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我正在尝试对 python 中的 3d 数组中的特定元素执行操作。下面是一个数组示例：

[[[   0.5         0.5        50.      ]
  [  50.5        50.5       100.      ]
  [   0.5       100.5        50.      ]
  [ 135.         90.         45.      ]]

 [[  50.5        50.5       100.      ]
  [ 100.5         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.      ]]

我需要做的一个例子是获取数组中看到的三个值，即 0.5、0.5、50。并从第 4 行获取第一个元素，即 135。并将这四个元素发送到一个函数中。然后该函数返回需要放入数组的 3 个元素的新值。

我对python很陌生，所以我很难让它工作。我应该做一个循环吗？或者是其他东西？

谢谢尼克

尝试解决方案：

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])
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1 回答 1

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您可以使您的VectMath.rotate_x函数旋转向量数组，然后使用 slice 来获取 & 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，如果我们可以在没有 python for 循环的情况下旋转所有向量，可能会更快。

这里我通过展开矩阵乘积来进行旋转计算：

x' = x
y' = cos(t)*y - sin(t)*z
z' = sin(t)*y + cos(t)*z

所以我们可以向量化这些公式：

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)
于 2013-07-26T05:40:27.760 回答