[英]Dynamically create matrix from a vectors in numpy
I'm trying to create a matrix of shape Nx3 where N is not known at first.我正在尝试创建一个形状为 Nx3 的矩阵,其中 N 一开始是未知的。 This is what I'm basically trying to do:
这就是我基本上想要做的:
F = np.array([[],[],[]])
for contact in contacts:
xp,yp,theta = contact
# Create vectors for points and normal
P = [xp, yp, 0]
N = [np.cos(theta), np.sin(theta), 0]
# Calculate vector product
cross_PN = np.cross(P,N)
# f = [mz, fx, fi]
mz = cross_PN[2]
fx = N[0]
fy = N[1]
f = np.array([mz, fx, fy])
F = np.vstack([F, f])
But this code doesn't work.但是这段代码不起作用。 I can do similar thing in Matlab very easily, but that is not the case in Python using numpy.
我可以很容易地在 Matlab 中做类似的事情,但在使用 numpy 的 Python 中情况并非如此。
Any help is greatly appreciated.任何帮助是极大的赞赏。 Thank you
谢谢
I would like to create a matrix by adding new rows, but in the beginning the matrix is empty.我想通过添加新行来创建一个矩阵,但一开始矩阵是空的。 That is why I receive the error: "along dimension 1, the array at index 0 has size 0 and the array at index 1 has size 3"
这就是为什么我收到错误消息:“沿着维度 1,索引 0 处的数组大小为 0,索引 1 处的数组大小为 3”
Based on your comments and suggestion, I guess the correct and more efficient way to do this is the following:根据您的意见和建议,我想正确且更有效的方法如下:
L = len(contacts)
F = np.zeros((L, 3))
for ii in range(L):
xp,yp,theta = contacts[ii]
# Create vectors for points and normal
P = [xp, yp, 0]
N = [np.cos(theta), np.sin(theta), 0]
# Calculate vector product
cross_PN = np.cross(P,N)
# f = [mz, fx, fi]
mz = cross_PN[2]
fx = N[0]
fy = N[1]
fi = np.array([mz, fx, fy])
F[ii] = fi
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