I want to detect words in text, i.e. I need to know which characters in a given text are letters, that is they can be part of a (spoken) word and which are, on the other hand, punctuation and such.
For example, in the above sentence, "I", "want" and "i" and "e" are words in this regard, while spaces, "." and comma are not.
The difficulty in this is that I want to be able to read any kind of script that's based on Unicode. E.g., the german word "schön" is one word. But what about greek, arabic or japanese?
So, what I need is a table or list specifying all ranges of characters that can form words. Optionally, I also like to know which chars are digits that can form numbers (assuming other scripts have similar numbering schemes as the arabic numbers do).
I need this for Mac OS X, Windows and Linux. I'll write a C app, so it needs to be either a OS library or a complete code/data solution that I could translate into C.
I know that Mac OS (Cocoa) offers functions for this purpose, but I am not sure if there are similar solutions for Win and Linux (gtk based, probably?).
Alternatively, I could write my own code if I had the complete tables.
I have found the unicode charts (http://unicode.org/charts/index.html#scripts) but that's not coming in one convenient form I could use in programming.
So, can someone tell me if there are functions for Windows and Linux for this purpose, or where I can find a complete table/list of word characters in unicode?
You can try to use the Unicode character category to figure out what the word separators may be, but be aware that some languages (e.g. Japanese) do not even have word separators.
If you are familiar with Python at all, the Natural Language Toolkit provides chunkers/ lexical tools that will do this across languages. I'd pretend to be smart here and tell you more, but everything I know is out of this book, which I highly recommend. I realize you could code up a technical solution with a regex that would get you 80% of the way to where you want to be, but why reinvent the wheel?
the c-runtime has
ispunct() is a punctuation character
isctrl() is a control character.
In Java, there is static int java.lang.Character.getType(int codePoint) which can be compared to the constants provided in the same class, like this:
switch(Character.getType(codePoint)) {
case Character.UPPERCASE_LETTER:
case Character.LOWERCASE_LETTER:
case Character.TITLECASE_LETTER:
case Character.MODIFIER_LETTER:
case Character.OTHER_LETTER:
// you found a letter
break;
case Character.NON_SPACING_MARK:
// you found a combining diacritical mark
// see: https://en.wikipedia.org/wiki/Combining_character
break;
default:
// you found other symbols
break;
}
Related
I have a dataset which mixes use of unicode characters \u0421, 'С' and \u0043, 'C'. Is there some sort of unicode comparison which considers those two characters the same? So far I've tried several ICU collations, including the Russian one.
There is no Unicode comparison that treats characters as the same on the basis of visual identity of glyphs. However, Unicode Technical Standard #39, Unicode Security Mechanisms, deals with “confusables” – characters that may be confused with each other due to visual identity or similarity. It includes a data file of confusables as well as “intentionally confusable” pairs, i.e. “characters whose glyphs in any particular typeface would probably be designed to be identical in shape when using a harmonized typeface design”, which mainly consists of pairs of Latin and Cyrillic or Greek letters, like C and С. You would probably need to code your own use of this data, as ICU does not seem to have anything related to the confusable concept.
when you take a look at http://www.unicode.org/Public/UCD/latest/ucd/UnicodeData.txt, you will see that some code positions are annotated for codepoints that are similar in use; however, i'm not aware of any extensive list that covers visual similarities across scripts. you might want to search for URL spoofing using intentional misspellings, which was discussed when they came up with punycode. other than that, your best bet might be to search the data for characters outside the expected using regular expressions, and compile a series of ad-hoc text fixers like text = text.replace /с/, 'c'.
What is the subset of Unicode characters that are normally used in writing — such as those that would be typically found in a newspaper article?
For example, in English, the characters in the range [a-zA-Z0-9], plus some punctuation characters, would be sufficient for most writing.
But I want to support languages that use characters that fall outside the ASCII range, while excluding the non-printing or decorative characters.
