points

问题 Is there a way to add bullet points to an HTML textarea? I want to add a simple feature where a bullet point is added for every line in a text area (similar to a list bullet points). 回答1: You can't do that but there is another way. delete the textarea. '<section id="textarea" contenteditable="true"> <ul> <li>List item here</li> <li>List item here</li> <li>List item here</li> <li>List item here</li> </ul> </section>' 回答2: I think, you will not able to add bullet points (i.e. rich text) in text

问题 I have a list of points that make a curve, and I would like to reduce the number of points, but still keep the overall shape of the curve. Basically, I want to go from this: To this: So the algorithm would remove the points that are redundant but preserve those that really define the shape (like the points at the bottom of the curve). Is there any known algorithm to do that? I expect there is but I'm not sure what to search for on Google. Any help would be appreciated. 回答1: Consider Douglas

问题 I have a single series of values (i.e. one column of data), and I would like to create a plot with the range of data values on the x-axis and the frequency that each value appears in the data set on the y-axis. What I would like is very close to a Kernel Density Plot: # Kernel Density Plot d <- density(mtcars$mpg) # returns the density data plot(d) # plots the results and Frequency distribution in R on stackoverflow. However, I would like frequency (as opposed to density) on the y-axis.

问题 I have a 3D plot and two points coordinates A(0,0,0) and B(13,-11,19). I just want to plot a visible line connecting this two points ... I tried plot3(0,0,0, 13,-11,19) and other stuff but everything i tried failed miserably. 回答1: Here's how: % Your two points P1 = [0,0,0]; P2 = [13,-11,19]; % Their vertial concatenation is what you want pts = [P1; P2]; % Because that's what line() wants to see line(pts(:,1), pts(:,2), pts(:,3)) % Alternatively, you could use plot3: plot3(pts(:,1), pts(:,2),

问题 My current implementation doesn't show anything on the form even though the collections I thought bounded have data (I checked in debug). Here's some code: public event PropertyChangedEventHandler PropertyChanged; PointCollection imagePoints; public PointCollection ImagePoints { get { return this.imagePoints; } set { if (this.imagePoints != value) { this.imagePoints = value; if (this.PropertyChanged != null) { PropertyChanged(this, new PropertyChangedEventArgs("ImagePoints")); } } } } and the

问题 I need to find the 2 points of the visual horizon , of a curved face. I have: XYZ of the 4 corner points XYZ of the 2 curved edge bezier points And I need to calculate either: XY of the 2 horizon points XYZ of the 2 horizon points Note: I got a solution the last time I asked this question, but it only found the extrema of the curves, not the horizon points , which changes based on the position and rotation of both curves in respect to each other. 回答1: You don't say how your surface is defined

问题 I want to draw the lines between points on the view, and then pull those points upto desired positions even the shape will change. i know how to draw the line between two points canvas.drawLine(10, 10, 90, 10, paint); by using this i can draw the lines between points. EDIT : here i am attaching image for clear explanation, from Paul answer now i am able to draw the lines between points, still have the problem of pulling points... 回答1: Here's how it's done. Suppose you have your points, make

问题 I think I may have stumbled onto a limitation of Flot, but I'm not sure. I'm trying to represent a single data series over time. The items' "State" is represented on the Y-Axis (there are 5 of them), and time is on the X-Axis (items can change states over time). I want the graph to have points and lines connecting those points for each data series. In addition to tracking an item's State over time, I'd also like to represent it's "Status" at any of the particular points. This I would like to

问题 I have the same problem as here: how to order vertices in a simple, non-convex polygon but there is no solutions I can use. I have coordinates of points and need to find some polygon. Does not matter that there is more solutions for one list of dots. I need some algorithm to find one of them. Does not matter which one. I really don't know how to solve this. (I have stored coordinates in array and I want to use some algorithm in Javascript) Thanks a lot. 回答1: First, find the center of the

问题 I have a set K of randomly selected pixels in a 2D image. For every other pixel in the image I need to find out which pixel in set K is closest to it (using the standard sqrt(dx^2 + dy^2) measure of distance). I am aware that there may be more than one solution for each pixel. Obviously it can be done by brute force against every pixel in the set, but I'd rather avoid this as it's not efficient. Any other good suggestions? Cheers. 回答1: Don't forget that you don't need to bother with the