如何在Python中将向量投影到由其正交向量定义的平面上? [英] How may I project vectors onto a plane defined by its orthogonal vector in Python?

问题描述

我有一个平面，平面A ，由其正交矢量定义，例如(a，b，c).

I have a plane, plane A, defined by its orthogonal vector, say (a, b, c).

(即向量(a，b，c)与 A平面正交)

我希望将向量(d，e，f)投影到 A平面上.

I wish to project a vector (d, e, f) onto plane A.

如何在Python中完成此操作?我认为必须有一些简单的方法.

How can I do it in Python? I think there must be some easy ways.

推荐答案

取(d，e，f)并减去其在垂直于该平面的法线上的投影(在您的情况下)(a，b，c)).所以:

Take (d, e, f) and subtract off the projection of it onto the normalized normal to the plane (in your case (a, b, c)). So:

v = (d, e, f)
        - sum((d, e, f) *. (a, b, c)) * (a, b, c) / sum((a, b, c) *. (a, b, c))

在这里，用 *.是指按组件划分的产品.所以这意味着:

Here, by *. I mean the component-wise product. So this would mean:

sum([x * y for x, y in zip([d, e, f], [a, b, c])])

或

d * a + e * b + f * c

如果您只想保持清醒而又学究的话

if you just want to be clear but pedantic

，对于(a，b，c)*同样.(a，b，c).因此，在Python中:

and similarly for (a, b, c) *. (a, b, c). Thus, in Python:

from math import sqrt

def dot_product(x, y):
    return sum([x[i] * y[i] for i in range(len(x))])

def norm(x):
    return sqrt(dot_product(x, x))

def normalize(x):
    return [x[i] / norm(x) for i in range(len(x))]

def project_onto_plane(x, n):
    d = dot_product(x, n) / norm(n)
    p = [d * normalize(n)[i] for i in range(len(n))]
    return [x[i] - p[i] for i in range(len(x))]

那么你可以说:

p = project_onto_plane([3, 4, 5], [1, 2, 3])

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