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
Related
I have some unique codes that are generated from strings (ex: website host names) in various independent components of my application.
These codes are meant to be used by machines only so i would like to keep them as short as possible.
The below algorithm would be applied to every word in the string. The output words would be concatenated with a dash to generate the unique code.
The current algorithm I have used:
- Skip word if length is less than 6
- Leave first character as is
- Remove every wowel in the word from the second character onwards
architectural digest eu => archtctrl-dgst-eu
arizona foothills magazine => arzn-fthlls-mgzn
Is there a better way to shorten an English word leaving it as recognisable as possible to a human reader?
The output should be deterministic and produce the same shortened version whenever it is run on the same input.
A good algorithm should also minimise the number of clashes for similarly spelt words.
I have some unique codes that are generated from strings
I am afraid that is not true. There are many English words that will reduce to the same 'code word' when stripped of their vowels. For example, 'leaving' -> 'living' Given, this is fairly rare, it could still cause issues.
How important is it that these 'code words' remain human-readable if as you say, they are meant to be used by machines only? If its not that important, I'd suggest looking into some simpler compression algorithms like Huffman Coding or LZW Compression. Then if the user needs to see the translation of the code word, just uncompress it.
If you must keep it human-readable, I'm not sure that there is much more you can do to shorten it. You could take a look at specific latin + greek roots, and determine if you can shorten those any more by hand, and then just substitute those out automatically.
Alternatively, you could turn to a phonetic approach. Automatically search the pronunciation of the word, and then see if that is any shorter (or itself can be compressed, taking 'cee' to 'C', or 'kay' to 'K'). This would be much more time and CPU intensive, but its still an option if you really, really need short but yet readable codes.
What you're generating sounds like what's called a "slug". There are many libraries to handle this for blogs or site generators that should suit your purposes. Here's a usage example from a Python library called slugify:
txt = "___This is a test ---"
r = slugify(txt)
self.assertEqual(r, "this-is-a-test")
Slug libraries generally work like this:
replacing non-ascii linguistic characters via a mapping (ex: 影師嗎 -> ying-shi-ma)
replace accented latin letters with ascii equivalents via a mapping (ex: C'est déjà l'été. -> c-est-deja-l-ete)
remove beginning and trailing spaces/punctuation
convert remaining spaces and punctuation to dashes, collapsing multiple dashes in a row to a single dash
If you want to make slugs shorter you could remove vowels or, more simply, use a maximum length.
I have some text generated by some lousy OCR software.
The output contains mixture of words and space-separated characters, which should have been grouped into words. For example,
Expr e s s i o n Syntax
S u m m a r y o f T e r minology
should have been
Expression Syntax
Summary of Terminology
What algorithms can group characters into words?
If I program in Python, C#, Java, C or C++, what libraries provide the implementation of the algorithms?
Thanks.
Minimal approach:
In your input, remove the space before any single letter words. Mark the final words created as part of this somehow (prefix them with a symbol not in the input, for example).
Get a dictionary of English words, sorted longest to shortest.
For each marked word in your input, find the longest match and break that off as a word. Repeat on the characters left over in the original "word" until there's nothing left over. (In the case where there's no match just leave it alone.)
More sophisticated, overkill approach:
The problem of splitting words without spaces is a real-world problem in languages commonly written without spaces, such as Chinese and Japanese. I'm familiar with Japanese so I'll mainly speak with reference to that.
Typical approaches use a dictionary and a sequence model. The model is trained to learn transition properties between labels - part of speech tagging, combined with the dictionary, is used to figure out the relative likelihood of different potential places to split words. Then the most likely sequence of splits for a whole sentence is solved for using (for example) the Viterbi algorithm.
Creating a system like this is almost certainly overkill if you're just cleaning OCR data, but if you're interested it may be worth looking into.
A sample case where the more sophisticated approach will work and the simple one won't:
input: Playforthefunofit
simple output: Play forth efunofit (forth is longer than for)
sophistiated output: Play for the fun of it (forth efunofit is a low-frequency - that is, unnatural - transition, while for the is not)
You can work around the issue with the simple approach to some extent by adding common short-word sequences to your dictionary as units. For example, add forthe as a dictionary word, and split it in a post processing step.
Hope that helps - good luck!
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I don't really understand regular expressions. Can you explain them to me in an easy-to-follow manner? If there are any online tools or books, could you also link to them?
