Python OpenCV Get Angle Direction of fitLine - python

Using Python OpenCv, How does one find the Angle of fitLine through an element?
I am trying to use debugger and locate, since I do not see it here in documentation ,
rows,cols = img.shape[:2]
[vx,vy,x,y] = cv2.fitLine(cnt, cv2.DIST_L2,0,0.01,0.01)
lefty = int((-x*vy/vx) + y)
righty = int(((cols-x)*vy/vx)+y)
img = cv2.line(img,(cols-1,righty),(0,lefty),(0,255,0),2)
Fitline References:
Fitline Documentation
Open CV Tutorial
Not sure which items to acquire for slope in debugger

As you already have a unit vector that defines the direction of your line [vx, vy] from,
[vx,vy,x,y] = cv2.fitLine(cnt, cv2.DIST_L2,0,0.01,0.01)
Then if you are looking for the angle between your fitted line & the x axis you can use the dot product to get the angle,
import numpy as np
x_axis = np.array([1, 0]) # unit vector in the same direction as the x axis
your_line = np.array([vx, vy]) # unit vector in the same direction as your line
dot_product = np.dot(x_axis, your_line)
angle_2_x = np.arccos(dot_product)

Related

How to use vtkOBBTree and IntersectWithLine to find the intersection of a line and a PolyDataFilter in python using vtk?

I have an oriented cylinder generated with vtkCylinderSource and some transformations are applied on it to get the orientation that i want. Here is the code for creating this oriented-cylinder:
def cylinder_object(startPoint, endPoint, radius, my_color="DarkRed", opacity=1):
colors = vtk.vtkNamedColors()
# Create a cylinder.
# Cylinder height vector is (0,1,0).
# Cylinder center is in the middle of the cylinder
cylinderSource = vtk.vtkCylinderSource()
cylinderSource.SetRadius(radius)
cylinderSource.SetResolution(50)
# Generate a random start and end point
# startPoint = [0] * 3
# endPoint = [0] * 3
rng = vtk.vtkMinimalStandardRandomSequence()
rng.SetSeed(8775070) # For testing.8775070
# Compute a basis
normalizedX = [0] * 3
normalizedY = [0] * 3
normalizedZ = [0] * 3
# The X axis is a vector from start to end
vtk.vtkMath.Subtract(endPoint, startPoint, normalizedX)
length = vtk.vtkMath.Norm(normalizedX)
vtk.vtkMath.Normalize(normalizedX)
# The Z axis is an arbitrary vector cross X
arbitrary = [0] * 3
for i in range(0, 3):
rng.Next()
arbitrary[i] = rng.GetRangeValue(-10, 10)
vtk.vtkMath.Cross(normalizedX, arbitrary, normalizedZ)
vtk.vtkMath.Normalize(normalizedZ)
# The Y axis is Z cross X
vtk.vtkMath.Cross(normalizedZ, normalizedX, normalizedY)
matrix = vtk.vtkMatrix4x4()
# Create the direction cosine matrix
matrix.Identity()
for i in range(0, 3):
matrix.SetElement(i, 0, normalizedX[i])
matrix.SetElement(i, 1, normalizedY[i])
matrix.SetElement(i, 2, normalizedZ[i])
# Apply the transforms
transform = vtk.vtkTransform()
transform.Translate(startPoint) # translate to starting point
transform.Concatenate(matrix) # apply direction cosines
transform.RotateZ(-90.0) # align cylinder to x axis
transform.Scale(1.0, length, 1.0) # scale along the height vector
transform.Translate(0, .5, 0) # translate to start of cylinder
# Transform the polydata
transformPD = vtk.vtkTransformPolyDataFilter()
transformPD.SetTransform(transform)
transformPD.SetInputConnection(cylinderSource.GetOutputPort())
cylinderSource.Update()
# Create a mapper and actor for the arrow
mapper = vtk.vtkPolyDataMapper()
actor = vtk.vtkActor()
if USER_MATRIX:
mapper.SetInputConnection(cylinderSource.GetOutputPort())
actor.SetUserMatrix(transform.GetMatrix())
else:
mapper.SetInputConnection(transformPD.GetOutputPort())
actor.SetMapper(mapper)
actor.GetProperty().SetColor(colors.GetColor3d(my_color))
actor.GetProperty().SetOpacity(opacity)
return actor, transformPD
Now i want to ray cast a line with this oriented cylinder. unfortunately, using the vtkCylinderSource as the dataset for vtkOBBTree produces the wrong points as the result. how can i use ray-casting with a PolyDataFilter?
