vectorization

问题 I have a matrix: A = [1 1 1 2 2 2 3 3 3] Is there a vectorized way of obtaining: B = [1 0 0 0 1 0 0 0 1 2 0 0 0 2 0 0 0 2 3 0 0 0 3 0 0 0 3] 回答1: Here's another way using sparse and repmat: A = [1 2 3; 4 5 6; 7 8 9]; A = A.'; B = full(sparse(1:numel(A), repmat(1:size(A,1),1,size(A,2)), A(:))); The original matrix is in A , and I transpose it so I can unroll the rows of each matrix properly for the next step. I use sparse to declare what is non-zero in a matrix. Specifically, we see that there

问题 I need to solve a system of linear equations Lx=b, where x is always a vector (3x1 array), L is an Nx3 array, and b is an Nx1 vector. N usually ranges from 4 to something like 10. I have no problems solving this using scipy.linalg.lstsq(L,b) However, I need to do this many times (something like 200x200=40000 times) as x is actually something associated with each pixel in an image. So x is actually stored in an PxQx3 array where P and Q is something like 200-300, and the last number '3' refers

问题 I have an array A whose shape is (N, N, K) and I would like to compute another array B with the same shape where B[:, :, i] = np.linalg.inv(A[:, :, i]) . As solutions, I see map and for loops but I am wondering if numpy provides a function to do this (I have tried np.apply_over_axes but it seems that it can only handle 1D array). with a for loop: B = np.zeros(shape=A.shape) for i in range(A.shape[2]): B[:, :, i] = np.linalg.inv(A[:, :, i]) with map : B = np.asarray(map(np.linalg.inv, np

问题 I have an array A whose shape is (N, N, K) and I would like to compute another array B with the same shape where B[:, :, i] = np.linalg.inv(A[:, :, i]) . As solutions, I see map and for loops but I am wondering if numpy provides a function to do this (I have tried np.apply_over_axes but it seems that it can only handle 1D array). with a for loop: B = np.zeros(shape=A.shape) for i in range(A.shape[2]): B[:, :, i] = np.linalg.inv(A[:, :, i]) with map : B = np.asarray(map(np.linalg.inv, np

问题 I need to generate a large number of random poker card decks. Speed is important so everything has to be in numpy matrix form. I understand I can generate two cards from a deck as follows: np.random.choice(12*4,2, replace=False) How can I execute the same query so that a 2d array is created without a for loop? The difficulty is that each round will need to distribute from the original stack, so replace is only true for rows but False for columns. I've also tried it with originalDeck=np.arange

问题 The question: What is the most efficient sequence to generate a stride-3 gather of 32-bit elements from memory? If the memory is arranged as: MEM = R0 G0 B0 R1 G1 B1 R2 G2 B2 R3 G3 B3 ... We want to obtain three YMM registers where: YMM0 = R0 R1 R2 R3 R4 R5 R6 R7 YMM1 = G0 G1 G2 G3 G4 G5 G6 G7 YMM2 = B0 B1 B2 B3 B4 B5 B6 B7 Motivation and discussion The scalar C code is something like template <typename T> T Process(const T* Input) { T Result = 0; for (int i=0; i < 4096; ++i) { T R = Input[3

问题 The question: What is the most efficient sequence to generate a stride-3 gather of 32-bit elements from memory? If the memory is arranged as: MEM = R0 G0 B0 R1 G1 B1 R2 G2 B2 R3 G3 B3 ... We want to obtain three YMM registers where: YMM0 = R0 R1 R2 R3 R4 R5 R6 R7 YMM1 = G0 G1 G2 G3 G4 G5 G6 G7 YMM2 = B0 B1 B2 B3 B4 B5 B6 B7 Motivation and discussion The scalar C code is something like template <typename T> T Process(const T* Input) { T Result = 0; for (int i=0; i < 4096; ++i) { T R = Input[3

问题 How can I avoid for loops and nested if sentences and be more Pythonic? At first glance this may seem like a "please do my all of my work for me" question. I can assure you that it is not. I'm trying to learn some real Python, and would like to discover ways of speeding up code based on a reproducible example and a pre-defined function. I'm calculating returns from following certain signals in financial markets using loads of for loops and nested if sentences. I have made several attempts,

问题 How can I avoid for loops and nested if sentences and be more Pythonic? At first glance this may seem like a "please do my all of my work for me" question. I can assure you that it is not. I'm trying to learn some real Python, and would like to discover ways of speeding up code based on a reproducible example and a pre-defined function. I'm calculating returns from following certain signals in financial markets using loads of for loops and nested if sentences. I have made several attempts,

问题 This piece of code works as I want it to, but in the spirit of good MATLAB code, is there a way to vectorize this (previous is a k x 1 vector): start = zeros(k,1); for i = 2:length(previous) if (previous(i-1) == -1) start(previous(i))= start(previous(i))+1; end end What, in general, is the intuitive way to go about vectorizing code in MATLAB? 回答1: You can do this without find , for maximum performance: I = [false; previous(1:end-1) == -1]; idx = previous(I); start(idx) = start(idx) + 1; This