I need to find all character in languages.
first I try to find a table that map language to a unicode range. but I can't find any thing useful.
next I think is it possible to change language or culture and by simulation press or keydown event find characters?
is it possible? if it is how and how I can engage shift key?
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I'm struggling with Microsoft Word, characters.
I'm write an article about digital electronic circuits, and I'm describing some Flip-Flop use.
I'm not finding how to write the outputs of the Flip-Flops: NOT-Q (it's a "Q" with a dash over it). I tried to find something into the character-map but I didn't find what I need
Here below in the screenshot from Wikipedia the character I'm looking for
Is there any way please?
I found the way, which is way cumbersome but it exists.
Insert > Equation > Insert New Equation > Accents
after a while, appear all the possible accents, included the ons I was looking for.
Writing any letter, will be accented as I asked for.
Or there is the Combined Diacritical Marks as described here:
http://wordfaqs.ssbarnhill.com/Overbar.htm
I've a french text with some words in english and I want to find those words and highlight all of them at once. Is there any program that can help me do that? Is it possible to do this with any other foreign language?
I'm using microsoft Word.
Word can do this IF the English words are formatted with the English language (and the rest in the French language). In that case, Word's FIND functionality advanced options are able to filter so that the language formatting is searched (instead of text).
Is there an unicode symbol for "n/a"? There are some fractions like ½, but a n/a symbol seems to be missing.
If there is none, what would be the most appropriate unicode symbol to use for n/a in a website (which should be contained in common fonts, to avoid needing a webfont)?
Looking at the Unicode code charts, I do not see a single N/A symbol. I do, however, see ⁿ (U+207F) and ₐ (U+2090), which you could separate with / (U+002F) eg: ⁿ/ₐ, or ̷ (U+0337), eg: ⁿ̷ₐ, or ̸ (U+0338), eg: ⁿ̸ₐ. Probably not what you are hoping for, though. And I don't know if "common" fonts implement them, either.
For future reference, the fastest way I know to answer questions like the OP's when I have them myself is to go to unicodelookup.com, because of the way it works: there's a search bar at the top, and you just type a string and it will return any and all unicode characters containing that string (this is also a great way to discover new and useful symbols). So in the OP's case, he could proceed like this:
first try entering "not" (without the quotes) in the search field
visually scan through the results... doing so would not reveal a "not
applicable" character in this case
try again but this time entering "applic" in the search field
again, doing so would not turn up anything along the lines of what he's
looking for
At that point he would be reasonably confident the current Unicode standard does not have a "n/a" symbol.
If you use Firefox you can define a keyword like "uni" to search that site from the URL bar, meaning any time the browser is open and regardless of what page or site is currently showing, you could do this:
hit [F6]... this moves the cursor to the URL bar at the top
type something like "uni applic" and hit [Enter]... this brings up the
unicodelookup.com website with the search results for "applic" already
showing
For the above to work you would need to define your keyword ("uni" or wtv you prefer) to point to location http://unicodelookup.com/#%s.
There's a Negative Acknowlege icon...
␕ symbol for negative acknowledge 022025 9237 0x2415 ␕
Found by searching negative on the Unicode Lookup site.
I'm not a fan, and for my purposes have just gone with __N/A__ (Markdown..)
I see lots of answers going head-on at the "Not Applicable" abbreviation, without exploring what a symbol is. A quick search for the equivalent phrase "out of scope" brings up a couple of variations on the No symbol: ⃠ – this seems to fit the bill (and since I was looking for a way to represent inapplicability, I'll be using it in my technical document).
Per the Wikipedia article, the Unicode codepoint U+20E0 is a combining character, so it is superimposed on the preceding character; e.g. ! ⃠ overlays an exclamation point. To get it to appear isolated, use a non-breaking space
If you don't want to bother with the combining symbol, the article mentions there's also an emoji U+1F6AB 🚫 but it's typically going to be colored red, or won't render!
There's actually a single character that could be repurposed for this: the "Square Na" character ㎁ (U+3381), which is used to represent the nanoampere in fullwidth (CJK) scripts.
What about the "SYMBOL FOR NULL" ␀ (U+2400)?
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.
Can someone please tell me how to determine the unicode character point of a multi-key combination that includes the "command" key? For example, if a user presses the "command" key and "1" key on the keyboard at the same time, what is the unicode character representation for that?
Maybe I'm searching on the wrong thing, but I am not able to locate this in the character maps, keyboard references, or unicode tables I find. I can sort out other key combinations (e.g. shift-1) as there is an obvious character output of "!" that I can look up and find that it is U+0021. When I go to character maps or applications the command key always seems to take an action rather than output a character result to screen.
My app is for iOS, which I would expect to be the same as Mac OS X in terms of the unicode code point. All of the iOS APIs that provide access to the keyboard see it as a source of Unicode characters. Thus the reason I am trying to detect keystrokes this way.
Thanks.
Keyboard codes are basically independent of character codes.
While (as you mention) many keys have standard mappings to standard ASCII codes, it is up to the application to decide what to do with them.
Some input API's may be widely used on a particular OS, and some applications (e.g., terminal emulators) may be used as a common input method for a class of tasks, but there is no universal standard.
Obligatory wikipedia link for Unicode input.
You can't. There simply are no Unicode codepoints that correspond to Command + some-other-character.
The same is true of Shift, by the way. The fact that your computer happens to map certain combinations to certain Unicode codepoints does not imply that Unicode specifies such mappings, or that mappings exist for every combination of keys, or that those mappings are the same for everyone else. I use two keyboards every day; one of them maps Shift+3 to #, the other maps it to £. This is decided by the operating system, not by Unicode. If you tried to detect a Shift+3 keypress by listening for #, your program would seem to me to be broken half the time.
This is a perfect example of an XY question. You don't really care about Unicode -- what you really want to know is how to detect keypresses with the Command modifier on iOS. You should just have asked how to do that! There is probably an API that does exactly what you need that you have simply missed, because you were concentrating on your assumption that the solution would involve Unicode -- and there are probably numerous iOS experts who have not bothered to read this question at all, because they thought your problem related to Unicode rather than iOS.
Simple answer: no.
You haven't told us what sort of computer you are using. Mapping a key press to a Unicode code point is operating system specific, and then it depends on the locale that is active.