Making a list of integers more human friendly - algorithm
This is a bit of a side project I have taken on to solve a no-fix issue for work. Our system outputs a code to represent a combination of things on another thing. Some example codes are:
9-9-0-4-4-5-4-0-2-0-0-0-2-0-0-0-0-0-2-1-2-1-2-2-2-4
9-5-0-7-4-3-5-7-4-0-5-1-4-2-1-5-5-4-6-3-7-9-72
9-15-0-9-1-6-2-1-2-0-0-1-6-0-7
The max number in one of the slots I've seen so far is about 150 but they will likely go higher.
When the system was designed there was no requirement for what this code would look like. But now the client wants to be able to type it in by hand from a sheet of paper, something the code above isn't suited for. We've said we won't do anything about it, but it seems like a fun challenge to take on.
My question is where is a good place to start loss-less compressing this code? Obvious solutions such as store this code with a shorter key are not an option; our database is read only. I need to build a two way method to make this code more human friendly.
1) I agree that you definately need a checksum - data entry errors are very common, unless you have really well trained staff and independent duplicate keying with automatic crosss-checking.
2) I suggest http://en.wikipedia.org/wiki/Huffman_coding to turn your list of numbers into a stream of bits. To get the probabilities required for this, you need a decent sized sample of real data, so you can make a count, setting Ni to the number of times number i appears in the data. Then I suggest setting Pi = (Ni + 1) / (Sum_i (Ni + 1)) - which smooths the probabilities a bit. Also, with this method, if you see e.g. numbers 0-150 you could add a bit of slack by entering numbers 151-255 and setting them to Ni = 0. Another way round rare large numbers would be to add some sort of escape sequence.
3) Finding a way for people to type the resulting sequence of bits is really an applied psychology problem but here are some suggestions of ideas to pinch.
3a) Software licences - just encode six bits per character in some 64-character alphabet, but group characters in a way that makes it easier for people to keep place e.g. BC017-06777-14871-160C4
3b) UK car license plates. Use a change of alphabet to show people how to group characters e.g. ABCD0123EFGH4567IJKL...
3c) A really large alphabet - get yourself a list of 2^n words for some decent sized n and encode n bits as a word e.g. GREEN ENCHANTED LOGICIAN... -
i worried about this problem a while back. it turns out that you can't do much better than base64 - trying to squeeze a few more bits per character isn't really worth the effort (once you get into "strange" numbers of bits encoding and decoding becomes more complex). but at the same time, you end up with something that's likely to have errors when entered (confusing a 0 with an O etc). one option is to choose a modified set of characters and letters (so it's still base 64, but, say, you substitute ">" for "0". another is to add a checksum. again, for simplicity of implementation, i felt the checksum approach was better.
unfortunately i never got any further - things changed direction - so i can't offer code or a particular checksum choice.
ps i realised there's a missing step i didn't explain: i was going to compress the text into some binary form before encoding (using some standard compression algorithm). so to summarize: compress, add checksum, base64 encode; base 64 decode, check checksum, decompress.
This is similar to what I have used in the past. There are certainly better ways of doing this, but I used this method because it was easy to mirror in Transact-SQL which was a requirement at the time. You could certainly modify this to incorporate Huffman encoding if the distribution of your id's is non-random, but it's probably unnecessary.
You didn't specify language, so this is in c#, but it should be very easy to transition to any language. In the lookup you'll see commonly confused characters are omitted. This should speed up entry. I also had the requirement to have a fixed length, but it would be easy for you to modify this.
static public class CodeGenerator
{
static Dictionary<int, char> _lookupTable = new Dictionary<int, char>();
static CodeGenerator()
{
PrepLookupTable();
}
private static void PrepLookupTable()
{
_lookupTable.Add(0,'3');
_lookupTable.Add(1,'2');
_lookupTable.Add(2,'5');
_lookupTable.Add(3,'4');
_lookupTable.Add(4,'7');
_lookupTable.Add(5,'6');
_lookupTable.Add(6,'9');
_lookupTable.Add(7,'8');
_lookupTable.Add(8,'W');
_lookupTable.Add(9,'Q');
_lookupTable.Add(10,'E');
_lookupTable.Add(11,'T');
_lookupTable.Add(12,'R');
_lookupTable.Add(13,'Y');
_lookupTable.Add(14,'U');
_lookupTable.Add(15,'A');
_lookupTable.Add(16,'P');
_lookupTable.Add(17,'D');
_lookupTable.Add(18,'S');
_lookupTable.Add(19,'G');
_lookupTable.Add(20,'F');
_lookupTable.Add(21,'J');
_lookupTable.Add(22,'H');
_lookupTable.Add(23,'K');
_lookupTable.Add(24,'L');
_lookupTable.Add(25,'Z');
_lookupTable.Add(26,'X');
_lookupTable.Add(27,'V');
_lookupTable.Add(28,'C');
_lookupTable.Add(29,'N');
_lookupTable.Add(30,'B');
}
public static bool TryPCodeDecrypt(string iPCode, out Int64 oDecryptedInt)
{
//Prep the result so we can exit without having to fiddle with it if we hit an error.
