Start with row 1. Go right until you hit the first 1. Then go down to row 2, but remain in the same column and repeat the process of going right until you hit a 1. Do this repeatedly. The row in which you last stepped right is your answer. This is an O(N+M) solution … Read more
numpy.dot: For 2-D arrays it is equivalent to matrix multiplication, and for 1-D arrays to inner product of vectors (without complex conjugation). For N dimensions it is a sum product over the last axis of a and the second-to-last of b: numpy.inner: Ordinary inner product of vectors for 1-D arrays (without complex conjugation), in higher … Read more
You can use numpy.apply_along_axis(). Assuming that your array is 2D, you can use it like: import numpy as np myarray = np.array([[11, 12, 13], [21, 22, 23], [31, 32, 33]]) def myfunction(x): return x[0] + x[1]**2 + x[2]**3 print(np.apply_along_axis(myfunction, axis=1, arr=myarray)) #[ 2352 12672 36992]
That works in numpy but only if the trailing axes have the same dimension. Here is an example of successfully subtracting a vector from a matrix: In [27]: print m; m.shape [[ 0 1 2] [ 3 4 5] [ 6 7 8] [ 9 10 11]] Out[27]: (4, 3) In [28]: print v; v.shape … Read more
The set-methods will replace the current Matrix with new values, disregarding whatever the Matrix contained before. The pre and post method will apply a new transformation before or after whatever the current Matrix contains. In this example, the rotation will be ignored since we are using the set method and the m will only contain … Read more
You can use the data() member function of the Eigen Matrix class. The layout by default is column-major, not row-major as a multidimensional C array (the layout can be chosen when creating a Matrix object). For sparse matrices the preceding sentence obviously doesn’t apply. Example: ArrayXf v = ArrayXf::LinSpaced(11, 0.f, 10.f); // vc is the … Read more
You could do ## Some data set.seed(123) mm <- matrix(rbinom(40, 20, 0.5), 8, 5) mm # [,1] [,2] [,3] [,4] [,5] # [1,] 9 10 8 11 11 # [2,] 12 10 6 11 12 # [3,] 9 14 9 10 6 # [4,] 13 10 14 11 10 # [5,] 13 11 13 9 … Read more
Many ways to do this. ismember is the first that comes to mind, since it is a set membership action you wish to take. Thus X = primes(20); ismember([15 17],X) ans = 0 1 Since 15 is not prime, but 17 is, ismember has done its job well here. Of course, find (or any) will … Read more