The objective is to restrict the user input to the application to codepoints that are legitimately used in written language. Because the user input will be saved and displayed, I do not want to allow pranksters to input text consisting entirely of things like diacritics, Unicode combining characters, Unicode flow control characters, etc.
Regrettably, I am not fluent in every single language found in Unicode. Has anyone compiled a list of all of the subset of Unicode characters that are normally used in writing?
The official list of Unicode code points is UnicodeData.txt. This is a plain text file with one line per code point; it's easily machine-readable. For example:
0022;QUOTATION MARK;Po;0;ON;;;;;N;;;;;
The third semicolon-delimited field is the abbreviated name of the "General Category". This is explained further in chapter 4 of the Unicode Standard, specifically in section 4.5; see the table on page 131 (page 12 of the PDF file). For example, "Lu" is uppercase letters, "Ll" is lowercase letters, Pc, Pd, Ps, et al are various kinds of punctuation. (The first letter of the two-letter abbreviation represents a higher-level category such as letter, digit, punctuation, etc.)
Note that some ranges of code points are not listed explicitly. For example, the range of CJK (Chinese, Japanese, Korean) ideographs is represented as:
4E00;<CJK Ideograph, First>;Lo;0;L;;;;;N;;;;;
9FCC;<CJK Ideograph, Last>;Lo;0;L;;;;;N;;;;;
I think there are other files on unicode.org that fill in these gaps.
I'm still not 100% clear on just what subset you're trying to define, but you can probably define it as a particular set of General Category values.
I do not want to allow pranksters to input text consisting entirely of things like diacritics, Unicode combining characters
Diacritics/combining characters will be used in normal written language. So if you want to stop 'pranksters' you're going to need something more sophisticated than just a list of permitted characters. You'll have to do some sort of linguistic analysis for every language you want to permit.
I'd recommend not bothering with this, because it's going to be hard and you won't succeed anyway. Just let people write what they want.
Try WGL4 (652 characters), MES-1 (335 characters) or MES-2 (1062 characters). Find these at Wikipedia.
You may wish to exclude characters IJijĸĿŀʼn˚―⅛⅜⅝⅞♪ from MES-1 if you want to use this set.
Edit: I realize this is a bad answer. Especially the removing characters from MES-1 part was total garbage. I shouldn't have posted this. I'm ashamed of whoever upvoted this.
If anything, use Subset1 (678 characters), Subset2 (1193 characters) and Subset3 (2823 characters). https://unicodesubsets.miraheze.org/wiki/User:PiotrGrochowski
I'm trying to gather a Unicode list of all the 'o' like shapes in the Hindi character-set. In fact, a list of any characters (in any language) that makes uses of separate characters to indicate an accent would be better.
I intend to use this unicode-list in a RegExp.
I been trying to edit a list of character-ranges by outputting them in an Input TextField, but editing this text causes weird issues (the keyboard-cursor isn't place on the correct character, selections suddenly dissappear / incorrectly warps... in other words... HINDI HELL!)
I've tried this with Notepad++ too, but although it was more responsive, it eventually crapped out on me like it did in the Flash Player textfield. This seems to occur especially while removing the [] block (nulls?) characters. Some of them trigger odd behaviors.
Anyways, all I want is a list of the accents.
An example of a few are in the image below (but I would need ALL accents):
Thanks!
You can find pdf's containing lists of unicode ranges, grouped by language, here: http://unicode.org/charts/
For Hindi, you probably want Devanagari or Devanagari Extended.
Here is the character class for Devanagari combining marks:
[\u901\u902\u903\u93c\u93e\u93f\u940\u941\u942\u943
\u944\u945\u946\u947\u948\u949\u94a\u94b\u94c\u94d
\u951\u952\u953\u954\u962\u963]
This is only the basic Devanagari block (not Devanagari Extended).
If you want the complete set (for all languages), you can do it problematically.