oDecryptedInt = 0;
if (iPCode.Length > 3)
{
Char[] Bits = iPCode.ToCharArray(0,iPCode.Length-2);
int CheckInt7 = 0;
int CheckInt3 = 0;
if (!int.TryParse(iPCode[iPCode.Length-1].ToString(),out CheckInt7) ||
!int.TryParse(iPCode[iPCode.Length-2].ToString(),out CheckInt3))
{
//Unsuccessful -- the last check ints are not integers.
return false;
}
//Adjust the CheckInts to the right values.
CheckInt3 -= 2;
CheckInt7 -= 2;
int COffset = iPCode.LastIndexOf('M')+1;
Int64 tempResult = 0;
int cBPos = 0;
while ((cBPos + COffset) < Bits.Length)
{
//Calculate the current position.
int cNum = 0;
foreach (int cKey in _lookupTable.Keys)
{
if (_lookupTable[cKey] == Bits[cBPos + COffset])
{
cNum = cKey;
}
}
tempResult += cNum * (Int64)Math.Pow((double)31, (double)(Bits.Length - (cBPos + COffset + 1)));
cBPos += 1;
}
if (tempResult % 7 == CheckInt7 && tempResult % 3 == CheckInt3)
{
oDecryptedInt = tempResult;
return true;
}
return false;
}
else
{
//Unsuccessful -- too short.
return false;
}
}
public static string PCodeEncrypt(int iIntToEncrypt, int iMinLength)
{
int Check7 = (iIntToEncrypt % 7) + 2;
int Check3 = (iIntToEncrypt % 3) + 2;
StringBuilder result = new StringBuilder();
result.Insert(0, Check7);
result.Insert(0, Check3);
int workingNum = iIntToEncrypt;
while (workingNum > 0)
{
result.Insert(0, _lookupTable[workingNum % 31]);
workingNum /= 31;
}
if (result.Length < iMinLength)
{
for (int i = result.Length + 1; i <= iMinLength; i++)
{
result.Insert(0, 'M');
}
}
return result.ToString();
}
}
Related
How to find a frequent character in a string written in pseudocode. Thanks
Most Frequent Character Design a program that prompts the user to enter a string, and displays the character that appears most frequently in the string. It is a homework question, but my teacher wasn't helpful and its driving me crazy i can't figure this out. Thank You in advance. This is what i have so far! Declare String str Declare Integer maxChar Declare Integer index Set maxChar = 0 Display “Enter anything you want.” Input str For index = 0 To length(str) – 1 If str[index] = And now im stuck. I dont think its right and i dont know where to go with it!
It seems to me that the way you want to do it is: "Go through every character in the string and remember the character we've seen most times". However, that won't work. If we only remember the count for a single character, like "the character we've seen most times is 'a' with 5 occurrences", we can't know if perhaps the character in the 2nd place doesn't jump ahead. So, what you have to do is this: Go through every character of the string. For every character, increase the occurrence count for that character. Yes, you have to save this count for every single character you encounter. Simple variables like string or int are not going to be enough here. When you're done, you're left with a bunch of data looking like "a"=5, "b"=2, "e"=7,... you have to go though that and find the highest number (I'm sure you can find examples for finding the highest number in a sequence), then return the letter which this corresponds to. Not a complete answer, I know, but that's all I'm going to say. If you're stuck, I suggest getting a pen and a piece of paper and trying to calculate it manually. Try to think - how would you do it without a computer? If your answer is "look and see", what if the text is 10 pages? I know it can be pretty confusing, but the point of all this is to get you used to a different way of thinking. If you figure this one out, the next time will be easier because the basic principles are always the same.
This is the code I have created to count all occurences in a string. String abc = "aabcabccc"; char[] x = abc.toCharArray(); String _array = ""; for(int i = 0; i < x.length; i++) //copy distinct data to a new string { if(_array.indexOf(x[i]) == -1) _array = _array+x[i]; } char[] y = _array.toCharArray(); int[] count1 = new int[_array.length()]; for(int j = 0; j<x.length;j++) //count occurences { count1[new String(String.valueOf(y)).indexOf(x[j])]++; } for(int i = 0; i<y.length;i++) //display { System.out.println(y[i] + " = " + count1[i]); }
Converting an if code into forloop statement
Right now i have to write a code that will print out "*" for each collectedDots, however if it doesn't collect any collectedDots==0 then it print out "". Using too many if statements look messy and i was wandering how you would implement the forloop in this case.
