
/(?<!\S)[A-Za-z](?!\S)/
But that leaves out I in May I. because it has a dot after it.
So, do you mean a single letter that isn't preceded by a printable character, and is followed by whitespace, a dot, or the end of the string (or a comma, a semicolon or a colon for good measure)? Then you want
/(?<!\S)[A-Za-z](?=(?:[\s.,;:]|\z))/
which finds exactly three I characters in your string.
I hope that helps.
I have gone through the docs for Atomic Grouping and rubyinfo and some questions came into my mind:
Why the name "Atomic grouping"? What "atomicity" does it have that general grouping doesn't?
How does atomic grouping differ to general grouping?
Why are atomic groups called non-capturing groups?
I tried the below code to understand but had confusion about the output and how differently they work on the same string as well?
irb(main):001:0> /a(?>bc|b)c/ =~ "abbcdabcc"
=> 5
irb(main):004:0> $~
=> #<MatchData "abcc">
irb(main):005:0> /a(bc|b)c/ =~ "abcdabcc"
=> 0
irb(main):006:0> $~
=> #<MatchData "abc" 1:"b">
A () has some properties (include those such as (?!pattern), (?=pattern), etc. and the plain (pattern)), but the common property between all of them is grouping, which makes the arbitrary pattern a single unit (unit is my own terminology), which is useful in repetition.
The normal capturing (pattern) has the property of capturing and group. Capturing means that the text matches the pattern inside will be captured so that you can use it with back-reference, in matching or replacement. The non-capturing group (?:pattern) doesn't have the capturing property, so it will save a bit of space and speed up a bit compared to (pattern) since it doesn't store the start and end index of the string matching the pattern inside.
Atomic grouping (?>pattern) also has the non-capturing property, so the position of the text matched inside will not be captured.
Atomic grouping adds property of atomic compared to capturing or non-capturing group. Atomic here means: at the current position, find the first sequence (first is defined by how the engine matches according to the pattern given) that matches the pattern inside atomic grouping and hold on to it (so backtracking is disallowed).
A group without atomicity will allow backtracking - it will still find the first sequence, then if the matching ahead fails, it will backtrack and find the next sequence, until a match for the entire regex expression is found or all possibilities are exhausted.
Example
Input string: bbabbbabbbbc
Pattern: /(?>.*)c/
The first match by .* is bbabbbabbbbc due to the greedy quantifier *. It will hold on to this match, disallowing c from matching. The matcher will retry at the next position to the end of the string, and the same thing happens. So nothing matches the regex at all.
Input string: bbabbbabbbbc
Pattern: /((?>.*)|b*)[ac]/, for testing /(((?>.*))|(b*))[ac]/
There are 3 matches to this regex, which are bba, bbba, bbbbc. If you use the 2nd regex, which is the same but with capturing groups added for debugging purpose, you can see that all the matches are result of matching b* inside.
You can see the backtracking behavior here.
Without the atomic grouping /(.*|b*)[ac]/, the string will have a single match which is the whole string, due to backtracking at the end to match [ac]. Note that the engine will go back to .* to backtrack by 1 character since it still have other possibilities.
Pattern: /(.*|b*)[ac]/
bbabbbabbbbc
^ -- Start matching. Look at first item in alternation: .*
bbabbbabbbbc
^ -- First match of .*, due to greedy quantifier
bbabbbabbbbc
X -- [ac] cannot match
-- Backtrack to ()
bbabbbabbbbc
^ -- Continue explore other possibility with .*
-- Step back 1 character
bbabbbabbbbc
^ -- [ac] matches, end of regex, a match is found
With the atomic grouping, all possibilities of .* is cut off and limited to the first match. So after greedily eating the whole string and fail to match, the engine have to go for the b* pattern, where it successfully finds a match to the regex.
Pattern: /((?>.*)|b*)[ac]/
bbabbbabbbbc
^ -- Start matching. Look at first item in alternation: (?>.*)
bbabbbabbbbc
^ -- First match of .*, due to greedy quantifier
-- The atomic grouping will disallow .* to be backtracked and rematched
bbabbbabbbbc
X -- [ac] cannot match
-- Backtrack to ()
-- (?>.*) is atomic, check the next possibility by alternation: b*
bbabbbabbbbc
^ -- Starting to rematch with b*
bbabbbabbbbc
^ -- First match with b*, due to greedy quantifier
bbabbbabbbbc
^ -- [ac] matches, end of regex, a match is found
The subsequent matches will continue on from here.
I recently had to explain Atomic Groups to someone else and I thought I'd tweak and share the example here.
Consider /the (big|small|biggest) (cat|dog|bird)/
Matches in bold
the big dog
the small bird
the biggest dog
the small cat
DEMO
For the first line, a regex engine would find the .
It would then proceed on to our adjectives (big, small, biggest), it finds big.
Having matched big, it proceeds and finds the space.
It then looks at our pets (cat, dog, bird), finds cat, skips it, and finds dog.
For the second line, our regex would find the .
It would proceed and look at big, skip it, look at and find small.
It finds the space, skips cat and dog because they don't match, and finds bird.
For the third line, our regex would find the ,
It continues on and finds big which matches the immediate requirement, and proceeds.
It can't find the space, so it backtracks (rewinds the position to the last choice it made).
It skips big, skips small, and finds biggest which also matches the immediate requirement.
It then finds the space.
It skips cat , and matches dog.
For the fourth line, our regex would find the .
It would proceed to look at big, skip it, look at and find small.
It then finds the space.
It looks at and matches cat.
Consider /the (?>big|small|biggest) (cat|dog|bird)/
Note the ?> atomic group on adjectives.
Matches in bold
the big dog
the small bird
the biggest dog
the small cat
DEMO
For the first line, second line, and fourth line, we'll get the same result.
For the third line, our regex would find the ,
It continues on and find big which matches the immediate requirement, and proceeds.
It can't find the space, but the atomic group, being the last choice the engine made, won't allow that choice to be re-examined (prohibits backtracking).
Since it can't make a new choice, the match has to fail, since our simple expression has no other choices.
This is only a basic summary. An engine wouldn't need to look at the entirety of cat to know that it doesn't match dog, merely looking at the c is enough. When trying to match bird, the c in cat and the d in dog are enough to tell the engine to examine other options.
However if you had ...((cat|snake)|dog|bird), the engine would also, of course, need to examine snake before it dropped to the previous group and examined dog and bird.
There are also plenty of choices an engine can't decide without going past what may not seem like a match, which is what results in backtracking. If you have ((red)?cat|dog|bird), The engine will look at r, back out, notice the ? quantifier, ignore the subgroup (red), and look for a match.
An "atomic group" is one where the regular expression will never backtrack past. So in your first example /a(?>bc|b)c/ if the bc alternation in the group matches, then it will never backtrack out of that and try the b alternation. If you slightly alter your first example to match against "abcdabcc" then you'll see it still matches the "abcc" at the end of the string instead of the "abc" at the start. If you don't use an atomic group, then it can backtrack past the bc and try the b alternation and end up matching the "abc" at the start.
As for question two, how it's different, that's just a rephrasing of your first question.
And lastly, atomic groups are not "called" non-capturing groups. That's not an alternate name for them. Non-capturing groups are groups that do not capture their content. Typically when you match a regular expression against a string, you can retrieve all the matched groups, and if you use a substitution, you can use backreferences in the substitution like \1 to insert the captured groups there. But a non-capturing group does not provide this. The classic non-capturing group is (?:pattern). An atomic group happens to also have the non-capturing property, hence why it's called a non-capturing group.