You start from the Unicode date file at ftp://ftp.unicode.org/Public/6.1.0/ucd/UnicodeData.txt, described by TR-44 (http://unicode.org/reports/tr44/#Property_Definitions)
You can use the Canonical_Combining_Class field (see at http://unicode.org/reports/tr44/#Canonical_Combining_Class_Values) to filter the exact characters you want.
Can't be more precise, because "accent" a bit vague :-)
You might even have to also look at General_Category to get the filter right (and exclude certain marks, or symbols, or punctuation).
And a script doing this would definitely be better than trying to mess with text editors.
One of the characteristics of combining characters is that they combine :-)
So you might get all kind of puzzling results (like this: http://www.siao2.com/2006/02/17/533929.aspx :-)
I'm trying to implement a word count function for my app that uses UITextView.
There's a space between two words in English, so it's really easy to count the number of words in an English sentence.
The problem occurs with Chinese and Japanese word counting because usually, there's no any space in the entire sentence.
I checked with three different text editors in iPad that have a word count feature and compare them with MS Words.
For example, here's a series of Japanese characters meaning the world's idea: 世界(the world)の('s)アイデア(idea)
世界のアイデア
1) Pages for iPad and MS Words count each character as one word, so it contains 7 words.
2) iPad text editor P*** counts the entire as one word --> They just used space to separate words.
3) iPad text editor i*** counts them as three words --> I believe they used CFStringTokenizer with kCFStringTokenizerUnitWord because I could get the same result)
I've researched on the Internet, and Pages and MS Words' word counting seems to be correct because each Chinese character has a meaning.
I couldn't find any class that counts the words like Pages or MS Words, and it would be very hard to implement it from scratch because besides Japanese and Chinese, iPad supports a lot of different foreign languages.
I think CFStringTokenizer with kCFStringTokenizerUnitWord is the best option though.
Is there a way to count words in NSString like Pages and MSWords?
Thank you
I recommend keep using CFStringTokenizer. Because it's platform feature, so will be upgraded by platform upgrade. And many people in Apple are working hardly to reflect real cultural difference. Which are hard to know for regular developers.
This is hard because this is not a programming problem essentially. This is a human cultural linguistic problem. You need a human language specialist for each culture. For Japanese, you need Japanese culture specialist. However, I don't think Japanese people needs word count feature seriously, because as I heard, the concept of word itself is not so important in the Japanese culture. You should define concept of word first.
And I can't understand why you want to force concept of word count into the character count. The Kanji word that you instanced. This is equal with counting universe as 2 words by splitting into uni + verse by meaning. Not even a logic. Splitting word by it's meaning is sometimes completely wrong and useless by the definition of word. Because definition of word itself are different by the cultures. In my language Korean, word is just a formal unit, not a meaning unit. The idea that each word is matching to each meaning is right only in roman character cultures.
Just give another feature like character counting for the users in east-asia if you think need it. And counting character in unicode string is so easy with -[NSString length] method.
I'm a Korean speaker, (so maybe out of your case :) and in many cases we count characters instead of words. In fact, I never saw people counting words in my whole life. I laughed at word counting feature on MS word because I guessed nobody would use it. (However now I know it's important in roman character cultures.) I have used word counting feature only once to know it works really :) I believe this is similar in Chinese or Japanese. Maybe Japanese users use the word counting because their basic alphabet is similar with roman characters which have no concept of composition. However they're using Kanji heavily which are completely compositing, character-centric system.
If you make word counting feature works greatly on those languages (which are using by people even does not feel any needs to split sentences into smaller formal units!), it's hard to imagine someone who using it. And without linguistic specialist, the feature should not correct.