As a general principle the kind of rearrangement you've done here is good. You have found a way to express the rule in a general way rather than as a sequence of special cases. This is much easier to reason about and to check, and it's obviously extensible to cases where you have more than 3 dots. You probably have made an error in confusing your target number and the iteration value, I assume that collectedDots contains the number of dots you have (as per your if statement) and so you need to introduce a variable to count up to that value for (int i =0; i <= collectedDots; i++) { stars = "*"; System.out.print(stars); }
Ok, so you already have a variable called collectedDots that is a number which tells you how many stars to print? So your loop would be something like for every collected dot print * But you can't just print it out, you need to return a string that will be printed out. So it's more like for every collected dot add a * to our string return the string They key difference between this and your attempt so far is that you were assigning a star to be your string each time through the loop, then at the end of it, you return that string–no matter how many times you assign a star to the string, the string will always just be one star. You also need a separate variable to keep track of your loop, this should do the trick: String stars = ""; for(int i = 0; i < collectedDots; i++) { stars = stars + "*"; } return stars;
You are almost correct. Just need to change range limit of looping. Looping initial value is set to 1. So whenever you have collectedDots = 0, it will not go in loop and will return "", as stars is intialized with "" before loop. String stars = ""; for (int i =1; i <= collectedDots; i++) { stars = "*"; System.out.print(stars); } return stars;
Generating nice looking BETA keys
I built a web application that is going to launch a beta test soon. I would really like to hand out beta invites and keys that look nice. i.e. A3E6-7C24-9876-235B This is around 16 character, hexadecimal digits. It looks like the typical beta key you might see. My question is what is a standard way to generate something like this and make sure that it is unique and that it will not be easy for someone to guess a beta key and generate their own. I have some ideas that would probably work for beta keys: MD5 is secure enough for this, but it is long and ugly looking and could cause confusion between 0 and O, or 1 and l. I could start off with a large hexadecimal number that is 16 digits in length. To prevent people from guessing what the next beta key might be increment the value by a random number each time. The range of numbers between 1111-1111-1111-1111 and eeee-eeee-eeee-eeee will have plenty of room to spare even if I am skipping large quantities of numbers. I guess I am just wondering if there is a standard way for doing this that I am not finding with google. Is there a better way?
The canonical "unique identifying number" is a uuid. There are various forms - you can generate one from random numbers (version 4) or from a hash of some value (user's email + salt?) (versions 3 and 5), for example. Libraries for java, python and a bunch more exist. PS I have to add that when I read your question title I thought you were looking for something cool and different. You might consider using an "interesting" word list and combining words with hyphens to encode a number (based on hash of email + salt). That would be much more attractive imho: "your beta code is secret-wombat-cookie-ninja" (I'm sure I read an article describing an example, but I can't find it now).
One way (C# but the code is simple enough to port to other languages): private static readonly Random random = new Random(Guid.NewGuid().GetHashCode()); static void Main(string[] args) { string x = GenerateBetaString(); } public static string GenerateBetaString() { const string alphabet = "ABCDEF0123456789"; string x = GenerateRandomString(16, alphabet); return x.Substring(0, 4) + "-" + x.Substring(4, 4) + "-" + x.Substring(8, 4) + "-" + x.Substring(12, 4); } public static string GenerateRandomString(int length, string alphabet) { int maxlen = alphabet.Length; StringBuilder randomChars = new StringBuilder(length); for (int i = 0; i < length; i++) { randomChars.Append(alphabet[random.Next(0, maxlen)]); } return randomChars.ToString(); } Output: 97A8-55E5-C6B8-959E 8C60-6597-B71D-5CAF 8E1B-B625-68ED-107B A6B5-1D2E-8D77-EB99 5595-E8DC-3A47-0605 Doing this way gives you precise control of the characters in the alphabet. If you need crypto strength randomness (unlikely) use the cryto random class to generate random bytes (possibly mod the alphabet length).
Computing power is cheap, take your idea of the MD5 and run an "aesthetic" of your own devising over the set. The code below generates 2000 unique keys almost instantaneously that do not have a 0,1,L,O character in them. Modify aesthetic to fit any additional criteria: import random, hashlib def potential_key(): x = random.random() m = hashlib.md5() m.update(str(x)) s = m.hexdigest().upper()[:16] return "%s-%s-%s-%s" % (s[:4],s[4:8],s[8:12],s[12:]) def aesthetic(s): bad_chars = ["0","1","L","O"] for b in bad_chars: if b in s: return False return True key_set = set() while len(key_set) < 2000: k = potential_key() if aesthetic(k): key_set.add(k) print key_set Example keys: '4297-CAC6-9DA8-625A', '43DD-2ED4-E4F8-3E8D', '4A8D-D5EF-C7A3-E4D5', 'A68D-9986-4489-B66C', '9B23-6259-9832-9639', '2C36-FE65-EDDB-2CF7', 'BFB6-7769-4993-CD86', 'B4F4-E278-D672-3D2C', 'EEC4-3357-2EAB-96F5', '6B69-C6DA-99C3-7B67', '9ED7-FED5-3CC6-D4C6', 'D3AA-AF48-6379-92EF', ...
