point

问题 I'm working with the image segmentation. I need to compute image gradient in the point, so I was trying to understand by myself (because I know how to calculate gradient when having an ordinary equation) but I failed. I was googling... and googling but to find wright answer I couldn't. Can anyone say or give some information how to compute image gradient in the point step by step? 回答1: The gradient at each pixel of an image I(x, y) is simply the 2D vector (dI/dx,dI/dy)(x, y) . Approximate dI

问题 I'm looking to build a browser multiplayer game using rollback netcode that runs a deterministic simulation on the clients. I prototyped the netcode in Flash already before I ran into the floating point roadblock. Basically, from what I understand, integer math in Flash is done by casting int s to Number s, doing the math, then casting back to int . It's faster apparently, but it means that there's no chance of deterministic math across different computer architectures. Before I dump all my

问题 I'm looking to build a browser multiplayer game using rollback netcode that runs a deterministic simulation on the clients. I prototyped the netcode in Flash already before I ran into the floating point roadblock. Basically, from what I understand, integer math in Flash is done by casting int s to Number s, doing the math, then casting back to int . It's faster apparently, but it means that there's no chance of deterministic math across different computer architectures. Before I dump all my

问题 Suppose I have data in the following form: library(ggplot2) Data <- data.frame( "ID" = c("ABC111", "ABC111", "ABC111", "ABC111", "ABC112", "ABC112", "ABC112", "ABC113", "ABC113", "ABC114", "ABC115"), "color" = c("red", "red", "red", "red", "blue", "blue", "blue", "green", "green", "black", "yellow"), "start_date" = c("2005/01/01", "2006/01/01", "2007/01/01", "2008/01/01", "2009/01/01", "2010/01/01", "2011/01/01", "2012/01/01", "2013/01/01", "2014/01/01", "2015/01/01"), "end_date" = c("2005/09

问题 I want to create a real-time, point plotting GUI. I am using the Scanse Sweep LiDAR, and at each sweep of this LiDAR (working between 1 - 10Hz) I receive approximately 1000 points (x, y) describing the LiDARs surrounding. This is a 2D LiDAR. I have looked everywhere and tried countless of code snippets for pyqtgraph, but either it crashes, is super slow or doesn't work at all. Is there a straight-forward way of creating a plotter window and upon each new scan/data delivery, push those points

问题 I want to create a real-time, point plotting GUI. I am using the Scanse Sweep LiDAR, and at each sweep of this LiDAR (working between 1 - 10Hz) I receive approximately 1000 points (x, y) describing the LiDARs surrounding. This is a 2D LiDAR. I have looked everywhere and tried countless of code snippets for pyqtgraph, but either it crashes, is super slow or doesn't work at all. Is there a straight-forward way of creating a plotter window and upon each new scan/data delivery, push those points

来源： https://stackoverflow.com/questions/49579960/how-to-get-random-point-near-edges-of-a-square-in-javascript

来源： https://stackoverflow.com/questions/49579960/how-to-get-random-point-near-edges-of-a-square-in-javascript

问题 I guess I need to add some explanation, that I want to ask this question, because too short question doesn't quality standards... funny... So, here is the question: How I can get the 'y' coordinate of the point on svg path at a specific 'x' coordinate? 回答1: Well this is not straightforward, because a path could have multiple points with the specified x coordinate. There is no built-in function in the SVG DOM to do this. One solution is to step along the path segments and do the maths yourself

问题 In Java, I have a class Line that has two variables : m and b , such that the line follows the formula mx + b . I have two such lines. How am I to find the x and y coordinates of the intersection of the two lines? (Assuming the slopes are different) Here is class Line : import java.awt.Graphics; import java.awt.Point; public final class Line { public final double m, b; public Line(double m, double b) { this.m = m; this.b = b; } public Point intersect(Line line) { double x = (this.b - line.b)