I came up with a solution where i export my oriented-cylinder to a .stl file and then read it again to implement the ray-casting algorithm using IntersectWithLine. The problem is i have thousands of these oriented-cylinders and this method (exporting and reading) makes my code extremely slow.
def ray_cast(filename, p_source, p_target):
'''
:param filename: STL file to perform ray casting on.
:param p_source: first point
:param p_target: second point
:return: code --> 0 : No intersection.
:return: code --> +1 : p_source lies OUTSIDE the closed surface.
:return; code --> -1 : p_source lies INSIDE closed surface
'''
reader = vtk.vtkSTLReader()
reader.SetFileName(filename)
reader.Update()
mesh = reader.GetOutput()
obbtree = vtk.vtkOBBTree()
obbtree.SetDataSet(mesh)
obbtree.BuildLocator()
pointsVTKIntersection = vtk.vtkPoints()
code = obbtree.IntersectWithLine(p_source, p_target, pointsVTKIntersection, None)
# Extracting data
pointsVTKIntersectionData = pointsVTKIntersection.GetData()
noPointsVTKIntersection = pointsVTKIntersectionData.GetNumberOfTuples()
pointsIntersection = []
for idx in range(noPointsVTKIntersection):
_tup = pointsVTKIntersectionData.GetTuple3(idx)
pointsIntersection.append(_tup)
return code, pointsIntersection, noPointsVTKIntersection
Below image shows the desired result using export-stl method. (the green spheres are intersection points)
I would appreciate any suggestion and help..
With vedo:
from vedo import *
cyl = Cylinder() # vtkActor
cyl.alpha(0.5).pos(3,3,3).orientation([2,1,1])
p1, p2 = (0,0,0), (4,4,5)
ipts_coords = cyl.intersectWithLine(p1, p2)
print('hit coords are', ipts_coords)
pts = Points(ipts_coords, r=10).color("yellow")
# print(pts.polydata()) # is the vtkPolyData object
origin = Point()
ln = Line(p1,p2)
show(origin, cyl, ln, pts, axes=True)

Rotating an image without external function

I have an image which I read using scipy.misc.imread.
The pivot ([0,0]) is in the top left side of the image.
I want to flip it by 90 degrees clockwise.
The result I got was:
I hope you can find the mistake and how do I fix it:
For point (x,y) and image size h=Image_Hieght-1 and w=Image_Width-1, I changed the pivot to be at the center of the graph. (x,y) -> (x-w/2,y-h/2).
Now I can use the linear transformation of rotating a point in 90 degrees clockwise:
(1,0) -> (0,-1)
(0,-1) -> (1,0)
Conclustion: (x1,y1) -> (y1,-x1) ==> (x - w/2,y - h/2) -> (y-h/2,w/2 - x)
Now, I need to move back the point to the original graph (before I did (x,y) -> (x-w/2,y-h/2)): (y-h/2,w/2 - x) -> (y-h/2+h/2,w/2 - x+w/2) = (y,w-x)
In the code I'm using i,j (i = row, j= column), so (y,w-x) == (-j+w,i)
Conclustion:
def movePointBy90(hieght, width, i, j):
iNew = -j + width - 1
jNew = i
return int(iNew), int(jNew)
The rest of the code:
image1 = imread('image.jpg')
image2 = np.zeros([image1.shape[1], image1.shape[0], image1.shape[2]])
print(image1.shape)
print(image2.shape)
for c in range(2):
for i in range(image1.shape[0]-1):
for j in range(image1.shape[1]-1):
newPoint = moveBy90(image1.shape[0], image1.shape[1], i, j)
image2[newPoint[0], newPoint[1], c]=image1[i,j,c]
plt.subplot(1, 2, 1)
plt.imshow(image1)
plt.subplot(1, 2, 2)
plt.imshow(image2)
plt.show()
Update:
After changing to (I thought the range is a close range):
for c in range(3):
for i in range(image1.shape[0]):
for j in range(image1.shape[1]):
How many channels your image have? When I change range of c variable in for loop it works correctly. From
for c in range(2):
I think that is an issue of numpy/matplotlib with pixel values of your image. If your intensity values are in range [0,255] instead of [0,1], you should first perform
image1 = image1 / 255
after reading your image and it will be displayed correctly.