How to convert Chinese characters to Pinyin [closed]
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. We don’t allow questions seeking recommendations for books, tools, software libraries, and more. You can edit the question so it can be answered with facts and citations. Closed last year. The community reviewed whether to reopen this question 4 months ago and left it closed: Original close reason(s) were not resolved Improve this question For sorting Chinese language text, I want to convert Chinese characters to Pinyin, properly separating each Chinese character and grouping successive characters together. Can you please help me in this task by providing the logic or source code for doing this? Please let me know if any open source or lib already present for this.
Short answer: you don't. Long answer: There is no one-to-one mapping for 汉字 to 汉语拼音. Just some quick examples: 把 can be "ba" in the third tone or fourth tone. 了 can be "le" toneless or "liao" third tone. 乐 can be "le" or "yue", both in the fourth tone. 落 can be "luo", "la" or "lao", all in the fourth tone. And so on. I have a beginners' book on this topic that has 207 examples. I stress that this is a beginners' book and is by no means complete. Each one has a page or two of examples of use and conditions under which you choose the appropriate pronunciation. It is not something that could be easily programmed (if at all). And this doesn't even address the other slippery thing you want to deal with: the separation of characters into grouped words. The very notion of a word is a bit slippery in Chinese. (There's two terms that correspond, roughly to "word" in Chinese for example: 字 and 词. The first is the character, the second groups of characters that are put together into one concept. (I frequently get asked by Chinese speakers how many "words" I can read when they really mean "characters".) While in some cases the distinction is clear (the 词 "乌鸦", for example, is "crow" -- the two 字 must be together to express the idea properly and it would be incorrect to translate it as "black crow"), in others it is not so clear. What does "你好" translate to? Is it one word meaning, idiomatically, "hello"? Or is it two words translating literally to "you good"? Each of the characters involved stands alone or in groups with other words, but together they mean something entirely different from their individual meanings. Given this, how, precisely, do you plan to group the 汉语拼音 transliterations (which are difficult to impossible to get right in the first place!) into "words"?
While #JUST MY correct OPINION's answer addresses some of the difficulties of converting characters into pinyin, it is not an impossible problem to solve. I have written a library (pinyinify) that solves this task with decent accuracy. Even though there is not a one-to-one mapping between characters and pinyin, my library can usually decide which pronunciation is correct. For example, "我受不了了" correctly converts to "wǒ shòubùliǎo le", with two different pronunciations of 了. My approach to solving the problem is pretty simple: First segment the text into words. For example, 我喜欢旅游 would be divided into three words: 我 喜欢 旅游. This is also not a simple process, but there are many libraries for it. jieba is one of the more popular libraries for this purpose. Use a dictionary to convert the words into pinyin. If the word is not in the dictionary, fall back to converting the individual characters to pinyin using their most common pronunciation.
CoreFoundation provides certain method to do the conversion: CFMutableStringRef string = CFStringCreateMutableCopy(NULL, 0, CFSTR("中文")); CFStringTransform(string, NULL, kCFStringTransformMandarinLatin, NO); CFStringTransform(string, NULL, kCFStringTransformStripDiacritics, NO); NSLog(#"%#", string); The output is zhong wen
