Unicode have categories of characters. Some are alpha numeric. Some are punctuation.
What about if I want to know whether a word belongs to keyword or not
For example,
A,a,b,c, tend to belong to words. So is Ƈ,Ǝ,ǟ, so are all chinese characters.
Sentences like
Hello World, I "like" (to) eat ƇƎǟ and 款开源 ©
Have keywords:
Hello
World
I
like
to
eat
ƇƎǟ
款
开
源
Here, , (),© are not word characters and hence should just be ignored and use.
© doesn't count as punctuation either. '©'.IsPunctuation returns false in vb.net but I want to get rid of that too.
Now I want to make a program that can split sentences into keywords. For that I need to know which characters are word characters and which one is not.
Is there a vb.net function for that?
Do it the other way round: use IsLetter for your test. Or better yet, use regular expressions to split your string by words:
Dim str = "Hello World, I ""like"" (to) eat ƇƎǟ and 款开源 ©"
Dim wordPattern As New Regex("\p{L}+")
For Each match in wordPattern.Matches(str))
Console.WriteLine(match)
Next
Here, \p{L} matches any word character. However, the above matches “款开源” in a single rather than in separate matches since there is no separator between the characters.
u need to deal with "keycodes"
like if u only want letters [a-z]
then
for(c>='a' && c<='z'){
}
or
for(c>=97 && C<=122){
}
Related
Hi may i know what does the below query means?
REGEXP_REPLACE(number,'[^'' ''-/0-9:-#A-Z''[''-`a-z{-~]', 'xy') ext_number
part 1
In terms of explaining what the function function call is doing:
It is a function call to analyse an input string 'number' with a regex (2nd argument) and replace any parts of the string which match a specific string. As for the name after the parenthesis I am not sure, but the documentation for the function is here
part 2
Sorry to be writing a question within an answer here but I cannot respond in comments yet (not enough rep)
Does this regex work? Unless sql uses different syntax this would appear to be a non-functional regex. There are some red flags, e.g:
The entire regex is wrapped in square parenthesis, indicating a set of characters but seems to predominantly hold an expression
There is a range indicator between a single quote and a character (invalid range: if a dash was required in the match it should be escaped with a '\' (backslash))
One set of square brackets is never closed
After some minor tweaks this regex is valid syntax:
^'' ''\-\/0-9:-#A-Z''[''-a-z{-~]`, but does not match anything I can think of, it is important to know what string is being examined/what the context is for the program in order to identify what the regex might be attempting to do
It seems like it is meant to replaces all ASCII control characters in the column or variable number with xy.
[] encloses a class of characters. Any character in that class matches. [^] negates that, hence all characters match, that are not in the class.
- is a range operator, e.g. a-z means all characters from a to z, like abc...xyz.
It seams like characters enclosed in ' should be escaped (The second ' is to escape the ' in the string itself.) At least this would make some sense. (But for none of the DBMS I found having a regexp_replace() function (Postgres, Oracle, DB2, MariaDB, MySQL), I found something in the docs, that would indicate this escape mechanism. They all use \, but maybe I missed something? Unfortunately you didn't tag which DBMS you're actually using!)
Now if you take an ASCII table you'll see, that the ranges in the expression make up all printable characters (counting space as printable) in groups from space to /, 0 to 9, : to #, etc.. Actually it might have been shorter to express it as '' ''-~, space to ~.
Given the negation, all these don't match. The ones left are from NUL to US and DEL. These match and get replaced by xy one by one.
Before posting, I tried the hive sentences function and did some search but couldn't get a clear understanding, my question is based on what delimiter hive sentences function breaks each sentence? hive manual says "appropriate boundary" what does that mean? Below is an example of my tries, I tried adding period (.) and exclamatory sign(!) at different points of the sentence. I'm getting different outputs, can someone explain on this?
with period (.)
select sentences('Tokenizes a string of natural language text into words and sentences. where each sentence is broken at the appropriate sentence boundary and returned as an array of words.') from dummytable
output - 1 array
[["Tokenizes","a","string","of","natural","language","text","into","words","and","sentences","where","each","sentence","is","broken","at","the","appropriate","sentence","boundary","and","returned","as","an","array","of","words"]]
with '!'
select sentences('Tokenizes a string of natural language text into words and sentences! where each sentence is broken at the appropriate sentence boundary and returned as an array of words.') from dummytable
output - 2 arrays
[["Tokenizes","a","string","of","natural","language","text","into","words","and","sentences"],["where","each","sentence","is","broken","at","the","appropriate","sentence","boundary","and","returned","as","an","array","of","words"]]
If you understand the functionality of sentences()..it clears your doubt.
