Here's an interesting puzzle, and I'd welcome any thoughts that anyone has. I don't think that there's a right or wrong answer here.
My program is loading a file which contains (amongst other things) string data structures. It helpfully declares what type the structure is (UTF8, UTF16 etc), and its length (of course) before the structure, so my program knows how to handle the data. Up to now this has worked perfectly every time.
Now I have been given a data file to load which has rubbish on the end of it - and, when I say rubbish, I mean that it looks like UTF8 at the end of a structure which is declared as UTF16.
D·a·v·e···E·d·m·u·n·d·s·o·n·d`upbqp!c°rÞPrpupÎÐAgâh(28.RïSÿ
The Dave Edmundson part is fine - it's everything after that which needs to be trimmed in this case. The fly in the ointment is that I still need to be able to handle legitimate UTF16 extended characters (like Korean, Chinese etc).
I could just fling my hands up in the air and say, "this data is corrupt", and spit out an error. But I'd like to be able to clean it if at all possible. Any ideas that anyone might have would be very welcome!
This is a logical issue, hence no code - if anyone is interested, I'm using Objective-C, but all I'm really after is some intelligent conversation on how I might approach this issue. I don't need code to be written for me!
Is there a way to change how google assistant reads out numbers?
For example, 108 is a number, and assistant reads it out as "one hundred eight." Here, instead of "one hundred eight" I want the assistant to speak like "one oh eight".
Probably. While there are some ways you can change how it reads out numbers using the <say-as> ssml tag, you still don't have complete control over how it pronounces each number.
So
<speak>
<say-as interpret-as="cardinal">108</say-as>
</speak>
speaks it the default way - "one hundred eight" (at least for US English. Different locales may speak it differently).
<speak>
<say-as interpret-as="characters">108</say-as>
</speak>
says it the way you want (again, in US English): "one oh eight".
If you wanted it as "one zero eight", however, you'd be out of luck.
I plan to have a socket reading data until it gets to the \n character (In order to read individual messages from a data stream). What happens, however, if the message you're sending happens to have bytes that match the \n character? Won't that end reading prematurely and mess up everything? How do people usually read until a certain part in their data?
Ok, Joe provided a lot of good alternatives to reading till the "\n" character. (Now that I think of it, \n is probably only used for text based things)
Quote Joe: There's a lot of choices here. 1) substitute something for the delimiter when it's part of the content. 2) pick a less likely delimiter than \n. 3) include a message length ahead of each message that you parse out. And
#3 seems like the best way to accomplish what I'm trying to do.
Does anyone have a sample code for parsing the CEDICT file? CEDICT is a Chinese-English Dictionary. For instance, currently, if I open it in a text editor, a line in the CEDICT file looks like:
不 不 [bu4] /(negative prefix)/not/no/
I would like to see it as:
不 不 [bu4] /(negative prefix)/not/no/
I found Textwrangler to do this for me as a text editor. What I now need is sample code that achieves the same.
The thing is, it's just an encoding problem. If the line looks like
不 不 [bu4] /(negative prefix)/not/no/
It's because the text editor doesn't know/realize that the text is encoded as UTF-8. Text Wrangler, or its big brother BBEdit, are very good at guessing encoding, and can even be asked to display text in a specific encoding.
Since we don't know what you want, in the end, to achieve, it's hard to tell you exactly what has to be done, specifically. What I can say is that your app (which language are you using anyway?) needs to be Unicode aware (and be able to read/manipulate UTF strings).
I wrote a couple of apps based on the CEDICT, one for Mac OS X, one for Android. Parsing and indexing the CEDICT is not very hard.
UPDATE
Regarding the parsing itself of the CEDICT, it's nothing complicated. I don't do Objective-C, never have, never will, but the process would be the same in any language:
Read a line. Say your own example: 不 不 [bu4] /(negative prefix)/not/no/
You have four fields: Trad. Ch., Simp. Ch., Reading, Meaning(s).
These fields are space separated. Of course the 4th field may contain spaces, so be careful.
Store (I used an sqlite db) the 4 fields in to db.
You might want to remove the slashes from the definition field, replace them with something else.
Loop
You have now converted the CEDICT to a database. That's the easy part. As for tokenizing Chinese, good luck with that, mate. Better minds than mine are still banging their heads on this one.
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.