在Numpy中将矩阵乘以另一个矩阵的每一行 [英] Multiply matrix by each row of another matrix in Numpy

问题描述

我有一个大小为(4x4)的齐次变换矩阵，大小为(nx3)的轨迹.该轨迹的每一行都是一个向量.

我想将均匀变换矩阵乘以轨迹的每一行.下面是代码:

#append zero column at last
trajectory = np.hstack((trajectory, np.zeros((trajectory.shape[0], 1)))) #(nx3)->(nx4)

trajectory_new = np.zeros((1, 3)) #(1x3)
for row in trajectory:
    vect = row.reshape((-1,1)) #convert (1x4) to (4x1)
    vect = np.dot(HTM, vect) #(4x4) x (4x1) = (4x1)
    vect = vect.T #(1x4)
    vect = np.delete(vect, -1, axis=1) #remove last element from vector
    trajectory_new = np.vstack((trajectory_new, vect)) #(nx3)

trajectory_new = np.delete(trajectory_new, 0, axis=0)#remove first row

上面的代码有效.但是，我正在寻找更简单的解决方案，例如:

trajectory_new = np.apply_along_axis(np.multiply, 0, trajectory, HTM)

请帮助.

答案:

trajectory = np.hstack((trajectory, np.ones((trajectory.shape[0], 1))))#(nx3)->(nx4)
trajectory_new = trajectory.dot(HTM.T)[:,:-1]

解决方案

您可以简单地将矩阵乘法与 解决方案

You can simply use matrix-multiplication with np.dot on the input before stacking zeros -

trajectory.dot(HTM[:,:3].T)[:,:3]

Approaches -

def dot_based(trajectory):
    return trajectory.dot(HTM[:,:3].T)[:,:3]

def original_app(trajectory):
    # append zero column at last
    traj_stacked = np.hstack((trajectory, np.zeros((trajectory.shape[0], 1))))

    trajectory_new = np.zeros((1, 3)) #(1x3)
    for row in traj_stacked:
        vect = row.reshape((-1,1)) #convert (1x4) to (4x1)
        vect = np.dot(HTM, vect) #(4x4) x (4x1) = (4x1)
        vect = vect.T #(1x4)
        vect = np.delete(vect, -1, axis=1) #remove last element from vector
        trajectory_new = np.vstack((trajectory_new, vect)) #(nx3)

    trajectory_new = np.delete(trajectory_new, 0, axis=0)#remove first row
    return trajectory_new

Sample run -

In [37]: n = 5
    ...: trajectory = np.random.rand(n,3)
    ...: HTM = np.random.rand(4,4)
    ...: 

In [38]: np.allclose(dot_based(trajectory), original_app(trajectory))
Out[38]: True

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