Update: I run code below with your image and got shown result, the important point is to specify the data type of your new initialized array.
import numpy as np
import matplotlib.pyplot as plt
def moveBy90(hieght, width, i, j):
iNew = -j + width - 1
jNew = i
return int(iNew), int(jNew)
image1 = plt.imread('3PWin.png')
image2 = np.zeros([image1.shape[1], image1.shape[0], image1.shape[2]], dtype=image1.dtype)
for c in range(4):
for i in range(image1.shape[0]-1):
for j in range(image1.shape[1]-1):
newPoint = moveBy90(image1.shape[0], image1.shape[1], i, j)
try:
image2[newPoint[0], newPoint[1], c]=image1[i,j,c]
except IndexError as error:
print(error)
plt.subplot(1, 2, 1)
plt.imshow(image1)
plt.subplot(1, 2, 2)
plt.imshow(image2)
plt.show()
The procedure to rotate an image seems rather complicated. Just using numpy means you may rotate 3 times by 90 degrees to achieve the equivalent of a 90 degrees clockwise rotation
import numpy as np
import matplotlib.pyplot as plt
from scipy import misc
data_orig = misc.face()
data_rotated = np.rot90(data_orig, k=3)
plt.imshow(data_rotated)
plt.show()
Or, if you want to use scipy,
import matplotlib.pyplot as plt
from scipy import misc
from scipy.ndimage import rotate
data_orig = misc.face() # or use data_orig = imread('image.jpg')
data_rotated = rotate(data_orig, 270)
plt.imshow(data_rotated)
plt.show()
In both cases the image is rotated 90° clockwise:

Interpolate between two images

I'm trying to interpolate between two images in Python.
Images are of shapes (188, 188)
I wish to interpolate the image 'in-between' these two images. Say Image_1 is at location z=0 and Image_2 is at location z=2. I want the interpolated image at location z=1.
I believe this answer (MATLAB) contains a similar problem and solution.
Creating intermediate slices in a 3D MRI volume with MATLAB
I've tried to convert this code to Python as follows:
from scipy.interpolate import interpn
from scipy.interpolate import griddata
# Construct 3D volume from images
# arr.shape = (2, 182, 182)
arr = np.r_['0,3', image_1, image_2]
slices,rows,cols = arr.shape
# Construct meshgrids
[X,Y,Z] = np.meshgrid(np.arange(cols), np.arange(rows), np.arange(slices));
[X2,Y2,Z2] = np.meshgrid(np.arange(cols), np.arange(rows), np.arange(slices*2));
# Run n-dim interpolation
Vi = interpn([X,Y,Z], arr, np.array([X1,Y1,Z1]).T)
However, this produces an error:
ValueError: The points in dimension 0 must be strictly ascending
I suspect I am not constructing my meshgrid(s) properly but am kind of lost on whether or not this approach is correct.
Any ideas?
---------- Edit -----------
Found some MATLAB code that appears to solve this problem:
Interpolating Between Two Planes in 3d space
I attempted to convert this to Python:
from scipy.ndimage.morphology import distance_transform_edt
from scipy.interpolate import interpn
def ndgrid(*args,**kwargs):
"""
Same as calling ``meshgrid`` with *indexing* = ``'ij'`` (see
``meshgrid`` for documentation).
"""
kwargs['indexing'] = 'ij'
return np.meshgrid(*args,**kwargs)
def bwperim(bw, n=4):
"""
perim = bwperim(bw, n=4)
Find the perimeter of objects in binary images.
A pixel is part of an object perimeter if its value is one and there
is at least one zero-valued pixel in its neighborhood.
By default the neighborhood of a pixel is 4 nearest pixels, but
if `n` is set to 8 the 8 nearest pixels will be considered.
Parameters
----------
bw : A black-and-white image
n : Connectivity. Must be 4 or 8 (default: 8)
Returns
-------
perim : A boolean image
From Mahotas: http://nullege.com/codes/search/mahotas.bwperim
"""
if n not in (4,8):
raise ValueError('mahotas.bwperim: n must be 4 or 8')
rows,cols = bw.shape
# Translate image by one pixel in all directions
north = np.zeros((rows,cols))
south = np.zeros((rows,cols))
west = np.zeros((rows,cols))
east = np.zeros((rows,cols))
north[:-1,:] = bw[1:,:]
south[1:,:] = bw[:-1,:]
west[:,:-1] = bw[:,1:]
east[:,1:] = bw[:,:-1]
idx = (north == bw) & \
(south == bw) & \
(west == bw) & \
(east == bw)
if n == 8:
