orthogonal

问题 I am trying to find a way to determine the rectilinear polygon from a set of integer points (indicated by the red dots in the pictures below). The image below shows what I would like to achieve: 1. I need only the minimal set of points that define the boundary of the rectilinear polygon. Most hull algorithms I can find do not satisfy the orthogonal nature of this problem, such as the gift-wrapping algorithm, which produce the following result (which is not what I want)... 2. How can I get the

问题 I have my world rendered. Based on some specific requirements it includes (some time) some lights on the floor. I am rendering these lights using triangle primitives. Right now I have the following code to zoom and limit the rendering area: if(aspect>1) gl.glOrtho(-scale*aspect, scale*aspect, -scale, scale, 0, 2); else gl.glOrtho(-scale, scale, -scale/aspect, scale/aspect, 0, 2); As you can see in this image, the far plane is cutting the light throught a line (parallel to the line on the

问题 I am fitting data with weights using scipy.odr but I don't know how to obtain a measure of goodness-of-fit or an R squared. Does anyone have suggestions for how to obtain this measure using the output stored by the function? 回答1: The res_var attribute of the Output is the so-called reduced Chi-square value for the fit, a popular choice of goodness-of-fit statistic. It is somewhat problematic for non-linear fitting, though. You can look at the residuals directly ( out.delta for the X residuals

问题 I've been looking around a bit and can't seem to find just what I"m looking for. I've found "canonical formulas," but what's the best way to use these? Do I have to scale every single vertex down? Or is there a better way? A formula would really help me out, but I'm also looking for an explanation about the near and far z planes relative the viewer's position 回答1: Here is a reasonable source that derives an orthogonal project matrix: Consider a few points: First, in eye space, your camera is

问题 I'm using R to create a linear regression model having orthogonal polynomial. My model is: fit=lm(log(UFB2_BITRATE_REF3) ~ poly(QPB2_REF3,2) + B2DBSA_REF3,data=UFB) UFB2_FPS_REF1= 29.98 27.65 26.30 25.69 24.68 23.07 22.96 22.16 21.51 20.75 20.75 26.15 24.59 22.91 21.02 19.59 18.80 18.21 17.07 16.74 15.98 15.80 QPB2_REF1 = 36 34 32 30 28 26 24 22 20 18 16 36 34 32 30 28 26 24 22 20 18 16 B2DBSA_REF1 = DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF DOFFSOFF

问题 I need to get the svg path of a circle projected in a orthogonal space. Example What I want to do is create a function(in js) that has the following parameters: the position of the circle the radius and what panel is the circle parallel to axes inclination This is the function I use to create a simple circle (without perspective) function getPath(cx,cy,r){ return "M" + cx + "," + cy + "m" + (-r) + ",0a" + r + "," + r + " 0 1,0 " + (r * 2) + ",0a" + r + "," + r + " 0 1,0 " + (-r * 2) + ",0"; }

问题 I am currentliy using an orthogonal map, not an isometric. But I would like to get an perspective look. the first picuture shows, what I currently have, the second shows my goal. Isn't it possible to do it with a transform matrix somehow? I did only manage to scale the whole image.. 回答1: You can use MapRenderer.setView(...) to achieve that. Create a PerspectiveCamera and set it up to have the perspective that you've shown in your second picture. Note that the tiled map will be rendered to the

问题 I need to work on drawing a hierarchical tree structure (HV-Drawing – Binary Tree) with orthogonal lines(straight rectangular connecting lines) between root and children ( like the following: http://lab.kapit.fr/display/visualizationlayouts/Hierarchical+Tree+layout ). I want to know if there are any open source examples of the algorithm of drawing trees like that so that I can implement the same algorithm in actionscript. Thanks Palash 回答1: I did a C# version and put the code up on

问题 Essentially the same question was asked here, but in a non-programming context. A suggested solution is take { y, -x, 0 }. This would work for all vectors that have an x or y component, but fails if the vector is equal to + or - { 0, 0, 1 }. In this case we would get { 0, 0, 0 }. My current solution (in c++): // floating point comparison utilizing epsilon bool is_equal(float, float); // ... vec3 v = /* some unit length vector */ // ... // Set as a non-parallel vector which we will use to find

问题 I used a linear regression on data I have, using the lm function. Everything works (no error message), but I'm somehow surprised by the result: I am under the impression R "misses" a group of points, i.e. the intercept and slope are not the best fit. For instance, I am referring to the group of points at coordinates x=15-25,y=0-20. My questions: is there a function to compare fit with "expected" coefficients and "lm-calculated" coefficients? have I made a silly mistake when coding, leading