rounding

问题 I have the following function that rounds a number to the nearest number ending with the digits of $nearest , and I was wondering if there is a more elegant way of doing the same. /** * Rounds the number to the nearest digit(s). * * @param int $number * @param int $nearest * @return int */ function roundNearest($number, $nearest, $type = null) { $result = abs(intval($number)); $nearest = abs(intval($nearest)); if ($result <= $nearest) { $result = $nearest; } else { $ceil = $nearest - substr(

问题 I am just trying to use the ceil or round functions in Swift but am getting a compile time error: Ambiguous reference to member 'ceil'. I have already imported the Foundation and UIKit modules. I have tried to compile it with and without the import statements but no luck. Does anyone one have any idea what I am doing wrong? my code is as follow; import UIKit @IBDesignable class LineGraphView: GraphView { override func setMaxYAxis() { self.maxYAxis = ceil(yAxisValue.maxElement()) } } 回答1: This

问题 I am trying to add round corners to a jpeg file, but the problem is that after adding round corners, I am getting a black background color. Somehow I am not able to change it to any other color (white, transparent, red). It just simply shows black background where the image has rounded corners. The code that I am using is: <?php $image = new Imagick('example.jpg'); $image->setBackgroundColor("red"); $image->setImageFormat("jpg"); $image->roundCorners(575,575); $image->writeImage("rounded.jpg"

问题 How can I round up a complex number (e.g. 1.9999999999999998-2j ) as 2-2j ? When I tried using print(round(x,2)) it showed Traceback (most recent call last): File "C:\Python34\FFT.py", line 22, in <module> print(round(x,2)) TypeError: type complex doesn't define __round__ method 回答1: Round real part and imaginary part separately and combine them: >>> num = 1.9999999999999998-2j >>> round(num.real, 2) + round(num.imag, 2) * 1j (2-2j) 回答2: If all you want to do is represent the value rounded as

问题 There is something that baffles me with integer arithmetic in tutorials. To be precise, integer division. The seemingly preferred method is by casting the divisor into a float, then rounding the float to the nearest whole number, then cast that back into integer: #include <cmath> int round_divide_by_float_casting(int a, int b){ return (int) std::roundf( a / (float) b); } Yet this seems like scratching your left ear with your right hand. I use: int round_divide (int a, int b){ return a / b + a

问题 I have a value like this: 421.18834 And I have to round it mathematical correctly with a mask which can look like this: 0.05 0.04 0.1 For example, if the mask is 0.04, i have to get the value 421.20 , because .18 is nearer at .20 than .16. All functions that I found using Google didn't work. Can you please help me? 回答1: double initial = 421.18834; double range = 0.04; int factor = Math.round(initial / range); // 10530 - will round to correct value double result = factor * range; // 421.20 回答2

问题 Are there any drawbacks to this code, which appears to be a faster (and correct) version of java.lang.Math.round ? public static long round(double d) { if (d > 0) { return (long) (d + 0.5d); } else { return (long) (d - 0.5d); } } It takes advantage of the fact that, in Java, truncating to long rounds in to zero. 回答1: There are some special cases which the built in method handles, which your code does not handle. From the documentation: If the argument is NaN , the result is 0. If the argument

问题 I want to know if there is a simple function that I can use such this sample. I have a float value = 1.12345; I want to round it with calling something like float value2 = [roundFloat value:value decimal:3]; NSLog(@"value2 = %f", value2); And I get "1.123" Is there any Library or default function for that or I should write a code block for this type of calculations? thank for your help in advance 回答1: Using NSLog(@"%f", theFloat) always outputs six decimals, for example: float theFloat = 1;

问题 Say I have an array of integers: arr = [0,5,7,8,11,16] and I have another integer: n = 6 I need a function that rounds down to the nearest number from the array: foo(n) #=> 5 As you can see, the numbers do not have a fixed pattern. What's an elegant way to do this? Thanks 回答1: Use select followed by max: arr = [0,5,7,8,11,16] puts arr.select{|item| item < 6}.max Result: 5 This runs in linear time and doesn't require that the array is sorted. 回答2: If you are using relatively small arrays (and

问题 Say I have an array of integers: arr = [0,5,7,8,11,16] and I have another integer: n = 6 I need a function that rounds down to the nearest number from the array: foo(n) #=> 5 As you can see, the numbers do not have a fixed pattern. What's an elegant way to do this? Thanks 回答1: Use select followed by max: arr = [0,5,7,8,11,16] puts arr.select{|item| item < 6}.max Result: 5 This runs in linear time and doesn't require that the array is sorted. 回答2: If you are using relatively small arrays (and