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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.

Fill polygon with custom hatch with matplotlib

I want to fill a bunch of polygons with line hatch. The lines must have a specific angle with respect to x-axis. I found that matplotlib already suppots some hatch classes and one can define a custom class (like How to fill a polygon with a custom hatch in matplotlib?). I tried to generate a custom hatch but when I append it to the list of hatches the init function doesn't know the angle. I tried with the following class:
class AngularHatch(HatchPatternBase):
def __init__(self, hatch, density, angle):
self.num_lines = int((hatch.count('{'))*density*3)
self.num_vertices = self.num_lines * 2
self.R = np.array([[np.cos(angle), -np.sin(angle)],
[np.sin(angle), np.cos(angle)]])
def set_vertices_and_codes(self, vertices, codes):
steps, stepsize = np.linspace(0.0, 1.0, self.num_lines, False,
retstep=True)
steps += stepsize / 2.
vertices[0::2, 0] = 0
vertices[0::2, 1] = steps
vertices[1::2, 0] = 1
vertices[1::2, 1] = steps
for i, v in enumerate(vertices):
vertices[i] = self.R.dot(v)
codes[0::2] = Path.MOVETO
codes[1::2] = Path.LINETO
Then I add this class to the list of available classes for hatching. However this will not generate the correct lines since the code is modified from the HorizontalHatch source code here and I think this generates lines in the unit square. Moreover I need to generate this patch for a specific angle for each polygon I want to render. ¿Any ideas on how to give the correct angle to this class per polygon?

The following does not solve this issue. It just solves part of the problem and shows at which point the approach fails. I am currently convinced that hatching with arbitrary angles is not possible with matplotlib, because the size of the unit cell is fixed.
To overcome the problem of setting the angle, one may define a custom format from which to take the angle information. E.g. "{angle}{factor}", such that "{45}{2}" would produce a hatching with an angle of 45° and a density factor of 2.
I then do not completely understand the attempt of calculating the vertices. To replicate the behaviour of the hatches which are built-in, one may rotate them directly.
The problem is that this way the line hatches work only for angles of 45°. This is because the lines at the edges of the unit cell do not align well. See the following:
import numpy as np
import matplotlib.hatch
import matplotlib.path
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse, Rectangle
class AngularHatch(matplotlib.hatch.HatchPatternBase):
def __init__(self, hatch, density):
self.num_lines=0
self.num_vertices=0
if hatch[0] == "{":
h = hatch.strip("{}").split("}{")
angle = np.deg2rad(float(h[0])-45)
d = float(h[1])
self.num_lines = int(density*d)
self.num_vertices = (self.num_lines + 1) * 2
self.R = np.array([[np.cos(angle), -np.sin(angle)],
[np.sin(angle), np.cos(angle)]])
def set_vertices_and_codes(self, vertices, codes):
steps = np.linspace(-0.5, 0.5, self.num_lines + 1, True)
vertices[0::2, 0] = 0.0 + steps
vertices[0::2, 1] = 0.0 - steps
vertices[1::2, 0] = 1.0 + steps
vertices[1::2, 1] = 1.0 - steps
codes[0::2] = matplotlib.path.Path.MOVETO
codes[1::2] = matplotlib.path.Path.LINETO
vertices[:,:] = np.dot((vertices-0.5),self.R)+0.5
matplotlib.hatch._hatch_types.append(AngularHatch)
fig = plt.figure()
ax = fig.add_subplot(111)
ellipse = ax.add_patch(Rectangle((0.1, 0.1), 0.4, 0.8, fill=False))
ellipse.set_hatch('{45}{1}')
ellipse.set_color('red')
ellipse = ax.add_patch(Rectangle((0.55, 0.1), 0.4, 0.8, fill=False))
ellipse.set_hatch('{22}{1}')
ellipse.set_color('blue')
plt.show()

How to plot 3D ellipsoid with Mayavi

I would like to plot diffusion tensors(ellipsoid) in diffusion MRI. The data have three Eigenvalues of the corresponding diffusion tensor. I want to draw an 3D Ellipsoid with its semi-axes lengths corresponding to those three Eigenvalues.
How to do it with Mayavi?

