This is because you try to access an array out of its boundary. I will show you how you can debug such errors. I set options(error=recover) I run reach_full_in <- reachability(krack_full, ‘in’) I get : reach_full_in <- reachability(krack_full, ‘in’) Error in reach_mat[i, alter] = 1 : subscript out of bounds Enter a frame number, or … Read more
Just use matrix: matrix(vec,nrow = 7,ncol = 7) One advantage of using matrix rather than simply altering the dimension attribute as Gavin points out, is that you can specify whether the matrix is filled by row or column using the byrow argument in matrix.
Math.NET Numerics is very nice, if it supports the operations you want. The older Math.Net Iridium still supports more options. Also, dnAnalytics is quite nice, but no longer being developed. (It, as well as Iridium, are being merged into Math.NET Numerics.) On the commercial side, there are some very good, robust options. The Extreme Optimization … Read more
If you can afford to symmetrize the matrix just before doing calculations, the following should be reasonably fast: def symmetrize(a): “”” Return a symmetrized version of NumPy array a. Values 0 are replaced by the array value at the symmetric position (with respect to the diagonal), i.e. if a_ij = 0, then the returned array … Read more
The I attribute only exists on matrix objects, not ndarrays. You can use numpy.linalg.inv to invert arrays: inverse = numpy.linalg.inv(x) Note that the way you’re generating matrices, not all of them will be invertible. You will either need to change the way you’re generating matrices, or skip the ones that aren’t invertible. try: inverse = … Read more
Use torch.mm: torch.mm(a, b) torch.dot() behaves differently to np.dot(). There’s been some discussion about what would be desirable here. Specifically, torch.dot() treats both a and b as 1D vectors (irrespective of their original shape) and computes their inner product. The error is thrown because this behaviour makes your a a vector of length 6 and … Read more
I hope the following sequence of code will help you: Bitmap targetBitmap = Bitmap.createBitmap(targetWidth, targetHeight, config); Canvas canvas = new Canvas(targetBitmap); Matrix matrix = new Matrix(); matrix.setRotate(mRotation,source.getWidth()/2,source.getHeight()/2); canvas.drawBitmap(source, matrix, new Paint()); If you check the following method from ~frameworks\base\graphics\java\android\graphics\Bitmap.java public static Bitmap createBitmap(Bitmap source, int x, int y, int width, int height, Matrix m, boolean … Read more
You can use linear indexing to access each element. for idx = 1:numel(array) element = array(idx) …. end This is useful if you don’t need to know what i,j,k, you are at. However, if you don’t need to know what index you are at, you are probably better off using arrayfun()
You can also check if all the eigenvalues of matrix are positive, if so the matrix is positive definite: import numpy as np def is_pos_def(x): return np.all(np.linalg.eigvals(x) > 0)
To apply both scale and flip, just list both in your transform: transform=”scale(2,2) scale(-1,1)” Or simply combine the values: transform=”scale(-2,2)” Of course, the issue you have with negative scales is that the objects get flipped across the origin (top left) of the SVG, so they can go off the edge of the document. You need … Read more