north_east = np.zeros((rows, cols))
north_west = np.zeros((rows, cols))
south_east = np.zeros((rows, cols))
south_west = np.zeros((rows, cols))
north_east[:-1, 1:] = bw[1:, :-1]
north_west[:-1, :-1] = bw[1:, 1:]
south_east[1:, 1:] = bw[:-1, :-1]
south_west[1:, :-1] = bw[:-1, 1:]
idx &= (north_east == bw) & \
(south_east == bw) & \
(south_west == bw) & \
(north_west == bw)
return ~idx * bw
def signed_bwdist(im):
'''
Find perim and return masked image (signed/reversed)
'''
im = -bwdist(bwperim(im))*np.logical_not(im) + bwdist(bwperim(im))*im
return im
def bwdist(im):
'''
Find distance map of image
'''
dist_im = distance_transform_edt(1-im)
return dist_im
def interp_shape(top, bottom, num):
if num<0 and round(num) == num:
print("Error: number of slices to be interpolated must be integer>0")
top = signed_bwdist(top)
bottom = signed_bwdist(bottom)
r, c = top.shape
t = num+2
print("Rows - Cols - Slices")
print(r, c, t)
print("")
# rejoin top, bottom into a single array of shape (2, r, c)
# MATLAB: cat(3,bottom,top)
top_and_bottom = np.r_['0,3', top, bottom]
#top_and_bottom = np.rollaxis(top_and_bottom, 0, 3)
# create ndgrids
x,y,z = np.mgrid[0:r, 0:c, 0:t-1] # existing data
x1,y1,z1 = np.mgrid[0:r, 0:c, 0:t] # including new slice
print("Shape x y z:", x.shape, y.shape, z.shape)
print("Shape x1 y1 z1:", x1.shape, y1.shape, z1.shape)
print(top_and_bottom.shape, len(x), len(y), len(z))
# Do interpolation
out = interpn((x,y,z), top_and_bottom, (x1,y1,z1))
# MATLAB: out = out(:,:,2:end-1)>=0;
array_lim = out[-1]-1
out[out[:,:,2:out] >= 0] = 1
return out
I call this as follows:
new_image = interp_shape(image_1,image_2, 1)
Im pretty sure this is 80% of the way there but I still get this error when running:
ValueError: The points in dimension 0 must be strictly ascending
Again, I am probably not constructing my meshes correctly. I believe np.mgrid should produce the same result as MATLABs ndgrid though.
Is there a better way to construct the ndgrid equivalents?
I figured this out. Or at least a method that produces desirable results.
Based on: Interpolating Between Two Planes in 3d space
def signed_bwdist(im):
'''
Find perim and return masked image (signed/reversed)
'''
im = -bwdist(bwperim(im))*np.logical_not(im) + bwdist(bwperim(im))*im
return im
def bwdist(im):
'''
Find distance map of image
'''
dist_im = distance_transform_edt(1-im)
return dist_im
def interp_shape(top, bottom, precision):
'''
Interpolate between two contours
Input: top
[X,Y] - Image of top contour (mask)
bottom
[X,Y] - Image of bottom contour (mask)
precision
float - % between the images to interpolate
Ex: num=0.5 - Interpolate the middle image between top and bottom image
Output: out
[X,Y] - Interpolated image at num (%) between top and bottom
'''
if precision>2:
print("Error: Precision must be between 0 and 1 (float)")
top = signed_bwdist(top)
bottom = signed_bwdist(bottom)
# row,cols definition
r, c = top.shape
# Reverse % indexing
precision = 1+precision
# rejoin top, bottom into a single array of shape (2, r, c)
top_and_bottom = np.stack((top, bottom))
# create ndgrids
points = (np.r_[0, 2], np.arange(r), np.arange(c))
xi = np.rollaxis(np.mgrid[:r, :c], 0, 3).reshape((r**2, 2))
xi = np.c_[np.full((r**2),precision), xi]
# Interpolate for new plane
out = interpn(points, top_and_bottom, xi)
out = out.reshape((r, c))
# Threshold distmap to values above 0
out = out > 0
return out
# Run interpolation
out = interp_shape(image_1,image_2, 0.5)
Example output:
I came across a similar problem where I needed to interpolate the shift between frames where the change did not merely constitute a translation but also changes to the shape itself . I solved this problem by :
Using center_of_mass from scipy.ndimage.measurements to calculate the center of the object we want to move in each frame
Defining a continuous parameter t where t=0 first and t=1 last frame
Interpolate the motion between two nearest frames (with regard to a specific t value) by shifting the image back/forward via shift from scipy.ndimage.interpolation and overlaying them.