This is a really hard problem if your string doesn't contain tokens identifying word breaks (like spaces). One way I know derived from attempting to solve anagrams is this:
At the start of the string you start with one character. Is it a word? It could be a word like "A" but it could also be a part of a word like "AN" or "ANALOG". So the decision about what is a word has to be made considering all of the string. You would consider the next characters to see if you can make another word starting with the first character following the first word you think you might have found. If you decide the word is "A" and you are left with "NALOG" then you will soon find that there are no more words to be found. When you start finding words in the dictionary (see below) then you know you are making the right choices about where to break the words. When you stop finding words you know you have made a wrong choice and you need to backtrack.
A big part of this is having dictionaries sufficient to contain any word you might encounter. The English resource would be TWL06 or SOWPODS or other scrabble dictionaries, containing many obscure words. You need a lot of memory to do this because if you check the words against a simple array containing all of the possible words your program will run incredibly slow. If you parse your dictionary, persist it as a plist and recreate the dictionary your checking will be quick enough but it will require a lot more space on disk and more space in memory. One of these big scrabble dictionaries can expand to about 10MB with the actual words as keys and a simple NSNumber as a placeholder for value - you don't care what the value is, just that the key exists in the dictionary, which tells you that the word is recognised as valid.
If you maintain an array as you count you get to do [array count] in a triumphal manner as you add the last word containing the last characters to it, but you also have an easy way of backtracking. If at some point you stop finding valid words you can pop the lastObject off the array and replace it at the start of the string, then start looking for alternative words. If that fails to get you back on the right track pop another word.
I would proceed by experimentation, looking for a potential three words ahead as you parse the string - when you have identified three potential words, take the first away, store it in the array and look for another word. If you find it is too slow to do it this way and you are getting OK results considering only two words ahead, drop it to two. If you find you are running up too many dead ends with your word division strategy then increase the number of words ahead you consider.
Another way would be to employ natural language rules - for example "A" and "NALOG" might look OK because a consonant follows "A", but "A" and "ARDVARK" would be ruled out because it would be correct for a word beginning in a vowel to follow "AN", not "A". This can get as complicated as you like to make it - I don't know if this gets simpler in Japanese or not but there are certainly common verb endings like "ma su".
(edit: started a bounty, I'd like to know the very best way to do this if my way isn't it.)
If you are using iOS 4, you can do something like
__block int count = 0;
[string enumerateSubstringsInRange:range
options:NSStringEnumerationByWords
usingBlock:^(NSString *word,
NSRange wordRange,
NSRange enclosingRange,
BOOL *stop)
{
count++;
}
];
More information in the NSString class reference.
There is also WWDC 2010 session, number 110, about advanced text handling, that explains this, around minute 10 or so.
I think CFStringTokenizer with kCFStringTokenizerUnitWord is the best option though.
That's right, you have to iterate through text and simply count number of word tokens encontered on the way.
Not a native chinese/japanese speaker, but here's my 2cents.
Each chinese character does have a meaning, but concept of a word is combination of letters/characters to represent an idea, isn't it?
In that sense, there's probably 3 words in "sekai no aidia" (or 2 if you don't count particles like NO/GA/DE/WA, etc). Same as english - "world's idea" is two words, while "idea of world" is 3, and let's forget about the required 'the' hehe.
That given, counting word is not as useful in non-roman language in my opinion, similar to what Eonil mentioned. It's probably better to count number of characters for those languages.. Check around with Chinese/Japanese native speakers and see what they think.
If I were to do it, I would tokenize the string with spaces and particles (at least for japanese, korean) and count tokens. Not sure about chinese..
With Japanese you can create a grammar parser and I think it is the same with Chinese. However, that is easier said than done because natural language tends to have many exceptions, but it is not impossible.
Please note it won't really be efficient since you have to parse each sentence before being able to count the words.
I would recommend the use of a parser compiler rather than building one yourself as well to start at least you can concentrate on doing the grammar than creating the parser yourself. It's not efficient, but it should get the job done.
Also have a fallback algorithm in case your grammar didn't parse the input correctly (perhaps the input really didn't make sense to begin with) you can use the length of the string to make it easier on you.
