Maybe this? <?php $num = ‘49.82’; $new_num = $num; $hundredth = substr($num, -1, 1); switch($hundredth) { case ‘0’: break; case ‘1’: $new_num = ($new_num – 0.01); break; case ‘2’: $new_num = ($new_num – 0.02); break; case ‘3’: $new_num = ($new_num – 0.03); break; case ‘4’: $new_num = ($new_num – 0.04); break; case ‘5’: break; case … Read more
Binary floating point math is like this. In most programming languages, it is based on the IEEE 754 standard. The crux of the problem is that numbers are represented in this format as a whole number times a power of two; rational numbers (such as 0.1, which is 1/10) whose denominator is not a power … Read more
Your code is equivalent to: double x1 = 0.9; float xf = (float)x1; // discard some precision double x2 = (double)xf; if (x2 >= x1) { // … } The problem is that, after you discarded some precision from x1 by converting it to float, it can never be recovered; casting it back to double … Read more
Nothing to stop this float fun(float a) { cout << fixed << setprecision(1); return a; } But that’s stupid code for a stupid puzzle. What does this have to do with real programming? 3 solved How to add decimal to floating point number with non-decimal value [closed]
Suppose x is the number you wish to reduce the precision of and bits is the number of significant bits you wish to retain. When bits is sufficiently large and the order of magnitude of x is sufficiently close to 0, then x * (1L << (bits – Math.getExponent(x))) will scale x so that the … Read more
You have to flatten your lists as shown here, and then use classification_report from scikit-learn: correct = [[‘*’,’*’],[‘*’,’PER’,’*’,’GPE’,’ORG’],[‘GPE’,’*’,’*’,’*’,’ORG’]] predicted = [[‘PER’,’*’],[‘*’,’ORG’,’*’,’GPE’,’ORG’],[‘PER’,’*’,’*’,’*’,’MISC’]] target_names = [‘PER’,’ORG’,’MISC’,’LOC’,’GPE’] # leave out ‘*’ correct_flat = [item for sublist in correct for item in sublist] predicted_flat = [item for sublist in predicted for item in sublist] from sklearn.metrics import classification_report print(classification_report(correct_flat, … Read more
Something like as follows. Hope that its understandable. Output:https://www.ideone.com/EnP40j #include <iostream> #include <iomanip> #include <cmath> int main() { float num1 = 95.345f; float num2 = 95.344f; num1 = roundf(num1 * 100) / 100; //rounding two decimals num2 = roundf(num2 * 100) / 100; //rounding two decimals std::cout << std::setprecision(2) << std::fixed << num1 << ” … Read more
You could add std::fixed like so: float f; std::cin >> f; std::cout << std::setw(10) << std::right << std::fixed << std::setprecision(3) << f << “\n”; 1 solved Float formatting C++
Do not use the value pi = 22 / 7 if you want accuracy. Please use M_PI which is defined in math.h. In MSVC you also need #define _USE_MATH_DEFINES to make the following values available. #define M_E 2.71828182845904523536 // e #define M_LOG2E 1.44269504088896340736 // log2(e) #define M_LOG10E 0.434294481903251827651 // log10(e) #define M_LN2 0.693147180559945309417 // ln(2) … Read more