问题描述
我有一组(n)个几何形状，这些形状由固定数量(p)的2D点定义.这些形状是独立的，但出于效率原因，我将时间存储在单个numpy数组中.缩放或平移这些形状很容易，但是我想旋转它们，但我不确定该怎么做.我怀疑np.tensordot是我的朋友，但我找不到正确使用它的方法.
I've a set of (n) geometrical shapes that are defined by a fixed number (p) of 2D points. Those shapes are independent but for efficiency reason, I stored time in a single numpy array. Scaling or translating those shapes are easy, but I would like to rotate them and I'm not sure how to do that. I suspect np.tensordot is my friend but I can't find the way to properly use it.
n = 100 # Number of shape
p = 4 # Points per shape
P = np.random.uniform(0, 1, (n, p, 2))
# Scaling
S = 0.5*np.ones(n)
P *= S
# Translating
T = np.random.uniform(0, 1, (n, 1, 2))
P += T
# Rotating
A = np.random.uniform(0, 2*np.pi, n)
cosA, sinA = np.cos(A), np.sin(A)
R = np.empty((n,2,2))
R[:,0,0] = cosA
R[:,1,0] = sinA
R[:,0,1] = -sinA
R[:,1,1] = cosA
np.tensordot(P, R, axes=???)
推荐答案
似乎您要保持两个数组之间的第一个轴-P和R对齐，并且sum-reducing分别与其余数组分开输入数组.因此，我们可以使用 np.einsum 将允许我们使用轴对齐标准.
It seems you are keeping the first axis between the two arrays - P and R aligned and sum-reducing one each off the remaining axes from the input arrays. So, we can use np.einsum as it will allow us the axis-alignment criteria.
您正在使用P中的最后一个轴进行求和.现在，根据R的哪条轴，您将失去sum-reduction的旋转计算，其中之一应该可以完成工作-
You are using the last axis from P for the sum-reduction. Now, depending on which axis of R you are losing with sum-reduction for the rotation calculation, one of these should do the job -
np.einsum('ijk,ilk->ijl',P,R) # Using last dim of R for sum-reduction
np.einsum('ijk,ikl->ijl',P,R) # Using second dim of R for sum-reduction
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