Here is the code:
def inter(images,t):
#input:
# images: list of arrays/frames ordered according to motion
# t: parameter ranging from 0 to 1 corresponding to first and last frame
#returns: interpolated image
#direction of movement, assumed to be approx. linear
a=np.array(center_of_mass(images[0]))
b=np.array(center_of_mass(images[-1]))
#find index of two nearest frames
arr=np.array([center_of_mass(images[i]) for i in range(len(images))])
v=a+t*(b-a) #convert t into vector
idx1 = (np.linalg.norm((arr - v),axis=1)).argmin()
arr[idx1]=np.array([0,0]) #this is sloppy, should be changed if relevant values are near [0,0]
idx2 = (np.linalg.norm((arr - v),axis=1)).argmin()
if idx1>idx2:
b=np.array(center_of_mass(images[idx1])) #center of mass of nearest contour
a=np.array(center_of_mass(images[idx2])) #center of mass of second nearest contour
tstar=np.linalg.norm(v-a)/np.linalg.norm(b-a) #define parameter ranging from 0 to 1 for interpolation between two nearest frames
im1_shift=shift(images[idx2],(b-a)*tstar) #shift frame 1
im2_shift=shift(images[idx1],-(b-a)*(1-tstar)) #shift frame 2
return im1_shift+im2_shift #return average
if idx1<idx2:
b=np.array(center_of_mass(images[idx2]))
a=np.array(center_of_mass(images[idx1]))
tstar=np.linalg.norm(v-a)/np.linalg.norm(b-a)
im1_shift=shift(images[idx2],-(b-a)*(1-tstar))
im2_shift=shift(images[idx1],(b-a)*(tstar))
return im1_shift+im2_shift
Result example
I don't know the solution to your problem, but I don't think it's possible to do this with interpn.
I corrected the code that you tried, and used the following input images:
But the result is:
Here's the corrected code:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy import interpolate
n = 8
img1 = np.zeros((n, n))
img2 = np.zeros((n, n))
img1[2:4, 2:4] = 1
img2[4:6, 4:6] = 1
plt.figure()
plt.imshow(img1, cmap=cm.Greys)
plt.figure()
plt.imshow(img2, cmap=cm.Greys)
points = (np.r_[0, 2], np.arange(n), np.arange(n))
values = np.stack((img1, img2))
xi = np.rollaxis(np.mgrid[:n, :n], 0, 3).reshape((n**2, 2))
xi = np.c_[np.ones(n**2), xi]
values_x = interpolate.interpn(points, values, xi, method='linear')
values_x = values_x.reshape((n, n))
print(values_x)
plt.figure()
plt.imshow(values_x, cmap=cm.Greys)
plt.clim((0, 1))
plt.show()
I think the main difference between your code and mine is in the specification of xi. interpn tends to be somewhat confusing to use, and I've explained it in greater detail in an older answer. If you're curious about the mechanics of how I've specified xi, see this answer of mine explaining what I've done.
This result is not entirely surprising, because interpn just linearly interpolated between the two images: so the parts which had 1 in one image and 0 in the other simply became 0.5.
Over here, since one image is the translation of the other, it's clear that we want an image that's translated "in-between". But how would interpn interpolate two general images? If you had one small circle and one big circle, is it in any way clear that there should be a circle of intermediate size "between" them? What about interpolating between a dog and a cat? Or a dog and a building?
I think you are essentially trying to "draw lines" connecting the edges of the two images and then trying to figure out the image in between. This is similar to sampling a moving video at a half-frame. You might want to check out something like optical flow, which connects adjacent frames using vectors. I'm not aware if and what python packages/implementations are available though.

Crop a pattern-based area of multiple images in python

I have a big number of screenshots that need to be cropped. All the images look similar - there is a rectangular window with blue border, containing some graphical elements inside. This window is contained inside another one but I need to crop only the inner window. Across all images the dimensions of the inner window are different and so is the content. The content in most cases includes elements with rectangular form and sometimes - blue border, the same border as the inner window. I am mentioning this because I am thinking of the following flow:
A script that goes through all images in the target directory. For each of them:
Find the area to be cropped (inner window)
Crop the area
Save the file
How can this be done? Python is not compulsory, can be any other too also.
It's not straightforward but this is a possible recipe:
import matplotlib.pyplot as plt
import numpy as np
def synthimage():
w,h = 300,200
im = np.random.randint(0,255,(w,h,3))/255
xa = np.random.randint(50,w-60)
xb = xa + np.random.randint(50,90)
ya = np.random.randint(50,h-60)
yb = ya + np.random.randint(20,50)
im[xa:xb,ya] = np.array([1,0,0])
im[xa:xb,yb] = np.array([1,0,0])
im[xa,ya:yb] = np.array([1,0,0])
im[xb,ya:yb] = np.array([1,0,0])
return im
def getRectPoints(im):
x,y = [],[]
for i in range(im.shape[0]):
for j in range(im.shape[1]):
if (im[i,j]-np.array([1,0,0])).sum()==0:
x.append(i)
y.append(j)
return np.array(x),np.array(y)
def denoise(x,y):
nx,ny = [],[]
for i in range(x.shape[0]):
d = np.sqrt((x[i]-x)**2+(y[i]-y)**2)
m = d<2
if len(m.nonzero()[0])>2:
nx.append(x[i])
ny.append(y[i])
return np.array(nx),np.array(ny)
im = synthimage()
plt.imshow(np.swapaxes(im,0,1),origin='lower',interpolation='nearest')
plt.show()
x,y = getRectPoints(im)
plt.scatter(x,y,c='red')
plt.xlim(0,300)
plt.ylim(0,200)
plt.show()
nx,ny = denoise(x,y)
plt.scatter(nx,ny,c='red')
plt.xlim(0,300)
plt.ylim(0,200)
plt.show()
#Assuming rectangle has no rotation (otherwise check Scipy ConveHull)
xmi = nx.min()
xma = nx.max()
ymi = ny.min()
yma = ny.max()
new = np.ones(im.shape)
new[xmi:xma,ymi:yma] = im[xmi:xma,ymi:yma]
plt.imshow(np.swapaxes(new,0,1),origin='lower',interpolation='nearest')
plt.show()
, the name of the functions should be self-explaining. Synthetic data was generated for the purpose of this exercise. The results are (in order):
Obviously each one of this steps can be changed depending on the requirements but this would be a functional solution for the majority of case-studies.