The most important part is the concepts. Once you understand how the building blocks work, differences in syntax amount to little more than mild dialects. A layer on top of your regular expression engine's syntax is the syntax of the programming language you're using. Languages such as Perl remove most of this complication, but you'll have to keep in mind other considerations if you're using regular expressions in a C program.
If you think of regular expressions as building blocks that you can mix and match as you please, it helps you learn how to write and debug your own patterns but also how to understand patterns written by others.
Start simple
Conceptually, the simplest regular expressions are literal characters. The pattern N matches the character 'N'.
Regular expressions next to each other match sequences. For example, the pattern Nick matches the sequence 'N' followed by 'i' followed by 'c' followed by 'k'.
If you've ever used grep on Unix—even if only to search for ordinary looking strings—you've already been using regular expressions! (The re in grep refers to regular expressions.)
Order from the menu
Adding just a little complexity, you can match either 'Nick' or 'nick' with the pattern [Nn]ick. The part in square brackets is a character class, which means it matches exactly one of the enclosed characters. You can also use ranges in character classes, so [a-c] matches either 'a' or 'b' or 'c'.
The pattern . is special: rather than matching a literal dot only, it matches any character†. It's the same conceptually as the really big character class [-.?+%$A-Za-z0-9...].
Think of character classes as menus: pick just one.
Helpful shortcuts
Using . can save you lots of typing, and there are other shortcuts for common patterns. Say you want to match a digit: one way to write that is [0-9]. Digits are a frequent match target, so you could instead use the shortcut \d. Others are \s (whitespace) and \w (word characters: alphanumerics or underscore).
The uppercased variants are their complements, so \S matches any non-whitespace character, for example.
Once is not enough
From there, you can repeat parts of your pattern with quantifiers. For example, the pattern ab?c matches 'abc' or 'ac' because the ? quantifier makes the subpattern it modifies optional. Other quantifiers are
* (zero or more times)
+ (one or more times)
{n} (exactly n times)
{n,} (at least n times)
{n,m} (at least n times but no more than m times)
Putting some of these blocks together, the pattern [Nn]*ick matches all of
ick
Nick
nick
Nnick
nNick
nnick
(and so on)
The first match demonstrates an important lesson: * always succeeds! Any pattern can match zero times.
A few other useful examples:
[0-9]+ (and its equivalent \d+) matches any non-negative integer
\d{4}-\d{2}-\d{2} matches dates formatted like 2019-01-01
Grouping
A quantifier modifies the pattern to its immediate left. You might expect 0abc+0 to match '0abc0', '0abcabc0', and so forth, but the pattern immediately to the left of the plus quantifier is c. This means 0abc+0 matches '0abc0', '0abcc0', '0abccc0', and so on.
To match one or more sequences of 'abc' with zeros on the ends, use 0(abc)+0. The parentheses denote a subpattern that can be quantified as a unit. It's also common for regular expression engines to save or "capture" the portion of the input text that matches a parenthesized group. Extracting bits this way is much more flexible and less error-prone than counting indices and substr.
Alternation
Earlier, we saw one way to match either 'Nick' or 'nick'. Another is with alternation as in Nick|nick. Remember that alternation includes everything to its left and everything to its right. Use grouping parentheses to limit the scope of |, e.g., (Nick|nick).
For another example, you could equivalently write [a-c] as a|b|c, but this is likely to be suboptimal because many implementations assume alternatives will have lengths greater than 1.
Escaping
Although some characters match themselves, others have special meanings. The pattern \d+ doesn't match backslash followed by lowercase D followed by a plus sign: to get that, we'd use \\d\+. A backslash removes the special meaning from the following character.
Greediness
Regular expression quantifiers are greedy. This means they match as much text as they possibly can while allowing the entire pattern to match successfully.
For example, say the input is
"Hello," she said, "How are you?"
You might expect ".+" to match only 'Hello,' and will then be surprised when you see that it matched from 'Hello' all the way through 'you?'.
To switch from greedy to what you might think of as cautious, add an extra ? to the quantifier. Now you understand how \((.+?)\), the example from your question works. It matches the sequence of a literal left-parenthesis, followed by one or more characters, and terminated by a right-parenthesis.
If your input is '(123) (456)', then the first capture will be '123'. Non-greedy quantifiers want to allow the rest of the pattern to start matching as soon as possible.
(As to your confusion, I don't know of any regular-expression dialect where ((.+?)) would do the same thing. I suspect something got lost in transmission somewhere along the way.)
Anchors
Use the special pattern ^ to match only at the beginning of your input and $ to match only at the end. Making "bookends" with your patterns where you say, "I know what's at the front and back, but give me everything between" is a useful technique.
