orthogonal

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 set of points that defines the boundary shown in image 1. ? Updated: Figure 1. is no longer refereed

I have an n-sized collection of Rects, most of which intersect each other. I'd like to remove the intersections and reduce the intersecting Rects into smaller non-intersecting rects. I could easily brute force a solution, but I'm looking for an efficient algorithm. Here's a visualization: Original: Processed: Ideally the method signature would look like this: public static List<RectF> resolveIntersection(List<RectF> rects); the output would be greater or equal to the input, where the output resolves the above visual representation. this is a problem I solved in the past. The first thing it to

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? 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 and out.eps for the Y residuals). Implementing a cross-validation or bootstrap method for determining

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 the lm to do that? Following some answers: additionnal information on x and y x and y are both visual

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 DOFFSOFF DONSON DONSON DONSON DONSON DONSON DONSON DONSON DONSON DONSON DONSON DONSON Levels: DOFFSOFF

Is there a method that I can call to create a random orthonormal matrix in python? Possibly using numpy? Or is there a way to create a orthonormal matrix using multiple numpy methods? Thanks. hpaulj This is the rvs method pulled from the https://github.com/scipy/scipy/pull/5622/files , with minimal change - just enough to run as a stand alone numpy function. import numpy as np def rvs(dim=3): random_state = np.random H = np.eye(dim) D = np.ones((dim,)) for n in range(1, dim): x = random_state.normal(size=(dim-n+1,)) D[n-1] = np.sign(x[0]) x[0] -= D[n-1]*np.sqrt((x*x).sum()) # Householder