matrix-indexing

问题 Question Why are the numpy tuple indexing behaviors inconsistent? Please explain the rational or design decision behind these behaviors. In my understanding, Z[(0,2)] and Z[(0, 2), (0)] are both tuple indexing and expected the consistent behavior for copy/view. If this is incorrect, please explain, import numpy as np Z = np.arange(36).reshape(3, 3, 4) print("Z is \n{}\n".format(Z)) b = Z[ (0,2) # Select Z[0][2] ] print("Tuple indexing Z[(0,2)] is \n{}\nIs view? {}\n".format( b, b.base is not

问题 Continuation to the question numpy - why Z[(0,2)] is view but Z[(0, 2), (0)] is copy?. I got the answer and understood that comma triggering advanced index makes a totally different indexing. The answer also provided a way using __array_interface__ to understand copy/view behavior. However apparently I have not got to the bottom of this part of the answer. A view is returned if the new array can be described with shape, strides and all or part of the original data buffer Because I still

问题 I have a matrix A with m rows and I'd like to set a specific element of each row equal 1. The column index varies from row to row and is specified by a column vector a (with m values). That is, I want A_{i,a_i} = 1 . Is there a quick way to do this in Matlab (without a for-loop)? 回答1: I solved it using the sub2ind function: A(sub2ind(size(A), 1:numel(a), a')) = 1 来源： https://stackoverflow.com/questions/20957317/set-a-value-of-a-specific-column-for-each-row-of-a-matrix

问题 I have a matrix A with m rows and I'd like to set a specific element of each row equal 1. The column index varies from row to row and is specified by a column vector a (with m values). That is, I want A_{i,a_i} = 1 . Is there a quick way to do this in Matlab (without a for-loop)? 回答1: I solved it using the sub2ind function: A(sub2ind(size(A), 1:numel(a), a')) = 1 来源： https://stackoverflow.com/questions/20957317/set-a-value-of-a-specific-column-for-each-row-of-a-matrix

问题 For example, if I have a cell with arrays listed inside: C = {[1,2,3,4], [3,4], [2], [4,5,6], [4,5], [7]} I want to output: D = {[1,2,3,4], [4,5,6], [7]} What is the most efficient way to remove cell elements that are already included/subset in another larger element? My existing algorithm loops through each element, compares it to each element in the new cell list and updates the new cell list accordingly but it is extremely slow and inefficient (my original cell contains >200 array elements

问题 I'm trying to slice a PyTorch tensor using a logical index on the columns. I want the columns that correspond to a 1 value in the index vector. Both slicing and logical indexing are possible, but are they possible together? If so, how? My attempt keeps throwing the unhelpful error TypeError: indexing a tensor with an object of type ByteTensor. The only supported types are integers, slices, numpy scalars and torch.LongTensor or torch.ByteTensor as the only argument. MCVE Desired Output C =

问题 I have two matrix A and B,the first row of matrix A(1,:)=[1 2] refer to the number of row and column matrix B(1,2)=21,now I want to do this work for another rows of matrix A without loops? A=[1 2;2 3;1 3;3 3]; B=[1 21 34;45 65 87;4 55 66]; for i=1:4 d(i,:)=B(A(i,1),A(i,2)) end d =[21; 87;34;66] 回答1: An alternative to sub2ind is d = B(A(:,1)+ (A(:,2)-1)*size(B,1)); 回答2: Use sub2ind to get linear indices of the required values of B and then use these indices to retrieve those values. d = B

问题 Given a numpy ndarray and an index: a = np.random.randint(0,4,(2,3,4)) idx = (1,1,1) is there a clean way of retrieving the 0D subarray of a at idx ? Something equivalent to a[idx + (None,)].squeeze() but less hackish? Note that @filippo's clever a[idx][...] is not equivalent. First, it doesn't work for object arrays. But more seriously it does not return a subarray but a new array: b = a[idx][...] b[()] = 7 a[idx] == 7 # False 回答1: b = a[idx+(Ellipsis,)] I'm testing on one machine and

问题 Given a numpy ndarray and an index: a = np.random.randint(0,4,(2,3,4)) idx = (1,1,1) is there a clean way of retrieving the 0D subarray of a at idx ? Something equivalent to a[idx + (None,)].squeeze() but less hackish? Note that @filippo's clever a[idx][...] is not equivalent. First, it doesn't work for object arrays. But more seriously it does not return a subarray but a new array: b = a[idx][...] b[()] = 7 a[idx] == 7 # False 回答1: b = a[idx+(Ellipsis,)] I'm testing on one machine and

问题 I have three matrices, A, B and C. When B is larger than A, I want to saturate the value with A. It says that the number of elements in I (which is (B > A)) must be the same as the number of elements in A. I checked below and they are the same. >> A = [5 5 5; 5 5 5; 5 5 5]; >> B = [2 2 2; 2 2 2; 2 2 2]; >> C(B > A) = A In an assignment A(I) = B, the number of elements in B and I must be the same. >> numel(B > A) ans = 9 >> numel(A) ans = 9 >> numel(A>B) ans = 9 It is also strange that this