Say you want to match comments of the form
-- This is a comment --
you'd write ^--\s+(.+)\s+--$.
Build your own
Regular expressions are recursive, so now that you understand these basic rules, you can combine them however you like.
Tools for writing and debugging regexes:
RegExr (for JavaScript)
Perl: YAPE: Regex Explain
Regex Coach (engine backed by CL-PPCRE)
RegexPal (for JavaScript)
Regular Expressions Online Tester
Regex Buddy
Regex 101 (for PCRE, JavaScript, Python, Golang, Java 8)
I Hate Regex
Visual RegExp
Expresso (for .NET)
Rubular (for Ruby)
Regular Expression Library (Predefined Regexes for common scenarios)
Txt2RE
Regex Tester (for JavaScript)
Regex Storm (for .NET)
Debuggex (visual regex tester and helper)
Books
Mastering Regular Expressions, the 2nd Edition, and the 3rd edition.
Regular Expressions Cheat Sheet
Regex Cookbook
Teach Yourself Regular Expressions
Free resources
RegexOne - Learn with simple, interactive exercises.
Regular Expressions - Everything you should know (PDF Series)
Regex Syntax Summary
How Regexes Work
JavaScript Regular Expressions
Footnote
†: The statement above that . matches any character is a simplification for pedagogical purposes that is not strictly true. Dot matches any character except newline, "\n", but in practice you rarely expect a pattern such as .+ to cross a newline boundary. Perl regexes have a /s switch and Java Pattern.DOTALL, for example, to make . match any character at all. For languages that don't have such a feature, you can use something like [\s\S] to match "any whitespace or any non-whitespace", in other words anything.
I'd like to find good way to find some (let it be two) sentences in some text. What will be better - use regexp or split-method? Your ideas?
As requested by Jeremy Stein - there are some examples
Examples:
Input:
The first thing to do is to create the Comment model. We’ll create this in the normal way, but with one small difference. If we were just creating comments for an Article we’d have an integer field called article_id in the model to store the foreign key, but in this case we’re going to need something more abstract.
First two sentences:
The first thing to do is to create the Comment model. We’ll create this in the normal way, but with one small difference.
Input:
Mr. T is one mean dude. I'd hate to get in a fight with him.
First two sentences:
Mr. T is one mean dude. I'd hate to get in a fight with him.
Input:
The D.C. Sniper was executed was executed by lethal injection at a Virginia prison. Death was pronounced at 9:11 p.m. ET.
First two sentences:
The D.C. Sniper was executed was executed by lethal injection at a Virginia prison. Death was pronounced at 9:11 p.m. ET.
Input:
In her concluding remarks, the opposing attorney said that "...in this and so many other instances, two wrongs won’t make a right." The jury seemed to agree.
First two sentences:
In her concluding remarks, the opposing attorney said that "...in this and so many other instances, two wrongs won’t make a right." The jury seemed to agree.
Guys, as you can see - it's not so easy to determine two sentences from text. :(
As you've noticed, sentence tokenizing is a bit tricker than it first might seem. So you may as well take advantage of existing solutions. The Punkt sentence tokenizing algorithm is popular in NLP, and there is a good implementation in the Python Natural Language Toolkit which they describe the use of here. They also describe another approach here.
There's probably other implementations around, or you could also read the original paper describing the Punkt algorithm: Kiss, Tibor and Strunk, Jan (2006): Unsupervised Multilingual Sentence Boundary Detection. Computational Linguistics 32: 485-525.
You can also read another Stack Overflow question about sentence tokenizing here.
your_string = "First sentence. Second sentence. Third sentence"
sentences = your_string.split(".")
=> ["First sentence", " Second sentence", " Third sentence"]
No need to make simple code complicated.
Edit: Now that you've clarified that the real input is more complex that your initial example you should disregard this answer as it doesn't consider edge cases. An initial look at NLP should show you what you're getting into though.
Some of the edge cases that I've seen in the past to be a bit complicated are:
Dates: Some regions use dd.mm.yyyy
Quotes: While he was sighing — "Whatever, do it. Now. And by the way...". This was enough.
Units: He was going at 138 km. while driving on the freeway.
If you plan to parse these texts you should stay away from splits or regular expressions.
This will usually match sentences.