Google brought me here and to the answer. I found how to render an ellipsoid here: https://github.com/spyke/spyke/blob/master/demo/mayavi_test.py and combined it with the arrow from here https://stackoverflow.com/a/20109619/2389450 to produce something like: http://imageshack.com/a/img673/7664/YzbTHY.png
Cheers,
Max
Code:
from mayavi.api import Engine
from mayavi.sources.api import ParametricSurface
from mayavi.modules.api import Surface
from mayavi import mlab
from tvtk.tools import visual
import numpy as np
def Arrow_From_A_to_B(x1, y1, z1, x2, y2, z2,scale=None):
ar1=visual.arrow(x=x1, y=y1, z=z1)
ar1.length_cone=0.4
arrow_length=np.sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2)
if scale is None:
ar1.actor.scale=[arrow_length, arrow_length, arrow_length]
else:
ar1.actor.scale=scale
ar1.pos = ar1.pos/arrow_length
ar1.axis = [x2-x1, y2-y1, z2-z1]
return ar1
engine = Engine()
engine.start()
scene = engine.new_scene()
scene.scene.disable_render = True # for speed
visual.set_viewer(scene)
surfaces = []
for i in range(2):
source = ParametricSurface()
source.function = 'ellipsoid'
engine.add_source(source)
surface = Surface()
source.add_module(surface)
actor = surface.actor # mayavi actor, actor.actor is tvtk actor
#actor.property.ambient = 1 # defaults to 0 for some reason, ah don't need it, turn off scalar visibility instead
actor.property.opacity = 0.7
actor.property.color = (0,0,1) # tuple(np.random.rand(3))
actor.mapper.scalar_visibility = False # don't colour ellipses by their scalar indices into colour map
actor.property.backface_culling = True # gets rid of weird rendering artifact when opacity is < 1
actor.property.specular = 0.1
#actor.property.frontface_culling = True
actor.actor.orientation = np.array([1,0,0]) * 360 # in degrees
actor.actor.origin = np.array([0,0,0])
actor.actor.position = np.array([0,0,0])
actor.actor.scale = np.array([ 0.26490647, 0.26490647, 0.92717265])
actor.enable_texture=True
actor.property.representation = ['wireframe', 'surface'][i]
surfaces.append(surface)
Arrow_From_A_to_B(0,0,0, 0.26490647, 0, 0,np.array([0.26490647,0.4,0.4]))
Arrow_From_A_to_B(0,0,0, 0, 0.26490647, 0,np.array([0.4,0.26490647,0.4]))
Arrow_From_A_to_B(0,0,0, 0, 0, 0.92717265,np.array([0.4,0.4,0.92717265]))
source.scene.background = (1.0,1.0,1.0)
scene.scene.disable_render = False # now turn it on
# set the scalars, this has to be done some indeterminate amount of time
# after each surface is created, otherwise the scalars get overwritten
# later by their default of 1.0
for i, surface in enumerate(surfaces):
vtk_srcs = mlab.pipeline.get_vtk_src(surface)
print('len(vtk_srcs) = %d' % len(vtk_srcs))
vtk_src = vtk_srcs[0]
try: npoints = len(vtk_src.point_data.scalars)
except TypeError:
print('hit the TypeError on surface i=%d' % i)
npoints = 2500
vtk_src.point_data.scalars = np.tile(i, npoints)
# on pick, find the ellipsoid with origin closest to the picked coord,
# then check if that coord falls within that nearest ellipsoid, and if
# so, print out the ellispoid id, or pop it up in a tooltip
mlab.show()

Generate coordinates inside Polygon

I want to bin the values of polygons to a fine regular grid.
For instance, I have the following coordinates:
data = 2.353
data_lats = np.array([57.81000137, 58.15999985, 58.13000107, 57.77999878])
data_lons = np.array([148.67999268, 148.69999695, 148.47999573, 148.92999268])
My regular grid looks like this:
delta = 0.25
grid_lons = np.arange(-180, 180, delta)
grid_lats = np.arange(90, -90, -delta)
llx, lly = np.meshgrid( grid_lons, grid_lats )
rows = lly.shape[0]
cols = llx.shape[1]
grid = np.zeros((rows,cols))
Now I can find the grid pixel that corresponds to the center of my polygon very easily:
centerx, centery = np.mean(data_lons), np.mean(data_lats)
row = int(np.floor( centery/delta ) + (grid.shape[0]/2))
col = int(np.floor( centerx/delta ) + (grid.shape[1]/2))
grid[row,col] = data
However, there are probably a couple of grid pixels that still intersect with the polygon. Hence, I would like to generate a bunch of coordinates inside my polygon (data_lons, data_lats) and find their corresponding grid pixel as before. Do you a suggestion to generate the coordinates randomly or systematically? I failed, but am still trying.
Note: One data set contains around ~80000 polygons so it has to be really fast (a couple of seconds). That is also why I chose this approach, because it does not account the area of overlap... (like my earlier question Data binning: irregular polygons to regular mesh which is VERY slow)