the following code writing in C# can help you to simply convert chinese words that including in gb2312 encodec(just 2312 of often used Simplified-Chinese words) to pinyin.like convert "今天天气不错" to "JinTianTianQiBuCuo". sometimes a chinese word is not one to one map to a pinyin,it depends on the context we talk about.like the "行" in "自行车"(bike) is pronounced "Xing",but in "银行"(bank) it pronounced "Hang".so if you have problem with this,you may find more complex solution to handle this. sorry for my poor english.i hope this could give you a little help. public class ChineseToPinYin { private static int[] pyValue = new int[] { -20319,-20317,-20304,-20295,-20292,-20283,-20265,-20257,-20242,-20230,-20051,-20036, -20032,-20026,-20002,-19990,-19986,-19982,-19976,-19805,-19784,-19775,-19774,-19763, -19756,-19751,-19746,-19741,-19739,-19728,-19725,-19715,-19540,-19531,-19525,-19515, -19500,-19484,-19479,-19467,-19289,-19288,-19281,-19275,-19270,-19263,-19261,-19249, -19243,-19242,-19238,-19235,-19227,-19224,-19218,-19212,-19038,-19023,-19018,-19006, -19003,-18996,-18977,-18961,-18952,-18783,-18774,-18773,-18763,-18756,-18741,-18735, -18731,-18722,-18710,-18697,-18696,-18526,-18518,-18501,-18490,-18478,-18463,-18448, -18447,-18446,-18239,-18237,-18231,-18220,-18211,-18201,-18184,-18183, -18181,-18012, -17997,-17988,-17970,-17964,-17961,-17950,-17947,-17931,-17928,-17922,-17759,-17752, -17733,-17730,-17721,-17703,-17701,-17697,-17692,-17683,-17676,-17496,-17487,-17482, -17468,-17454,-17433,-17427,-17417,-17202,-17185,-16983,-16970,-16942,-16915,-16733, -16708,-16706,-16689,-16664,-16657,-16647,-16474,-16470,-16465,-16459,-16452,-16448, -16433,-16429,-16427,-16423,-16419,-16412,-16407,-16403,-16401,-16393,-16220,-16216, -16212,-16205,-16202,-16187,-16180,-16171,-16169,-16158,-16155,-15959,-15958,-15944, -15933,-15920,-15915,-15903,-15889,-15878,-15707,-15701,-15681,-15667,-15661,-15659, -15652,-15640,-15631,-15625,-15454,-15448,-15436,-15435,-15419,-15416,-15408,-15394, -15385,-15377,-15375,-15369,-15363,-15362,-15183,-15180,-15165,-15158,-15153,-15150, -15149,-15144,-15143,-15141,-15140,-15139,-15128,-15121,-15119,-15117,-15110,-15109, -14941,-14937,-14933,-14930,-14929,-14928,-14926,-14922,-14921,-14914,-14908,-14902, -14894,-14889,-14882,-14873,-14871,-14857,-14678,-14674,-14670,-14668,-14663,-14654, -14645,-14630,-14594,-14429,-14407,-14399,-14384,-14379,-14368,-14355,-14353,-14345, -14170,-14159,-14151,-14149,-14145,-14140,-14137,-14135,-14125,-14123,-14122,-14112, -14109,-14099,-14097,-14094,-14092,-14090,-14087,-14083,-13917,-13914,-13910,-13907, -13906,-13905,-13896,-13894,-13878,-13870,-13859,-13847,-13831,-13658,-13611,-13601, -13406,-13404,-13400,-13398,-13395,-13391,-13387,-13383,-13367,-13359,-13356,-13343, -13340,-13329,-13326,-13318,-13147,-13138,-13120,-13107,-13096,-13095,-13091,-13076, -13068,-13063,-13060,-12888,-12875,-12871,-12860,-12858,-12852,-12849,-12838,-12831, -12829,-12812,-12802,-12607,-12597,-12594,-12585,-12556,-12359,-12346,-12320,-12300, -12120,-12099,-12089,-12074,-12067,-12058,-12039,-11867,-11861,-11847,-11831,-11798, -11781,-11604,-11589,-11536,-11358,-11340,-11339,-11324,-11303,-11097,-11077,-11067, -11055,-11052,-11045,-11041,-11038,-11024,-11020,-11019,-11018,-11014,-10838,-10832, -10815,-10800,-10790,-10780,-10764,-10587,-10544,-10533,-10519,-10331,-10329,-10328, -10322,-10315,-10309,-10307,-10296,-10281,-10274,-10270,-10262,-10260,-10256,-10254 }; private static string[] pyName = new string[] { "A","Ai","An","Ang","Ao","Ba","Bai","Ban","Bang","Bao","Bei","Ben", "Beng","Bi","Bian","Biao","Bie","Bin","Bing","Bo","Bu","Ba","Cai","Can", "Cang","Cao","Ce","Ceng","Cha","Chai","Chan","Chang","Chao","Che","Chen","Cheng", "Chi","Chong","Chou","Chu","Chuai","Chuan","Chuang","Chui","Chun","Chuo","Ci","Cong", "Cou","Cu","Cuan","Cui","Cun","Cuo","Da","Dai","Dan","Dang","Dao","De", "Deng","Di","Dian","Diao","Die","Ding","Diu","Dong","Dou","Du","Duan","Dui", "Dun","Duo","E","En","Er","Fa","Fan","Fang","Fei","Fen","Feng","Fo", "Fou","Fu","Ga","Gai","Gan","Gang","Gao","Ge","Gei","Gen","Geng","Gong", "Gou","Gu","Gua","Guai","Guan","Guang","Gui","Gun","Guo","Ha","Hai","Han", "Hang","Hao","He","Hei","Hen","Heng","Hong","