/\S(?:(?![.?!]+\s).)*[.?!]+(?=\s|$)/m
For your example of two sentences, take the first two matches.
irb(main):005:0> a = "The first sentence. The second sentence. And the third"
irb(main):006:0> a.split(".")[0...2]
=> ["The first sentence", " The second sentence"]
irb(main):007:0>
EDIT: here's how you handle the "This is a sentence ...... and another . And yet another ..." case :
irb(main):001:0> a = "This is the first sentence ....... And the second. Let's not forget the third"
=> "This is the first sentence ....... And the second. Let's not forget the thir
d"
irb(main):002:0> a.split(/\.+/)
=> ["This is the first sentence ", " And the second", " Let's not forget the thi rd"]
And you can apply the same range operator ... to extract the first 2.
You will find tips and software links on the sentence boundary detection Wikipedia page.
If you know what sentences to search, Regex should do well searching for
((YOUR SENTENCE HERE)|(YOUR OTHER SENTENCE)){1}
Split would probably use up quite a lot of memory, as it also saves the things you don't need (the whole text that's not your sentence) as Regex only saves the sentence you searched (if it finds it, of course)
If you're segmenting a piece of text into sentences, then what you want to do is begin by determining which punction marks can separate sentences. In general, this is !, ? and . (but if all you care about is a . for the texts your processing, then just go with that).
Now since these can appear inside quotations, or as parts of abbreviations, what you want to do is find each occurrence of these punctuation marks and run some sort of machine learning classifier to determine whether that occurance starts a new sentence, or whether it does something else. This involves training data and a properly-constructed classifier. And it won't be 100% accurate, because there's probably no way to be 100% accurate.
I suggest looking in the literature for sentence segmentation techniques, and have a look at the various natural language processing toolkits that are out there. I haven't really found one for Ruby yet, but I happen to like OpenNLP (which is in Java).
Arising out of this question, I'm looking for an elegant (ruby) way to compute the word signature suggested in this answer.
The idea suggested is to sort the letters in the word, and also run length encode repeated letters. So, for example "mississippi" first becomes "iiiimppssss", and then could be further shortened by encoding as "4impp4s".
I'm relatively new to ruby and though I could hack something together, I'm sure this is a one liner for somebody with more experience of ruby. I'd be interested to see people's approaches and improve my ruby knowledge.
edit: to clarify, performance of computing the signature doesn't much matter for my application. I'm looking to compute the signature so I can store it with each word in a large database of words (450K words), then query for words which have the same signature (i.e. all anagrams of a given word, that are actual english words). Hence the focus on space. The 'elegant' part is just to satisfy my curiosity.
The fastest way to create a sorted list of the letters is this:
"mississippi".unpack("c*").sort.pack("c*")
It is quite a bit faster than split('') and join(). For comparison it is also best to pack the array back together into a String, so you dont have to compare arrays.
I'm not much of a Ruby person either, but as I noted on the other comment this seems to work for the algorithm described.
s = "mississippi"
s.split('').sort.join.gsub(/(.)\1{2,}/) { |s| s.length.to_s + s[0,1] }
Of course, you'll want to make sure the word is lowercase, doesn't contain numbers, etc.
As requested, I'll try to explain the code. Please forgive me if I don't get all of the Ruby or reg ex terminology correct, but here goes.
I think the split/sort/join part is pretty straightforward. The interesting part for me starts at the call to gsub. This will replace a substring that matches the regular expression with the return value from the block that follows it. The reg ex finds any character and creates a backreference. That's the "(.)" part. Then, we continue the matching process using the backreference "\1" that evaluates to whatever character was found by the first part of the match. We want that character to be found a minimum of two more times for a total minimum number of occurrences of three. This is done using the quantifier "{2,}".
If a match is found, the matching substring is then passed to the next block of code as an argument thanks to the "|s|" part. Finally, we use the string equivalent of the matching substring's length and append to it whatever character makes up that substring (they should all be the same) and return the concatenated value. The returned value replaces the original matching substring. The whole process continues until nothing is left to match since it's a global substitution on the original string.
I apologize if that's confusing. As is often the case, it's easier for me to visualize the solution than to explain it clearly.
I don't see an elegant solution. You could use the split message to get the characters into an array, but then once you've sorted the list I don't see a nice linear-time concatenate primitive to get back to a string. I'm surprised.
Incidentally, run-length encoding is almost certainly a waste of time. I'd have to see some very impressive measurements before I'd think it worth considering. If you avoid run-length encoding, you can anagrammatize any string, not just a string of letters. And if you know you have only letters and are trying to save space, you can pack them 5 bits to a letter.
---Irma Vep
EDIT: the other poster found join which I missed. Nice.