If you build it, there could be a market opportunity for you to use it as a natural language Domain Specific Language for Japanese/Chinese business rules as well.
Just use the length method:
[#"世界のアイデア" length]; // is 7
That being said, as a Japanese speaker, I think 3 is the right answer.
Where can I find a Unicode table showing only the simplified Chinese characters?
I have searched everywhere but cannot find anything.
UPDATE :
I have found that there is another encoding called GB 2312 -
http://en.wikipedia.org/wiki/GB_2312
- which contains only simplified characters.
Surely I can use this to get what I need?
I have also found this file which maps GB2312 to Unicode -
http://cpansearch.perl.org/src/GUS/Unicode-UTF8simple-1.06/gb2312.txt
- but I'm not sure if it's accurate or not.
If that table isn't correct maybe someone could point me to one that is, or maybe just a table of the GB2312 characters and some way to convert them?
UPDATE 2 :
This site also provides a GB/Unicode table and even a Java program to generate a file
with all the GB characters as well as the Unicode equivalents :
http://www.herongyang.com/gb2312/
The Unihan database contains this information in the file Unihan_Variants.txt. For example, a pair of traditional/simplified characters are:
U+673A kTraditionalVariant U+6A5F
U+6A5F kSimplifiedVariant U+673A
In the above case, U+6A5F is 機, the traditional form of 机 (U+673A).
Another approach is to use the CC-CEDICT project, which publishes a dictionary of Chinese characters and compounds (both traditional and simplified). Each entry looks something like:
宕機 宕机 [dang4 ji1] /to crash (of a computer)/Taiwanese term for 當機|当机[dang4 ji1]/
The first column is traditional characters, and the second column is simplified.
To get all the simplified characters, read this text file and make a list of every character that appears in the second column. Note that some characters may not appear by themselves (only in compounds), so it is not sufficient to look at single-character entries.
The OP doesn't indicate which language they're using, but if you're using Ruby, I've written a small library that can distinguish between simplified and traditional Chinese (plus Korean and Japanese as a bonus). As suggested in Greg's answer, it relies on a distilled version of Unihan_Variants.txt to figure out which chars are exclusively simplified and which are exclusively traditional.
https://github.com/jpatokal/script_detector
Sample:
p string
=> "我的氣墊船充滿了鱔魚."
> string.chinese?
=> true
> string.traditional_chinese?
=> true
> string.simplified_chinese?
=> false
But as the Unicode FAQ duly warns, this requires sizable fragments of text to work reliably, and will give misleading results for short strings. Consider the Japanese for Tokyo:
p string
=> "東京"
> string.chinese?
=> true
> string.traditional_chinese?
=> true
> string.japanese?
=> false
Since both characters happen to also be valid traditional Chinese, and there are no exclusively Japanese characters, it's not recognized correctly.
I'm not sure if that's easily done. The Han ideographs are unified in Unicode, so it's not immediately obvious how to do it. But the Unihan database (http://www.unicode.org/charts/unihan.html) might have the data you need.
Here is a regex of all simplified Chinese characters I made. For some reason Stackoverflow is complaining, so it's linked in a pastebin below.
https://pastebin.com/xw4p7RVJ
You'll notice that this list features ranges rather than each individual character, but also that these are utf-8 characters, not escaped representations. It's served me well in one iteration or another since around 2010. Hopefully everyone else can make some use of it now.
If you don't want the simplified chars (I can't imagine why, it's not come up once in 9 years), iterate over all the chars from ['一-龥'] and try to build a new list. Or run two regex's, one to check it is Chinese, but is not simplified Chinese
According to wikipedia simplified Chinese v. traditional, kanji, or other formats is left up to the font rendering in many cases. So while you could have a selection of simplified Chinese codepoints, this list would not be at all complete since many characters are no longer distinct.
I don't believe that there's a table with only simplified code points. I think they're all lumped together in the CJK range of 0x4E00 through 0x9FFF