Hou","Hu","Hua","Huai","Huan", "Huang","Hui","Hun","Huo","Ji","Jia","Jian","Jiang","Jiao","Jie","Jin","Jing", "Jiong","Jiu","Ju","Juan","Jue","Jun","Ka","Kai","Kan","Kang","Kao","Ke", "Ken","Keng","Kong","Kou","Ku","Kua","Kuai","Kuan","Kuang","Kui","Kun","Kuo", "La","Lai","Lan","Lang","Lao","Le","Lei","Leng","Li","Lia","Lian","Liang", "Liao","Lie","Lin","Ling","Liu","Long","Lou","Lu","Lv","Luan","Lue","Lun", "Luo","Ma","Mai","Man","Mang","Mao","Me","Mei","Men","Meng","Mi","Mian", "Miao","Mie","Min","Ming","Miu","Mo","Mou","Mu","Na","Nai","Nan","Nang", "Nao","Ne","Nei","Nen","Neng","Ni","Nian","Niang","Niao","Nie","Nin","Ning", "Niu","Nong","Nu","Nv","Nuan","Nue","Nuo","O","Ou","Pa","Pai","Pan", "Pang","Pao","Pei","Pen","Peng","Pi","Pian","Piao","Pie","Pin","Ping","Po", "Pu","Qi","Qia","Qian","Qiang","Qiao","Qie","Qin","Qing","Qiong","Qiu","Qu", "Quan","Que","Qun","Ran","Rang","Rao","Re","Ren","Reng","Ri","Rong","Rou", "Ru","Ruan","Rui","Run","Ruo","Sa","Sai","San","Sang","Sao","Se","Sen", "Seng","Sha","Shai","Shan","Shang","Shao","She","Shen","Sheng","Shi","Shou","Shu", "Shua","Shuai","Shuan","Shuang","Shui","Shun","Shuo","Si","Song","Sou","Su","Suan", "Sui","Sun","Suo","Ta","Tai","Tan","Tang","Tao","Te","Teng","Ti","Tian", "Tiao","Tie","Ting","Tong","Tou","Tu","Tuan","Tui","Tun","Tuo","Wa","Wai", "Wan","Wang","Wei","Wen","Weng","Wo","Wu","Xi","Xia","Xian","Xiang","Xiao", "Xie","Xin","Xing","Xiong","Xiu","Xu","Xuan","Xue","Xun","Ya","Yan","Yang", "Yao","Ye","Yi","Yin","Ying","Yo","Yong","You","Yu","Yuan","Yue","Yun", "Za", "Zai","Zan","Zang","Zao","Ze","Zei","Zen","Zeng","Zha","Zhai","Zhan", "Zhang","Zhao","Zhe","Zhen","Zheng","Zhi","Zhong","Zhou","Zhu","Zhua","Zhuai","Zhuan", "Zhuang","Zhui","Zhun","Zhuo","Zi","Zong","Zou","Zu","Zuan","Zui","Zun","Zuo" }; /// <summary> /// 把汉字转换成拼音(全拼) /// </summary> /// <param name="hzString">汉字字符串</param> /// <returns>转换后的拼音(全拼)字符串</returns> public static string Convert(string hzString) { // 匹配中文字符 Regex regex = new Regex("^[\u4e00-\u9fa5]$"); byte[] array = new byte[2]; string pyString = ""; int chrAsc = 0; int i1 = 0; int i2 = 0; char[] noWChar = hzString.ToCharArray(); for (int j = 0; j < noWChar.Length; j++) { // 中文字符 if (regex.IsMatch(noWChar[j].ToString())) { array = System.Text.Encoding.Default.GetBytes(noWChar[j].ToString()); i1 = (short)(array[0]); i2 = (short)(array[1]); chrAsc = i1 * 256 + i2 - 65536; if (chrAsc > 0 && chrAsc < 160) { pyString += noWChar[j]; } else { // 修正部分文字 if (chrAsc == -9254) // 修正“圳”字 pyString += "Zhen"; else { for (int i = (pyValue.Length - 1); i >= 0; i--) { if (pyValue[i] <= chrAsc) { pyString += pyName[i]; break; } } } } } // 非中文字符 else { pyString += noWChar[j].ToString(); } } return pyString; } }
You can use the following method: from __future__ import unicode_literals from pypinyin import lazy_pinyin hanzi_list = ['如何', '将', '汉字','转为', '拼音'] pinyin_list = [''.join(lazy_pinyin(_)) for _ in hanzi_list] Output: ['ruhe', 'jiang', 'hanzi', 'zhuanwei', 'pinyin']
i had this problem and i found a solution in PHP (which could be cleaner i suppose but it works). I had some troubles because the file given in this topic is from hexa unicode. 1) Import the data from ftp://ftp.cuhk.hk/pub/chinese/ifcss/software/data/Uni2Pinyin.gz (thanks pierr) to your database or whatever 2) Import your data in an array as $pinyinArray[$hexaUnicode] = $pinyin; 3) Use this code: /* * Decimal representation of $c * function found there: http://www.cantonese.sheik.co.uk/phorum/read.php?2,19594 */ function uniord($c) { $ud = 0; if (ord($c{0})>=0 && ord($c{0})<=127) $ud = $c{0}; if (ord($c{0})>=192 && ord($c{0})<=223) $ud = (ord($c{0})-192)*64 + (ord($c{1})-128); if (ord($c{0})>=224 && ord($c{0})<=239) $ud = (ord($c{0})-224)*4096 + (ord($c{1})-128)*64 + (ord($c{2})-128); if (ord($c{0})>=240 && ord($c{0})<=247) $ud = (ord($c{0})-240)*262144 + (ord($c{1})-128)*4096 + (ord($c{2})-128)*64 + (ord($c{3})-128); if (ord($c{0})>=248 && ord($c{0})<=251) $ud = (ord($c{0})-248)*16777216 + (ord($c{1})-128)*262144 + (ord($c{2})-128)*4096 + (ord($c{3})-128)*64 + (ord($c{4})-128); if (ord($c{0})>=252 && ord($c{0})<=253) $ud = (ord($c{0})-252)*1073741824 + (ord($c{1})-128)*16777216 + (ord($c{2})-128)*262144 + (ord($c{3})-128)*4096 + (ord($c{4})-128)*64 + (ord($c{5})-128); if (ord($c{0})>=254 && ord($c{0})<=255) //error $ud = false; return $ud; } /* * Translate the $string string of a single chinese charactere to unicode */ function chineseToHexaUnicode($string) { return strtoupper(dechex(uniord($string))); } /* * */ function convertChineseToPinyin($string,$pinyinArray) { $pinyinValue = ''; for ($i = 0; $i < mb_strlen($string);$i++) $pinyinValue.=$pinyinArray[chineseToHexaUnicode(mb_substr($string, $i, 1))]; return $pinyinValue; } $string = '龙江省五大'; echo convertChineseToPinyin($string,$pinyinArray); echo: (long2)(jiang1)(sheng3,xing3)(wu3)(da4,dai4) Of course, $pinyinArray is your array of data (hexoUnicode => pinyin) Hope it will help someone.
If you use Visual Studio, this might be an option: Microsoft.International.Converters.PinYinConverter How to install: First, download the Visual Studio International Pack 2.0, Official Download. Once the download is complete install the run file VSIPSetup.msi installation (x86 operating system on the default installation directory (C:\Program Files\Microsoft Visual Studio International Feature Pack 2.0). After installation, you need to add a reference in VS, respectively reference: C:\Program Files\Microsoft Visual Studio International Pack\Simplified Chinese Pin-Yin Conversion Library (Pinyin) and C:\Program Files\Microsoft Visual Studio International Pack\Traditional Chinese to Simplified Chinese Conversion Library and Add-In Tool (Traditional and Simplified Huzhuan to) How to use: public static string GetPinyin(string str) { string r = string.Empty; foreach (char obj in str) { try { ChineseChar chineseChar = new ChineseChar(obj); string t = chineseChar.Pinyins[0].ToString(); r += t.Substring(0, t.Length - 1); } catch { r += obj.ToString(); } } return r; } Source: http://www.programering.com/a/MzM3cTMwATA.html
How to split a string into words. Ex: "stringintowords" -> "String Into Words"?
What is the right way to split a string into words ? (string doesn't contain any spaces or punctuation marks) For example: "stringintowords" -> "String Into Words" Could you please advise what algorithm should be used here ? ! Update: For those who think this question is just for curiosity. This algorithm could be used to camеlcase domain names ("sportandfishing .com" -> "SportAndFishing .com") and this algo is currently used by aboutus dot org to do this conversion dynamically.
Let's assume that you have a function isWord(w), which checks if w is a word using a dictionary. Let's for simplicity also assume for now that you only want to know whether for some word w such a splitting is possible. This can be easily done with dynamic programming. Let S[1..length(w)] be a table with Boolean entries. S[i] is true if the word w[1..i] can be split. Then set S[1] = isWord(w[1]) and for i=2 to length(w) calculate S[i] = (isWord[w[1..i] or for any j in {2..i}: S[j-1] and isWord[j..i]). This takes O(length(w)^2) time, if dictionary queries are constant time. To actually find the splitting, just store the winning split in each S[i] that is set to true. This can also be adapted to enumerate all solution by storing all such splits.
As mentioned by many people here, this is a standard, easy dynamic programming problem: the best solution is given by Falk Hüffner. Additional info though: (a) you should consider implementing isWord with a trie, which will save you a lot of time if you use properly (that is by incrementally testing for words). (b) typing "segmentation dynamic programming" yields a score of more detail answers, from university level lectures with pseudo-code algorithm, such as this lecture at Duke's (which even goes so far as to provide a simple probabilistic approach to deal with what to do when you have words that won't be contained in any dictionary).
