在Python中修改3D数组
我试图在python中对3d数组中的特定元素执行操作。以下是该阵列的一个示例:在Python中修改3D数组,python,arrays,list,numpy,Python,Arrays,List,Numpy,我试图在python中对3d数组中的特定元素执行操作。以下是该阵列的一个示例: [[[ 0.5 0.5 50. ] [ 50.5 50.5 100. ] [ 0.5 100.5 50. ] [ 135. 90. 45. ]] [[ 50.5 50.5 100. ] [ 100.
[[[ 0.5 0.5 50. ]
[ 50.5 50.5 100. ]
[ 0.5 100.5 50. ]
[ 135. 90. 45. ]]
[[ 50.5 50.5 100. ]
[ 100.5 0.5 50. ]
[ 100.5 100.5 50. ]
[ 45. 90. 45. ]]
[[ 100.5 100.5 50. ]
[ 100.5 100.5 0. ]
[ 0.5 100.5 50. ]
[ 90. 0. 90. ]]
我需要的一个例子是取数组中的三个值,即0.5、0.5、50。从第四排取第一个元素,即135。并将这四个元素发送到一个函数中。然后,函数返回需要放入数组的3个元素的新值
我对python很陌生,所以很难让它正常工作。我应该做一个循环吗?还是别的什么
谢谢
尼克
解决方案的尝试:
b = shape(a)
triangles = b[0]
for k in range(0,triangles):
for i in range(0,2):
a[k,i,:] = VectMath.rotate_x(a[k,i,0],a[k,i,1],a[k,i,2],a[k,3,2])
您可以使用
VectMath.rotate\u x
函数旋转向量数组,然后使用slice获取并将数据放入a
:
a = np.array(
[[[ 0.5, 0.5, 50., ],
[ 50.5, 50.5, 100., ],
[ 0.5, 100.5, 50., ],
[ 135. , 90. , 45., ]],
[[ 50.5, 50.5, 100., ],
[ 100.5, 0.5, 50., ],
[ 100.5, 100.5, 50., ],
[ 45. , 90. , 45., ]],
[[ 100.5, 100.5, 50., ],
[ 100.5, 100.5, 0., ],
[ 0.5, 100.5, 50., ],
[ 90. , 0. , 90., ]]])
def rotate_x(v, deg):
r = np.deg2rad(deg)
c = np.cos(r)
s = np.sin(r)
m = np.array([[1, 0, 0],
[0, c,-s],
[0, s, c]])
return np.dot(m, v)
vectors = a[:, :-1, :]
angles = a[:, -1, 0]
for i, (vec, angle) in enumerate(zip(vectors, angles)):
vec_rx = rotate_x(vec.T, angle).T
a[i, :-1, :] = vec_rx
print a
输出:
[[[ 5.00000000e-01 -3.57088924e+01 -3.50017857e+01]
[ 5.05000000e+01 -1.06419571e+02 -3.50017857e+01]
[ 5.00000000e-01 -1.06419571e+02 3.57088924e+01]
[ 1.35000000e+02 9.00000000e+01 4.50000000e+01]]
[[ 5.05000000e+01 -3.50017857e+01 1.06419571e+02]
[ 1.00500000e+02 -3.50017857e+01 3.57088924e+01]
[ 1.00500000e+02 3.57088924e+01 1.06419571e+02]
[ 4.50000000e+01 9.00000000e+01 4.50000000e+01]]
[[ 1.00500000e+02 -5.00000000e+01 1.00500000e+02]
[ 1.00500000e+02 6.15385017e-15 1.00500000e+02]
[ 5.00000000e-01 -5.00000000e+01 1.00500000e+02]
[ 9.00000000e+01 0.00000000e+00 9.00000000e+01]]]
如果有很多三角形,如果我们可以在不使用python for循环的情况下旋转所有向量,速度可能会更快
在这里,我通过展开矩阵乘积进行旋转计算:
x' = x
y' = cos(t)*y - sin(t)*z
z' = sin(t)*y + cos(t)*z
因此,我们可以将这些公式矢量化:
a2 = np.array(
[[[ 0.5, 0.5, 50., ],
[ 50.5, 50.5, 100., ],
[ 0.5, 100.5, 50., ],
[ 135. , 90. , 45., ]],
[[ 50.5, 50.5, 100., ],
[ 100.5, 0.5, 50., ],
[ 100.5, 100.5, 50., ],
[ 45. , 90. , 45., ]],
[[ 100.5, 100.5, 50., ],
[ 100.5, 100.5, 0., ],
[ 0.5, 100.5, 50., ],
[ 90. , 0. , 90., ]]])
vectors = a2[:, :-1, :]
angles = a2[:, -1:, 0]
def rotate_x_batch(vectors, angles):