How can an almost arbitrary plane in a 3D dataset be plotted by matplotlib?

There is an array containing 3D data of shape e.g. (64,64,64), how do you plot a plane given by a point and a normal (similar to hkl planes in crystallography), through this dataset?
Similar to what can be done in MayaVi by rotating a plane through the data.
The resulting plot will contain non-square planes in most cases.
Can those be done with matplotlib (some sort of non-rectangular patch)?
Edit: I almost solved this myself (see below) but still wonder how non-rectangular patches can be plotted in matplotlib...?
Edit: Due to discussions below I restated the question.
This is funny, a similar question I replied to just today. The way to go is: interpolation. You can use griddata from scipy.interpolate:
Griddata
This page features a very nice example, and the signature of the function is really close to your data.
You still have to somehow define the points on you plane for which you want to interpolate the data. I will have a look at this, my linear algebra lessons where a couple of years ago
I have the penultimate solution for this problem. Partially solved by using the second answer to Plot a plane based on a normal vector and a point in Matlab or matplotlib :
# coding: utf-8
import numpy as np
from matplotlib.pyplot import imshow,show
A=np.empty((64,64,64)) #This is the data array
def f(x,y):
return np.sin(x/(2*np.pi))+np.cos(y/(2*np.pi))
xx,yy= np.meshgrid(range(64), range(64))
for x in range(64):
A[:,:,x]=f(xx,yy)*np.cos(x/np.pi)
N=np.zeros((64,64))
"""This is the plane we cut from A.
It should be larger than 64, due to diagonal planes being larger.
Will be fixed."""
normal=np.array([-1,-1,1]) #Define cut plane here. Normal vector components restricted to integers
point=np.array([0,0,0])
d = -np.sum(point*normal)
def plane(x,y): # Get plane's z values
return (-normal[0]*x-normal[1]*y-d)/normal[2]
def getZZ(x,y): #Get z for all values x,y. If z>64 it's out of range
for i in x:
for j in y:
if plane(i,j)<64:
N[i,j]=A[i,j,plane(i,j)]
getZZ(range(64),range(64))
imshow(N, interpolation="Nearest")
show()
It's not the ultimate solution since the plot is not restricted to points having a z value, planes larger than 64 * 64 are not accounted for and the planes have to be defined at (0,0,0).
For the reduced requirements, I prepared a simple example
import numpy as np
import pylab as plt
data = np.arange((64**3))
data.resize((64,64,64))
def get_slice(volume, orientation, index):
orientation2slicefunc = {
"x" : lambda ar:ar[index,:,:],
"y" : lambda ar:ar[:,index,:],
"z" : lambda ar:ar[:,:,index]
}
return orientation2slicefunc[orientation](volume)
plt.subplot(221)
plt.imshow(get_slice(data, "x", 10), vmin=0, vmax=64**3)
plt.subplot(222)
plt.imshow(get_slice(data, "x", 39), vmin=0, vmax=64**3)
plt.subplot(223)
plt.imshow(get_slice(data, "y", 15), vmin=0, vmax=64**3)
plt.subplot(224)
plt.imshow(get_slice(data, "z", 25), vmin=0, vmax=64**3)
plt.show()
This leads to the following plot:
The main trick is dictionary mapping orienations to lambda-methods, which saves us from writing annoying if-then-else-blocks. Of course you can decide to give different names,
e.g., numbers, for the orientations.
Maybe this helps you.
Thorsten
P.S.: I didn't care about "IndexOutOfRange", for me it's o.k. to let this exception pop out since it is perfectly understandable in this context.
I had to do something similar for a MRI data enhancement:
Probably the code can be optimized but it works as it is.
