digits

问题 I have to use SVM classifier on digits dataset. The dataset consists of images of digits 28x28 and a toal of 2000 images. I tried to use svmtrain but the matlab gave an error that svmtrain has been removed. so now i am using fitcsvm. My code is as below: labelData = zeros(2000,1); for i=1:1000 labelData(i,1)=1; end for j=1001:2000 labelData(j,1)=1; end SVMStruct =fitcsvm(trainingData,labelData) %where training data is the set of images of digits. I need to know how i can predict the outputs

问题 Given a string such as September 3 September 3 September 2 September 1 September 10 August 14 I want to have an output of September 03 September 03 September 02 September 01 September 10 August 14 Tried some regular expressions with no luck :/ 回答1: use this simple pattern \b(\d)\b and replace w/ 0$1 Demo # \b(\d)\b \b # <word boundary> ( # Capturing Group (1) \d # <digit 0-9> ) # End of Capturing Group (1) \b # <word boundary> 回答2: You could try finding only numbers which are single digit

问题 Given a string such as September 3 September 3 September 2 September 1 September 10 August 14 I want to have an output of September 03 September 03 September 02 September 01 September 10 August 14 Tried some regular expressions with no luck :/ 回答1: use this simple pattern \b(\d)\b and replace w/ 0$1 Demo # \b(\d)\b \b # <word boundary> ( # Capturing Group (1) \d # <digit 0-9> ) # End of Capturing Group (1) \b # <word boundary> 回答2: You could try finding only numbers which are single digit

问题 I have a file.txt with hundreds of numbers. They have many digits (max 20) after the point and I need to get them all without truncation, otherwise they introduce errors in the following computations. I made these numbers with matlab so it has a monstrous precision but now I must replicate this behaviour in my program. I've done this way: fstream in; in.open(file.txt, ios::in); long double number; in>>number; I also tried this in.precision(20); in>>number; before each ">>" operation but it is

问题 Am trying to create a well-optimised bit of code to create number of X-digits in length (where X is read from a runtime properties file), based on a DB-generated sequence number (Y), which is then used a folder-name when saving a file. I've come up with three ideas so far, the fastest of which is the last one, but I'd appreciate any advice people may have on this... 1) Instantiate a StringBuilder with initial capacity X. Append Y. While length < X, insert a zero at pos zero. 2) Instantiate a

问题 I have a lines like these: ORIGINAL sometext1 sometext2 word:A12 B34 C56 sometext3 sometext4 sometext5 sometext6 word:A123 B45 C67 sometext7 sometext8 sometext9 sometext10 anotherword:(someword1 someword2 someword3) sometext11 sometext12 EDITED asdjfkklj lkdsjfic kdiw:A12 B34 C56 lksjdfioe sldkjflkjd lknal niewoc kdiw:A123 B45 C678 oknes lkwid cnqule nkdal anotherword:(kdlklks inlqok mncvmnx) unqieo lksdnf Desired output: asdjfkklj lkdsjfic kdiw:A12-B34-C56 lksjdfioe sldkjflkjd lknal niewoc

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 4 years ago . Is there any efficient way (without converting the float into a string) to obtain the number of digits a floating-point number consists of (independent of its length and precision) ? On that way I can implement a fairly good, portable, problematic-less function for comparison/conditioning by

问题 This problem is really confusing me; we're given two integers A , B , we want to count occurrences of digits in the range [A, B] . I though that if we could count the number of digit occurrences in the range [0, A] and [0, B] , then the rest is trivial. So how can I count digit occurrences in a range [0, x] ? This isn't homework, this is actually a problem from SPOJ. The naive approach won't work, as A and B can be as large as 10 ^ 9. Here are some examples: Input: 1 10 Output: 1 2 1 1 1 1 1

问题 I have two arrays (dividend, divisor): dividend[] = {1,2,0,9,8,7,5,6,6}; divisor[] = {9,8}; I need the result (dividend/divisor) as: quotient[] = {1,2,3,4,5,6,7}; I did this using array subtraction: subtract divisor from dividend until dividend becomes 0 or less than divisor, each time incrementing quotient by 1, but it takes a huge time. Is there a better way to do this? 回答1: Is there a better way to do this? You can use long division. 回答2: Do long division. Have a temporary storage of size

问题 How can we find the number of numbers less than a given number with no repeating digits in it? For example the number of such numbers less than 100 is 90. (11, 22, 33,44, 55,66,77,88,99 have repeating digits so are excluded). Similarly for less than 1000, digits like 101, 110, 122, 202 etc have to be excluded. 回答1: Here is a way to make it quicker. Notice that there is a correlation between the number of digits in the max number and the solution (number of numbers which I will call NON ) 100