问题
I have a 3d array created with numpy, and I was wondering how I can rotate it by a custom angle, not just the rot90 function that numpy has. Can anyone help?
The 3d matrix represents an image (such as a cube, or some other shape) ie
0:
1 1 1
1 1
1 1 1
1:
1 1
1 1
2:
1 1 1
1 1
1 1 1
EDIT: Moved solution to answer
回答1:
Have a look at the scipy.ndimage.interpolation.rotate function.
The reason this is in scipy and not in numpy is that rotating an image 90 degrees is done by just chaning the indices of the array. However, if you want to rotate an image by some arbitrary degrees you have to deal with interpolation, which adds a whole new layer of complexity to the problem. This is because all the pixels in the original image "perfectly lines up with" pixels in the rotated image when you rotate it by a factor of 90 degrees. This is not the case in general when you rotate an image.
回答2:
After some trial and error I came up with some code for my purposes (0 means empty in the array, another number will mean a filled voxel.
def rotate(self, deg_angle, axis):
d = len(self.matrix)
h = len(self.matrix[0])
w = len(self.matrix[0][0])
min_new_x = 0
max_new_x = 0
min_new_y = 0
max_new_y = 0
min_new_z = 0
max_new_z = 0
new_coords = []
angle = radians(deg_angle)
for z in range(d):
for y in range(h):
for x in range(w):
new_x = None
new_y = None
new_z = None
if axis == "x":
new_x = int(round(x))
new_y = int(round(y*cos(angle) - z*sin(angle)))
new_z = int(round(y*sin(angle) + z*cos(angle)))
elif axis == "y":
new_x = int(round(z*sin(angle) + x*cos(angle)))
new_y = int(round(y))
new_z = int(round(z*cos(angle) - x*sin(angle)))
elif axis == "z":
new_x = int(round(x*cos(angle) - y*sin(angle)))
new_y = int(round(x*sin(angle) + y*cos(angle)))
new_z = int(round(z))
val = self.matrix.item((z, y, x))
new_coords.append((val, new_x, new_y, new_z))
if new_x < min_new_x: min_new_x = new_x
if new_x > max_new_x: max_new_x = new_x
if new_y < min_new_y: min_new_y = new_y
if new_y > max_new_y: max_new_y = new_y
if new_z < min_new_z: min_new_z = new_z
if new_z > max_new_z: max_new_z = new_z
new_x_offset = abs(min_new_x)
new_y_offset = abs(min_new_y)
new_z_offset = abs(min_new_z)
new_width = abs(min_new_x - max_new_x)
new_height = abs(min_new_y - max_new_y)
new_depth = abs(min_new_z - max_new_z)
rotated = np.empty((new_depth + 1, new_height + 1, new_width + 1))
rotated.fill(0)
for coord in new_coords:
val = coord[0]
x = coord[1]
y = coord[2]
z = coord[3]
if rotated[new_z_offset + z][new_y_offset + y][new_x_offset + x] == 0:
rotated[new_z_offset + z][new_y_offset + y][new_x_offset + x] = val
self.matrix = rotated
The way I use the above code is:
cube = Rect_Prism(20, 20, 20) # creates a 3d array similar to above example, just bigger
cube.rotate(20, "x")
cube.rotate(60, "y")
Rect_Prism creates a MxNxD matrix, but in this case NxNxN.
And result when printing:
# # # # # # # # # # # #
# # # # # # #
# # # # # # #
# # # # #
# # # # # # # #
# # # # # # #
# # # # # # # # # # # # #
# # # #
# # # #
# # # #
# # # #
# # # #
# # # #
# # # #
# # # #
# # # # # #
# # # #
# # # # # # # # # # # #
# # # # #
# # # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # #
# # # # # # # # # # # #
回答3:
You have to create a rotation matrix and multiply your this matrix for your array. Here the information
Wikipedea rotation matrix information
An example for 2d rotating
回答4:
I think you should consider a "vector" representation for your data, instead of the current "raster" representation.
A vector representation would mean that, instead of each "voxel" being defined by its position in a grid, you would have some sort of list of voxels, with actual 3D coordinates.
So instead of having an MxNxD matrix where each voxel is a "black/white" dot, you could have a Mx3 matrix, where each row is a point, with columns being X, Y and Z.
That way, you would multiply the list by a 3x3 rotation matrix, and get another list of transformed coordinates.
You would remain with the problem of "rendering" these vector points (or lines, better yet) to a raster matrix (either pixels or voxels, but your sample image looks like 3D info has been projected to 2D space). There are a lot of techniques for doing this.
来源：https://stackoverflow.com/questions/49097667/how-can-i-rotate-a-3d-array-nxnxn-by-x-degrees-around-x-y-and-z-axes