我将3D体积数据存储在一个3维数组中,背景值为0,体积值为1。现在,我想获得此体积的任意剖面。我在这里阅读答案:How can an almost arbitrary plane in a 3D dataset be plotted by matplotlib?
但是似乎接受的答案是错误的,它生成xoy平面的映射坐标,而不是切片坐标。那么如何获得切片平面的正确形状呢?是否有将映射的形状转换为原始形状的方法?谢谢!
该问题可能已过时,但看来following function中的scipy.ndimage
可能解决了您的问题。
scipy.ndimage.interpolation.rotate
所做的是,它将整个3d数组绕3个轴中的任意一个旋转一定角度,将存储的值插入到新的“单元格”中。它还会相应地调整(扩展)新数组的大小,并使用您指定的值填充新的空白单元格。之后,您可以像平常一样进行切片,例如:array[:,sy//2,:]
。
简而言之,这是在平行于z轴的对角线上的中心切口(为简单起见:]
sz, sy, sx = array.shape
array[:,sy//2,:] # this is a cut in the x-z plane ...
# ... passing through the middle of the y-axis
# rotating by 45 degrees around the z axis ...
# ... `(2,1)` is `(x,y)` which defines the rotation plane
array_rotated = scipy.ndimage.interpolation.rotate(array, angle=45, axes=(2,1))
sz, sy, sx = array_rotated.shape
# now you'll notice that `sz` is the same ...
# ... but `sx` and `sy` have increased because the diagonal is longer
array_rotated[:,sy//2,:] # this slice is now in the new x'-z' plane ...
# ... but in terms of the original array ...
# ... it passes through the diagonal of x-y along z
您实际上可以再想一想,并通过绕不同轴旋转几次将其扩展到任意切片。
PS。如果感觉花费了太多时间,可以通过传递order=0
(默认值为order=3
)来牺牲插值的质量。这将运行得更快。