My data is 3 dimension numpy array representing an MRI scanner. It has size [128,128,128] but the code can be modified to accept any dimensions. Also when the plane is outside the cube boundary you have to give the default values to the variable fill in the main function, in my case I choose: data_cube[0:5,0:5,0:5].mean()
def create_normal_vector(x, y,z):
normal = np.asarray([x,y,z])
normal = normal/np.sqrt(sum(normal**2))
return normal
def get_plane_equation_parameters(normal,point):
a,b,c = normal
d = np.dot(normal,point)
return a,b,c,d #ax+by+cz=d
def get_point_plane_proximity(plane,point):
#just aproximation
return np.dot(plane[0:-1],point) - plane[-1]
def get_corner_interesections(plane, cube_dim = 128): #to reduce the search space
#dimension is 128,128,128
corners_list = []
only_x = np.zeros(4)
min_prox_x = 9999
min_prox_y = 9999
min_prox_z = 9999
min_prox_yz = 9999
for i in range(cube_dim):
temp_min_prox_x=abs(get_point_plane_proximity(plane,np.asarray([i,0,0])))
# print("pseudo distance x: {0}, point: [{1},0,0]".format(temp_min_prox_x,i))
if temp_min_prox_x < min_prox_x:
min_prox_x = temp_min_prox_x
corner_intersection_x = np.asarray([i,0,0])
only_x[0]= i
temp_min_prox_y=abs(get_point_plane_proximity(plane,np.asarray([i,cube_dim,0])))
# print("pseudo distance y: {0}, point: [{1},{2},0]".format(temp_min_prox_y,i,cube_dim))
if temp_min_prox_y < min_prox_y:
min_prox_y = temp_min_prox_y
corner_intersection_y = np.asarray([i,cube_dim,0])
only_x[1]= i
temp_min_prox_z=abs(get_point_plane_proximity(plane,np.asarray([i,0,cube_dim])))
#print("pseudo distance z: {0}, point: [{1},0,{2}]".format(temp_min_prox_z,i,cube_dim))
if temp_min_prox_z < min_prox_z:
min_prox_z = temp_min_prox_z
corner_intersection_z = np.asarray([i,0,cube_dim])
only_x[2]= i
temp_min_prox_yz=abs(get_point_plane_proximity(plane,np.asarray([i,cube_dim,cube_dim])))
#print("pseudo distance z: {0}, point: [{1},{2},{2}]".format(temp_min_prox_yz,i,cube_dim))
if temp_min_prox_yz < min_prox_yz:
min_prox_yz = temp_min_prox_yz
corner_intersection_yz = np.asarray([i,cube_dim,cube_dim])
only_x[3]= i
corners_list.append(corner_intersection_x)
corners_list.append(corner_intersection_y)
corners_list.append(corner_intersection_z)
corners_list.append(corner_intersection_yz)
corners_list.append(only_x.min())
corners_list.append(only_x.max())
return corners_list
def get_points_intersection(plane,min_x,max_x,data_cube,shape=128):
fill = data_cube[0:5,0:5,0:5].mean() #this can be a parameter
extended_data_cube = np.ones([shape+2,shape,shape])*fill
extended_data_cube[1:shape+1,:,:] = data_cube
diag_image = np.zeros([shape,shape])
min_x_value = 999999
for i in range(shape):
for j in range(shape):
for k in range(int(min_x),int(max_x)+1):
current_value = abs(get_point_plane_proximity(plane,np.asarray([k,i,j])))
#print("current_value:{0}, val: [{1},{2},{3}]".format(current_value,k,i,j))
if current_value < min_x_value:
diag_image[i,j] = extended_data_cube[k,i,j]
min_x_value = current_value
min_x_value = 999999
return diag_image
The way it works is the following:
you create a normal vector:
for example [5,0,3]
normal1=create_normal_vector(5, 0,3) #this is only to normalize
then you create a point:
(my cube data shape is [128,128,128])
point = [64,64,64]
You calculate the plane equation parameters, [a,b,c,d] where ax+by+cz=d
plane1=get_plane_equation_parameters(normal1,point)
then to reduce the search space you can calculate the intersection of the plane with the cube:
corners1 = get_corner_interesections(plane1,128)
where corners1 = [intersection [x,0,0],intersection [x,128,0],intersection [x,0,128],intersection [x,128,128], min intersection [x,y,z], max intersection [x,y,z]]
With all these you can calculate the intersection between the cube and the plane:
image1 = get_points_intersection(plane1,corners1[-2],corners1[-1],data_cube)
Some examples:
normal is [1,0,0] point is [64,64,64]
normal is [5,1,0],[5,1,1],[5,0,1] point is [64,64,64]:
normal is [5,3,0],[5,3,3],[5,0,3] point is [64,64,64]:
normal is [5,-5,0],[5,-5,-5],[5,0,-5] point is [64,64,64]:
Thank you.
