vectorization

问题 Solve_ivp is an initial value problem solver function from Scipy. In a few words scipy.integrate.solve_ivp(fun, t_span, y0, method=’RK45’, t_eval=None, dense_output=False, events=None, vectorized=False, args=None, **options) Solve an initial value problem for a system of ODEs. This function numerically integrates a system of ordinary differential equations given an initial value. In the solve_ivp function documentation (Scipy reference guide 1.4.1 page 695) we have the following Parameters

问题 Solve_ivp is an initial value problem solver function from Scipy. In a few words scipy.integrate.solve_ivp(fun, t_span, y0, method=’RK45’, t_eval=None, dense_output=False, events=None, vectorized=False, args=None, **options) Solve an initial value problem for a system of ODEs. This function numerically integrates a system of ordinary differential equations given an initial value. In the solve_ivp function documentation (Scipy reference guide 1.4.1 page 695) we have the following Parameters

问题 Solve_ivp is an initial value problem solver function from Scipy. In a few words scipy.integrate.solve_ivp(fun, t_span, y0, method=’RK45’, t_eval=None, dense_output=False, events=None, vectorized=False, args=None, **options) Solve an initial value problem for a system of ODEs. This function numerically integrates a system of ordinary differential equations given an initial value. In the solve_ivp function documentation (Scipy reference guide 1.4.1 page 695) we have the following Parameters

问题 I'm trying to implement a vectorized version of the regularised logistic regression. I have found a post that explains the regularised version but I don't understand it. To make it easy I will copy the code below: hx = sigmoid(X * theta); m = length(X); J = (sum(-y' * log(hx) - (1 - y') * log(1 - hx)) / m) + lambda * sum(theta(2:end).^2) / (2*m); grad =((hx - y)' * X / m)' + lambda .* theta .* [0; ones(length(theta)-1, 1)] ./ m ; I understand the first part of the Cost equation, If I'm

问题 I have a time series of 3D vectors in a Python numpy array similar to the following: array([[-0.062, -0.024, 1. ], [-0.071, -0.03 , 0.98 ], [-0.08 , -0.035, 0.991], [-0.083, -0.035, 0.98 ], [-0.083, -0.035, 0.977], [-0.082, -0.035, 0.993], [-0.08 , -0.034, 1.006], [-0.081, -0.032, 1.008], ....... I want to rotate each vector around a specified axis through a specified angle theta . I have been using quaternions to achieve this for one vector as found here in henneray's answer. v1 = np.array (

问题 I have a time series of 3D vectors in a Python numpy array similar to the following: array([[-0.062, -0.024, 1. ], [-0.071, -0.03 , 0.98 ], [-0.08 , -0.035, 0.991], [-0.083, -0.035, 0.98 ], [-0.083, -0.035, 0.977], [-0.082, -0.035, 0.993], [-0.08 , -0.034, 1.006], [-0.081, -0.032, 1.008], ....... I want to rotate each vector around a specified axis through a specified angle theta . I have been using quaternions to achieve this for one vector as found here in henneray's answer. v1 = np.array (

问题 Consider a randomly generated __m256i vector. Is there a faster precise way to convert them into __m256 vector of floats between 0 (inclusively) and 1 (exclusively) than division by float(1ull<<32) ? Here's what I have tried so far, where iRand is the input and ans is the output: const __m256 fRand = _mm256_cvtepi32_ps(iRand); const __m256 normalized = _mm256_div_ps(fRand, _mm256_set1_ps(float(1ull<<32))); const __m256 ans = _mm256_add_ps(normalized, _mm256_set1_ps(0.5f)); 回答1: The version

问题 I am having some issues with the steps to follow for over-sampling a dataset. What I have done is the following: # Separate input features and target y_up = df.Label X_up = df.drop(columns=['Date','Links', 'Paths'], axis=1) # setting up testing and training sets X_train_up, X_test_up, y_train_up, y_test_up = train_test_split(X_up, y_up, test_size=0.30, random_state=27) class_0 = X_train_up[X_train_up.Label==0] class_1 = X_train_up[X_train_up.Label==1] # upsample minority class_1_upsampled =

问题 I created some example code that mimic what code I got: import numpy as np arr = np.random.random(100) arr2 = np.linspace(0, 1, 20) arr3 = np.zeros(20) # this is the array i want to store the result in for index, num in enumerate(list(arr2)): arr3[index] = np.mean(arr[np.abs(num - arr) < 0.2]) >>> arr3 array([0.10970893, 0.1132479 , 0.14687451, 0.17257954, 0.19401919, 0.23852137, 0.29151448, 0.35715096, 0.43273118, 0.45800796, 0.52940421, 0.60345354, 0.63969432, 0.67656363, 0.72921913, 0

问题 I created some example code that mimic what code I got: import numpy as np arr = np.random.random(100) arr2 = np.linspace(0, 1, 20) arr3 = np.zeros(20) # this is the array i want to store the result in for index, num in enumerate(list(arr2)): arr3[index] = np.mean(arr[np.abs(num - arr) < 0.2]) >>> arr3 array([0.10970893, 0.1132479 , 0.14687451, 0.17257954, 0.19401919, 0.23852137, 0.29151448, 0.35715096, 0.43273118, 0.45800796, 0.52940421, 0.60345354, 0.63969432, 0.67656363, 0.72921913, 0