计算两组向量numpy之间的叉积的有效方法 [英] Efficient way of computing the cross products between two sets of vectors numpy

问题描述

我分别有两组2000个3D矢量，并且我需要计算每个可能的对之间的叉积.我目前是这样的

I have two sets of 2000 3D vectors each, and I need to compute the cross product between each possible pair. I currently do it like this

for tx in tangents_x:
    for ty in tangents_y:
         cross = np.cross(tx, ty)
         (... do something with the cross variable...)

这有效，但是速度很慢.有没有办法使其更快?

This works, but it's pretty slow. Is there a way to make it faster?

如果我对基于元素的产品感兴趣，我可以做以下事情

If I was interested in the element-wise product, I could just do the following

# Define initial vectors
tx = np.array([np.random.randn(3) for i in range(2000)])
ty = np.array([np.random.randn(3) for i in range(2000)])
# Store them into matrices
X = np.array([tx for i in range(2000)])
Y = np.array([ty for i in range(2000)]).T
# Compute the element-wise product
ew = X * Y
# Use the element_wise product as usual
for i,tx in enumerate(tangents_x):
    for j,ty in enumerate(tangents_y):
        (... use the element wise product of tx and ty as ew[i,j])

如何将其应用到叉积而不是按元素的积?或者，您是否看到另一种选择?

How can I apply this to the cross product instead of the element-wise one? Or, do you see another alternative?

非常感谢:)

推荐答案

就像许多numpy函数cross支持广播一样，因此您可以轻松地做到:

Like many numpy functions cross supports broadcasting, therefore you can simply do:

np.cross(tangents_x[:, None, :], tangents_y)

或-更详细，但可能更容易阅读

or - more verbose but maybe easier to read

np.cross(tangents_x[:, None, :], tangents_y[None, :, :])

这会将tangents_x和tangents_y重塑为形状2000, 1, 3和1, 2000, 3.根据广播规则，这将被解释为两个形状为2000, 2000, 3的数组，其中tangents_x沿轴1重复，而tangents_y沿轴0重复.

This reshapes tangents_x and tangents_y to shapes 2000, 1, 3 and 1, 2000, 3. By the rules of broadcasting this will be interpreted like two arrays of shape 2000, 2000, 3 where tangents_x is repeated along axis 1 and tangents_y is repeated along axis 0.

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