math

问题 I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1. Is there a any way to count how many 1s and 0s a number has without iterating through the number? What I am currently doing is: def binary(num, length=4): return format(num, '#0{}b'.format(length + 2)).replace('0b', '') n = binary(112, 8) // '01110000' and then n.count('0') n.count('1') Is there any more efficient

问题 I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1. Is there a any way to count how many 1s and 0s a number has without iterating through the number? What I am currently doing is: def binary(num, length=4): return format(num, '#0{}b'.format(length + 2)).replace('0b', '') n = binary(112, 8) // '01110000' and then n.count('0') n.count('1') Is there any more efficient

问题 I have the fisher's linear discriminant that i need to use it to reduce my examples A and B that are high dimensional matrices to simply 2D, that is exactly like LDA, each example has classes A and B, therefore if i was to have a third example they also have classes A and B, fourth, fifth and n examples would always have classes A and B, therefore i would like to separate them in a simple use of fisher's linear discriminant. Im pretty much new to machine learning, so i dont know how to

问题 I have the fisher's linear discriminant that i need to use it to reduce my examples A and B that are high dimensional matrices to simply 2D, that is exactly like LDA, each example has classes A and B, therefore if i was to have a third example they also have classes A and B, fourth, fifth and n examples would always have classes A and B, therefore i would like to separate them in a simple use of fisher's linear discriminant. Im pretty much new to machine learning, so i dont know how to

问题 standard power operation ( ** ) in Python does not work for negative power! Sure I could write the formula otherwise, with divide and positive power. However, I am checking optimization routine result, and sometimes power is negative, sometimes it is positive. Here again a if statement could do, but I am wondering if there is a workarouns and a Python library where negative exposant is allowed. Thanks and Regards. 回答1: Which version of python are you using? Perfectly works for me in Python 2

问题 standard power operation ( ** ) in Python does not work for negative power! Sure I could write the formula otherwise, with divide and positive power. However, I am checking optimization routine result, and sometimes power is negative, sometimes it is positive. Here again a if statement could do, but I am wondering if there is a workarouns and a Python library where negative exposant is allowed. Thanks and Regards. 回答1: Which version of python are you using? Perfectly works for me in Python 2

问题 So I want to be able to parse, and evaluate, "dice expressions" in C#. A dice expression is defined like so: <expr> := <expr> + <expr> | <expr> - <expr> | [<number>]d(<number>|%) | <number> <number> := positive integer So e.g. d6+20-2d3 would be allowed, and should evaluate as rand.Next(1, 7) + 20 - (rand.Next(1, 4) + rand.Next(1, 4)) Also d% should be equivalent to d100 . I know I could hack together some solution, but I also know that this seems like a very typical computer-science type

问题 It sounds easy for me until I tried The implementation uses lib igl but the problem is not there but in my lack of math knowledge :-\ Eigen::MatrixXd p; const int size = 128; p.resize(size * size, 3); int index = 0; for (int y = 0; y < size; y++) { double ypos = (double(y) / double(size - 1)) - 0.5; for (int x = 0; x < size; x++) { double xpos = (double(x) / double(size - 1)) - 0.5; const double sphericalTheta = xpos * (pi / 2.0); const double sphericalPhi = ypos * (pi / 2.0); const double mX

问题 It sounds easy for me until I tried The implementation uses lib igl but the problem is not there but in my lack of math knowledge :-\ Eigen::MatrixXd p; const int size = 128; p.resize(size * size, 3); int index = 0; for (int y = 0; y < size; y++) { double ypos = (double(y) / double(size - 1)) - 0.5; for (int x = 0; x < size; x++) { double xpos = (double(x) / double(size - 1)) - 0.5; const double sphericalTheta = xpos * (pi / 2.0); const double sphericalPhi = ypos * (pi / 2.0); const double mX

问题 It sounds easy for me until I tried The implementation uses lib igl but the problem is not there but in my lack of math knowledge :-\ Eigen::MatrixXd p; const int size = 128; p.resize(size * size, 3); int index = 0; for (int y = 0; y < size; y++) { double ypos = (double(y) / double(size - 1)) - 0.5; for (int x = 0; x < size; x++) { double xpos = (double(x) / double(size - 1)) - 0.5; const double sphericalTheta = xpos * (pi / 2.0); const double sphericalPhi = ypos * (pi / 2.0); const double mX