There should be a fair bit in the academic literature on this. The key words you want to search for are word segmentation. This paper looks promising, for example. In general, you'll probably want to learn about markov models and the viterbi algorithm. The latter is a dynamic programming algorithm that may allow you to find plausible segmentations for a string without exhaustively testing every possible segmentation. The essential insight here is that if you have n possible segmentations for the first m characters, and you only want to find the most likely segmentation, you don't need to evaluate every one of these against subsequent characters - you only need to continue evaluating the most likely one.
If you want to ensure that you get this right, you'll have to use a dictionary based approach and it'll be horrendously inefficient. You'll also have to expect to receive multiple results from your algorithm. For example: windowsteamblog (of http://windowsteamblog.com/ fame) windows team blog window steam blog
Consider the sheer number of possible splittings for a given string. If you have n characters in the string, there are n-1 possible places to split. For example, for the string cat, you can split before the a and you can split before the t. This results in 4 possible splittings. You could look at this problem as choosing where you need to split the string. You also need to choose how many splits there will be. So there are Sum(i = 0 to n - 1, n - 1 choose i) possible splittings. By the Binomial Coefficient Theorem, with x and y both being 1, this is equal to pow(2, n-1). Granted, a lot of this computation rests on common subproblems, so Dynamic Programming might speed up your algorithm. Off the top of my head, computing a boolean matrix M such M[i,j] is true if and only if the substring of your given string from i to j is a word would help out quite a bit. You still have an exponential number of possible segmentations, but you would quickly be able to eliminate a segmentation if an early split did not form a word. A solution would then be a sequence of integers (i0, j0, i1, j1, ...) with the condition that j sub k = i sub (k + 1). If your goal is correctly camel case URL's, I would sidestep the problem and go for something a little more direct: Get the homepage for the URL, remove any spaces and capitalization from the source HTML, and search for your string. If there is a match, find that section in the original HTML and return it. You'd need an array of NumSpaces that declares how much whitespace occurs in the original string like so: Needle: isashort Haystack: This is a short phrase Preprocessed: thisisashortphrase NumSpaces : 000011233333444444 And your answer would come from: location = prepocessed.Search(Needle) locationInOriginal = location + NumSpaces[location] originalLength = Needle.length() + NumSpaces[location + needle.length()] - NumSpaces[location] Haystack.substring(locationInOriginal, originalLength) Of course, this would break if madduckets.com did not have "Mad Duckets" somewhere on the home page. Alas, that is the price you pay for avoiding an exponential problem.
This can be actually done (to a certain degree) without dictionary. Essentially, this is an unsupervised word segmentation problem. You need to collect a large list of domain names, apply an unsupervised segmentation learning algorithm (e.g. Morfessor) and apply the learned model for new domain names. I'm not sure how well it would work, though (but it would be interesting).
This is basically a variation of a knapsack problem, so what you need is a comprehensive list of words and any of the solutions covered in Wiki. With fairly-sized dictionary this is going to be insanely resource-intensive and lengthy operation, and you cannot even be sure that this problem will be solved.
Create a list of possible words, sort it from long words to short words. Check if each entry in the list against the first part of the string. If it equals, remove this and append it at your sentence with a space. Repeat this.
A simple Java solution which has O(n^2) running time. public class Solution { // should contain the list of all words, or you can use any other data structure (e.g. a Trie) private HashSet<String> dictionary; public String parse(String s) { return parse(s, new HashMap<String, String>()); } public String parse(String s, HashMap<String, String> map) { if (map.containsKey(s)) { return map.get(s); } if (dictionary.contains(s)) { return s; } for (int left = 1; left < s.length(); left++) { String leftSub = s.substring(0, left); if (!dictionary.contains(leftSub)) { continue; } String rightSub = s.substring(left); String rightParsed = parse(rightSub, map); if (rightParsed != null) { String parsed = leftSub + " " + rightParsed; map.put(s, parsed); return parsed; } } map.put(s, null); return null; } }
I was looking at the problem and thought maybe I could share how I did it. It's a little too hard to explain my algorithm in words so maybe I could share my optimized solution in pseudocode: string mainword = "stringintowords"; array substrings = get_all_substrings(mainword); /** this way, one does not check the dictionary to check for word validity * on every substring; It would only be queried once and for all, * eliminating multiple travels to the data storage */ string query = "select word from dictionary where word in " + substrings; array validwords = execute(query).getArray(); validwords = validwords.sort(length, desc); array segments = []; while(mainword != ""){ for(x = 0; x < validwords.length; x++){ if(mainword.startswith(validwords[x])) { segments.push(validwords[x]); mainword = mainword.remove(v); x = 0; } } /** * remove the first character if any of valid words do not match, then start again * you may need to add the first character to the result if you want to */ mainword = mainword.substring(1); } string result = segments.join(" ");