ruby-on-rails,double,range,digit

You could add something like this: "%02d" % number It will return for example "%02d" % 1 => "01" "%02d" % 51 => "51" For your example it could be: <%= f.select :credit, (0..500).map { |i| "%03d" % i } %> ...

c++,language-lawyer,c++14,digit

There is a previous paper, n3499, which tell us that although Bjarne himself suggested spaces as separators: While this approach is consistent with one common typeographic style, it suffers from some compatibility problems. It does not match the syntax for a pp-number, and would minimally require extending that syntax. More...

c,arrays,digit,kernighan-and-ritchie

TL;DR: a "char" is just a one-byte-long integer. I don't understand how can these values be used in arithmetical expressions if they are characters, is it because they are mapped to numerical values? In C, a char is the "smallest addressable unit of the machine that can contain basic character...

So far its working till digit 32, and then, an error occurs. Digit 31 is wrong too, it should be 5 not a 4, and digit 32 should be a 0. When you get a 10 digit you need to carry over 1 to the previous digit and change...

Anchors are must important here. ^.*? will do a non-greedy match of zero or more characters until (?P<number>[0-9]+) the last number. And the last number is captured by the second group. ^(?P<prefix>.*?)(?P<number>[0-9]+)$ DEMO...

c++,function,extract,palindrome,digit

There are many ways to solve this. I like most the solution that builds the mirror number and checks whether it is identical to the original (even though, it is arguably not the most efficient way). The code should be something like: bool isPalindrom(int n) { int original = n;...

python,string,integer,boolean,digit

str objects have an isdigit method which accomplishes one of your tasks. From a broader perspective, it's better to just try it and see: def digit(s): try: int(s) return True except ValueError: return False For example, " 1234 ".isdigit() is False (there are spaces), but python can convert it to...

Have a look on this code (works with g++) with -O3 #include<iostream> #include<cctype> #include<ctime> #include <time.h> #include <sys/time.h> using namespace std; static inline bool is_digit(char c) { return c>='0'&&c<='9'; } int main() { char c='8'; struct timeval tvSt, tvEn; time_t t1=clock(),t2,t3; gettimeofday(&tvSt, 0); for(int i=0;i<1e9;i++) is_digit(c); gettimeofday(&tvEn, 0); cout <<...

You can use the modulo operator n % 10. For example: 18 % 10 # => 8 9 % 10 # => 9 0 % 10 # => 0 ...

You can do it like this : First define a simple function : bool is_a_bad_line(const std::vector<int>& l) { return l.size() < 14; } Then : std::vector< std::vector<int> > data; std::ifstream f("inputfile.txt"); std::string line; while(std::getline(f,line)) { std::vector<int> line_data; std::istringstream iss(line); int value; while(iss >> value) line_data.push_back(value); data.push_back(line_data); } std::remove_if(data.begin(), data.end(), is_bad_line);...

The keyup event fires only after the keyboard key has been released, and then the character has made its way into the textarea/input element already. keydown on the other hand fires as soon as the key is pressed, and if you cancel the event then, the charater will not even...

computer-science,computer-architecture,digit,bcd,vlsi

A Mod 4 in the original equation gives the two least significant bits of the BCD digit A. So, let's consider the two most and least significant bits of A separately, like so: X = A AND 3 = A Mod 4 (two least significant bits) Y = A AND...

javascript,split,zipcode,digit

http://jsfiddle.net/jeffreyTang/8038whkt/ var re = /[0-9]{6}/; var str = '7 CITE VANEAU - 750007 PARIS'; // get the index of the dash var dash = str.indexOf('-'); // remove everything before the dash str = str.substring(dash); // execute the pattern match var m = re.exec(str); // this is your answer console.log(m[0]); ...

javascript,numbers,int,square,digit

function sq(n){ var nos = (n + '').split(''); var res=""; for(i in nos){ res+= parseInt(nos[i]) * parseInt(nos[i]); } return parseInt(res); } var result = sq(21); alert(result) ...

Possible error in your code is: you are trying to replace with int type data in the string directly. Also only checking numbers.isdigit() will modify all numbers irrespective of what you specified to modify. Possible corrections to your code to make it run properly: def incrementNumbers(statement): number1 = int(input('What number...

Here's one option of doing what @Ronald has proposed: specify a formatting vector, and then mapply it to your data frame. Note that all columns are now characters. sprintf_formats <- c(rep("%.2f", 3), rep("%.2e", 2), "%.2f") Ind_B_sprintf <- Ind_B Ind_B_sprintf[] <- mapply(sprintf, sprintf_formats, Ind_B) Ind_B_sprintf # logFC AveExpr t P.Value adj.P.Val...

java,function,recursion,sum,digit

You aren't actually recursing. You've setup your base case, but when solving a problem recursively you should have code within the function that calls the function itself. So you're missing a line that calls sumDigits, with a smaller argument. Consider two important parts of a recursive soutlion: The base case....

To complete my comment, you can create your own function by modifying a bit the original function, like this: draw.quintuple.venn_mod <- function (area1, area2, area3, area4, area5, n12, n13, n14, n15, n23, n24, n25, n34, n35, n45, n123, n124, n125, n134, n135, n145, n234, n235, n245, n345, n1234, n1235, n1245,...

Although personally I do not see the sense doing this. I would create a string variable to hold the leading zeros before performing math operations. initialNumber = (raw_input("Enter your 10 digit number: ")) while len(str(initialNumber)) != 10: print("Next time please enter a 10 digit numerical value excluding decimals...") time.sleep(1) initialNumber...

It's been a long time since I've touched TI-Basic, however, I do know that there is a mathematical means of counting digits of a number. Since each place is a multiple of 10, you should just be able to use the (base 10 log of your number) plus one. This...