The other answers here do not appear to be very efficient with explicit loops over pixels or using scipy.interpolate.griddata, which is designed for unstructured input data. Here is an efficient (vectorized) and generic solution.
There is a pure numpy implementation (for nearest-neighbor "interpolation") and one for linear interpolation, which delegates the interpolation to scipy.ndimage.map_coordinates. (The latter function probably didn't exist in 2013, when this question was asked.)
import numpy as np
from scipy.ndimage import map_coordinates
def slice_datacube(cube, center, eXY, mXY, fill=np.nan, interp=True):
"""Get a 2D slice from a 3-D array.
Copyright: Han-Kwang Nienhuys, 2020.
License: any of CC-BY-SA, CC-BY, BSD, GPL, LGPL
Reference: https://stackoverflow.com/a/62733930/6228891
Parameters:
- cube: 3D array, assumed shape (nx, ny, nz).
- center: shape (3,) with coordinates of center.
can be float.
- eXY: unit vectors, shape (2, 3) - for X and Y axes of the slice.
(unit vectors must be orthogonal; normalization is optional).
- mXY: size tuple of output array (mX, mY) - int.
- fill: value to use for out-of-range points.
- interp: whether to interpolate (rather than using 'nearest')
Return:
- slice: array, shape (mX, mY).
"""
center = np.array(center, dtype=float)
assert center.shape == (3,)
eXY = np.array(eXY)/np.linalg.norm(eXY, axis=1)[:, np.newaxis]
if not np.isclose(eXY[0] # eXY[1], 0, atol=1e-6):
raise ValueError(f'eX and eY not orthogonal.')
# R: rotation matrix: data_coords = center + R # slice_coords
eZ = np.cross(eXY[0], eXY[1])
R = np.array([eXY[0], eXY[1], eZ], dtype=np.float32).T
# setup slice points P with coordinates (X, Y, 0)
mX, mY = int(mXY[0]), int(mXY[1])
Xs = np.arange(0.5-mX/2, 0.5+mX/2)
Ys = np.arange(0.5-mY/2, 0.5+mY/2)
PP = np.zeros((3, mX, mY), dtype=np.float32)
PP[0, :, :] = Xs.reshape(mX, 1)
PP[1, :, :] = Ys.reshape(1, mY)
# Transform to data coordinates (x, y, z) - idx.shape == (3, mX, mY)
if interp:
idx = np.einsum('il,ljk->ijk', R, PP) + center.reshape(3, 1, 1)
slice = map_coordinates(cube, idx, order=1, mode='constant', cval=fill)
else:
idx = np.einsum('il,ljk->ijk', R, PP) + (0.5 + center.reshape(3, 1, 1))
idx = idx.astype(np.int16)
# Find out which coordinates are out of range - shape (mX, mY)
badpoints = np.any([
idx[0, :, :] < 0,
idx[0, :, :] >= cube.shape[0],
idx[1, :, :] < 0,
idx[1, :, :] >= cube.shape[1],
idx[2, :, :] < 0,
idx[2, :, :] >= cube.shape[2],
], axis=0)
idx[:, badpoints] = 0
slice = cube[idx[0], idx[1], idx[2]]
slice[badpoints] = fill
return slice
# Demonstration
nx, ny, nz = 50, 70, 100
cube = np.full((nx, ny, nz), np.float32(1))
cube[nx//4:nx*3//4, :, :] += 1
cube[:, ny//2:ny*3//4, :] += 3
cube[:, :, nz//4:nz//2] += 7
cube[nx//3-2:nx//3+2, ny//2-2:ny//2+2, :] = 0 # black dot
Rz, Rx = np.pi/6, np.pi/4 # rotation angles around z and x
cz, sz = np.cos(Rz), np.sin(Rz)
cx, sx = np.cos(Rx), np.sin(Rx)
Rmz = np.array([[cz, -sz, 0], [sz, cz, 0], [0, 0, 1]])
Rmx = np.array([[1, 0, 0], [0, cx, -sx], [0, sx, cx]])
eXY = (Rmx # Rmz).T[:2]
slice = slice_datacube(
cube,
center=[nx/3, ny/2, nz*0.7],
eXY=eXY,
mXY=[80, 90],
fill=np.nan,
interp=False
)
import matplotlib.pyplot as plt
plt.close('all')
plt.imshow(slice.T) # imshow expects shape (mY, mX)
plt.colorbar()
Output (for interp=False):
For this test case (50x70x100 datacube, 80x90 slice size) the run time is 376 µs (interp=False) and 550 µs (interp=True) on my laptop.

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