rad = np.deg2rad(angles)
c = np.cos(rad)
s = np.sin(rad)
x = vectors[:, :, 0]
y = vectors[:, :, 1]
z = vectors[:, :, 2]
yr = c*y - s*z
zr = s*y + c*z
vectors[:, :, 1] = yr
vectors[:, :, 2] = zr
rotate_x_batch(vectors, angles)
print np.allclose(a, a2)
您可以使用
VectMath.rotate\u x
函数旋转向量数组,然后使用slice获取并将数据放入a
:
a = np.array(
[[[ 0.5, 0.5, 50., ],
[ 50.5, 50.5, 100., ],
[ 0.5, 100.5, 50., ],
[ 135. , 90. , 45., ]],
[[ 50.5, 50.5, 100., ],
[ 100.5, 0.5, 50., ],
[ 100.5, 100.5, 50., ],
[ 45. , 90. , 45., ]],
[[ 100.5, 100.5, 50., ],
[ 100.5, 100.5, 0., ],
[ 0.5, 100.5, 50., ],
[ 90. , 0. , 90., ]]])
def rotate_x(v, deg):
r = np.deg2rad(deg)
c = np.cos(r)
s = np.sin(r)
m = np.array([[1, 0, 0],
[0, c,-s],
[0, s, c]])
return np.dot(m, v)
vectors = a[:, :-1, :]
angles = a[:, -1, 0]
for i, (vec, angle) in enumerate(zip(vectors, angles)):
vec_rx = rotate_x(vec.T, angle).T
a[i, :-1, :] = vec_rx
print a
输出:
[[[ 5.00000000e-01 -3.57088924e+01 -3.50017857e+01]
[ 5.05000000e+01 -1.06419571e+02 -3.50017857e+01]
[ 5.00000000e-01 -1.06419571e+02 3.57088924e+01]
[ 1.35000000e+02 9.00000000e+01 4.50000000e+01]]
[[ 5.05000000e+01 -3.50017857e+01 1.06419571e+02]
[ 1.00500000e+02 -3.50017857e+01 3.57088924e+01]
[ 1.00500000e+02 3.57088924e+01 1.06419571e+02]
[ 4.50000000e+01 9.00000000e+01 4.50000000e+01]]
[[ 1.00500000e+02 -5.00000000e+01 1.00500000e+02]
[ 1.00500000e+02 6.15385017e-15 1.00500000e+02]
[ 5.00000000e-01 -5.00000000e+01 1.00500000e+02]
[ 9.00000000e+01 0.00000000e+00 9.00000000e+01]]]
如果有很多三角形,如果我们可以在不使用python for循环的情况下旋转所有向量,速度可能会更快
在这里,我通过展开矩阵乘积进行旋转计算:
x' = x
y' = cos(t)*y - sin(t)*z
z' = sin(t)*y + cos(t)*z
因此,我们可以将这些公式矢量化:
a2 = np.array(
[[[ 0.5, 0.5, 50., ],
[ 50.5, 50.5, 100., ],
[ 0.5, 100.5, 50., ],
[ 135. , 90. , 45., ]],
[[ 50.5, 50.5, 100., ],
[ 100.5, 0.5, 50., ],
[ 100.5, 100.5, 50., ],
[ 45. , 90. , 45., ]],
[[ 100.5, 100.5, 50., ],
[ 100.5, 100.5, 0., ],
[ 0.5, 100.5, 50., ],
[ 90. , 0. , 90., ]]])
vectors = a2[:, :-1, :]
angles = a2[:, -1:, 0]
def rotate_x_batch(vectors, angles):
rad = np.deg2rad(angles)
c = np.cos(rad)
s = np.sin(rad)
x = vectors[:, :, 0]
y = vectors[:, :, 1]
z = vectors[:, :, 2]
yr = c*y - s*z
zr = s*y + c*z
vectors[:, :, 1] = yr
vectors[:, :, 2] = zr
rotate_x_batch(vectors, angles)
print np.allclose(a, a2)
您尝试的解决方案看起来不错。它有效吗?到目前为止,我只是不知道这是否是最好的方法。如果它有效,我会保留你所拥有的,因为你了解它。如果您确实发现性能是一个问题,那么您可能会发现使用numpy会有所帮助(例如,您可以将所有三角形旋转成一条直线,而不是在其上循环)。您尝试的解决方案看起来不错。它有效吗?到目前为止,我只是不知道这是否是最好的方法。如果它有效,我会保留你所拥有的,因为你了解它。如果您确实发现性能是一个问题,那么您可能会发现使用numpy会有所帮助(例如,您可以将所有三角形旋转成一行,而不是在它们上循环)。