Definition of sentences(str):
Splits str into arrays of sentences, where each sentence is an array
of words.
Example:
SELECT sentences('Hello there! I am a UDF.') FROM src LIMIT 1;
[ ["Hello", "there"], ["I", "am", "a", "UDF"] ]
SELECT sentences('review . language') FROM movies;
[["review","language"]]
An exclamation point is a type of punctuation mark that goes at the end of a sentence. Other examples of related punctuation marks include periods and question marks, which also go at the end of sentences.But as per the definition of sentences() ,Unnecessary punctuation, such as periods and commas in English, is automatically stripped.So,we are able to get two arrays of words with !. It completely involves java.util.Locale.java
I don't know the actual reason but observed after period(.) if you put space and next word first letter as capital then it is working.
Here I changed from where to Where it it worked. However this is not require for !
Tokenizes a string of natural language text into words and sentences. Where each sentence is broken at the appropriate sentence boundary and returned as an array of words.
And this is giving below output
[["Tokenizes","a","string","of","natural","language","text","into","words","and","sentences"],["Where","each","sentence","is","broken","at","the","appropriate","sentence","boundary","and","returned","as","an","array","of","words"]]
This is related to
What are the characters that count as the same character under collation of UTF8 Unicode? And what VB.net function can be used to merge them?
This is how I plan to do this:
Use http://msdn.microsoft.com/en-us/library/dd374126%28v=vs.85%29.aspx to turn the string into
KD form.
Basically it'll turn most variation such as superscript into the normal number. Also it decompose tilda and accent into 2 characters.
Next step would be to remove all characters whose sole purpose is tildaing or accenting character.
How do I know which characters are like that? Which characters are just "composing characters"
How do I find such characters? After I find those, how do I get rid of it? Should I scan character by character and remove all such "combining characters?"
For example:
Character from 300 to 362 can be gotten rid off.
Then what?
Combining characters are listed in UnicodeData.txt as having a nonzero Canonical_Combining_Class, and a General_Category of Mn (Mark, nonspacing).
For each character in the string, call GetUnicodeCategory and check the UnicodeCategory for NonSpacingMark, SpacingCombiningMark or EnclosingMark.
You may be able to do it more efficiently using regex, eg Regex.Replace(str, "\p{M}", "").
I'm trying to implement stuff similar to spell check, but I need to get the word that is limited by a space. EX: "HI HOW R U", I need to collect HI, HOW and so on as they type. i.e. After user hits HI and space I need to collect HI and do a spell check.
Check the documentation for NSString Here. You want the message componentsSepeparatedByString:.
I don't know objective-C, but I'm fairly sure it'll have a Regexp library - although it'd be straightforward to code it without one.
Regexp: \b([^\s])*\b
\b = word boundary (whitespace, comma, dot, exclamation-mark, etc.)
\s = whitespace character
[...] = character set
[^...] = negated character set (any character(s) EXCEPT ...)
() = grouping construct
* = zero or more times
So the suggested expression would start matching at any word boundary, then match every subsequent character that is not a whitespace character, then match a word boundary.
Your stated case is so simple you may just want to look for spaces (one char at a time) and get the substring, but RegExp is very widely used across a range of languages and platforms, and so it's fairly easy to find an expression when you need to - and one often does for common stuff like checking if zip codes, phone numbers, email addresses and so on are syntactically correct. So it's worth learning in any case. :)
How can I find words like and, or, to, a, no, with, for etc. in a sentence using VB.NET and remove them. Also where can I find all words list like above.
Note that unless you use Regex word boundaries you risk falling afoul of the Scunthorpe (Sfannythorpe) problem.
string pattern = #"\band\b";
Regex re = new Regex(pattern);
string input = "a band loves and its fans";
string output = re.Replace(input, ""); // a band loves its fans
Notice the 'and' in 'band' is untouched.
You can indeed replace your list of words using the .Replace function (as colithium described) ...
myString.Replace("and", "")
Edit:
... but indeed, a nicer way is to use Regular Expressions (as edg suggested) to avoid replacing parts of words.
As your question suggests that you would like to clean-up a sentence to keep meaningfull words, you have to do more than just remove two- and three letter words.
What you need is a list of stop-words:
http://en.wikipedia.org/wiki/Stop_word
A comma seperated list of stop-words for the English language can be found here:
http://www.textfixer.com/resources/common-english-words.txt
The easiest way is:
myString.Replace("and", "")
You'd loop over your word list and have a statement like the above. Google for a list of common English words?
List of English 2 Letter Words
List of English 3 Letter Words
You can match the words and remove them using regular expressions.