I worked on a quick and dirty solution by simply calculating the coordinates between corner pixels. Take a look:
dlats = np.zeros((data_lats.shape[0],4))+np.nan
dlons = np.zeros((data_lons.shape[0],4))+np.nan
idx = [0,1,3,2,0] #rearrange the corner pixels
for cc in range(4):
dlats[:,cc] = np.mean((data_lats[:,idx[cc]],data_lats[:,idx[cc+1]]), axis=0)
dlons[:,cc] = np.mean((data_lons[:,idx[cc]],data_lons[:,idx[cc+1]]), axis=0)
data_lats = np.column_stack(( data_lats, dlats ))
data_lons = np.column_stack(( data_lons, dlons ))
Thus, the red dots represent the original corners - the blue ones the intermediate pixels between them.
I can do this one more time and include the center pixel (geo[:,[4,9]])
dlats = np.zeros((data.shape[0],8))
dlons = np.zeros((data.shape[0],8))
for cc in range(8):
dlats[:,cc] = np.mean((data_lats[:,cc], geo[:,4]), axis=0)
dlons[:,cc] = np.mean((data_lons[:,cc], geo[:,9]), axis=0)
data_lats = np.column_stack(( data_lats, dlats, geo[:,4] ))
data_lons = np.column_stack(( data_lons, dlons, geo[:,9] ))
This works really nice, and I can assign each point directly to its corresponding grid pixel like this:
row = np.floor( data_lats/delta ) + (llx.shape[0]/2)
col = np.floor( data_lons/delta ) + (llx.shape[1]/2)
However the final binning now takes ~7sec!!! How can I speed this code up:
for ii in np.arange(len(data)):
for cc in np.arange(data_lats.shape[1]):
final_grid[row[ii,cc],col[ii,cc]] += data[ii]
final_grid_counts[row[ii,cc],col[ii,cc]] += 1

You'll need to test the following approach to see if it is fast enough. First, you should modify all your lats and lons into, to make them (possibly fractional) indices into your grid:
idx_lats = (data_lats - lat_grid_start) / lat_grid step
idx_lons = (data_lons - lon_grid_start) / lon_grid step
Next, we want to split your polygons into triangles. For any convex polygon, you could take the center of the polygon as one vertex of all triangles, and then the vertices of the polygon in consecutive pairs. But if your polygon are all quadrilaterals, it is going to be faster to divide them into only 2 triangles, using vertices 0, 1, 2 for the first, and 0, 2, 3 for the second.
To know if a certain point is inside a triangle, I am going to use the barycentric coordinates approach described here. This first function checks whether a bunch of points are inside a triangle:
def check_in_triangle(x, y, x_tri, y_tri) :
A = np.vstack((x_tri[0], y_tri[0]))
lhs = np.vstack((x_tri[1:], y_tri[1:])) - A
rhs = np.vstack((x, y)) - A
uv = np.linalg.solve(lhs, rhs)
# Equivalent to (uv[0] >= 0) & (uv[1] >= 0) & (uv[0] + uv[1] <= 1)
return np.logical_and(uv >= 0, axis=0) & (np.sum(uv, axis=0) <= 1)
Given a triangle by its vertices, you can get the lattice points inside it, by running the above function on the lattice points in the bounding box of the triangle:
def lattice_points_in_triangle(x_tri, y_tri) :
x_grid = np.arange(np.ceil(np.min(x_tri)), np.floor(np.max(x_tri)) + 1)
y_grid = np.arange(np.ceil(np.min(y_tri)), np.floor(np.max(y_tri)) + 1)
x, y = np.meshgrid(x_grid, y_grid)
x, y = x.reshape(-1), y.reshape(-1)
idx = check_in_triangle(x, y, x_tri, y_tri)
return x[idx], y[idx]
And for a quadrilateral, you simply call this last function twice :
def lattice_points_in_quadrilateral(x_quad, y_quad) :
return map(np.concatenate,
zip(lattice_points_in_triangle(x_quad[:3], y_quad[:3]),
lattice_points_in_triangle(x_quad[[0, 2, 3]],
y_quad[[0, 2, 3]])))
If you run this code on your example data, you will get two empty arrays returned: that's because the order of the quadrilateral points is a surprising one: indices 0 and 1 define one diagonal, 2 and 3 the other. My function above was expecting the vertices to be ordered around the polygon. If you really are doing things this other way, you need to change the second call to lattice_points_in_triangle inside lattice_points_in_quadrilateral so that the indices used are [0, 1, 3] instead of [0, 2, 3].
And now, with that change :
>>> idx_lats = (data_lats - (-180) ) / 0.25
>>> idx_lons = (data_lons - (-90) ) / 0.25
>>> lattice_points_in_quadrilateral(idx_lats, idx_lons)
[array([952]), array([955])]
If you change the resolution of your grid to 0.1:
>>> idx_lats = (data_lats - (-180) ) / 0.1
>>> idx_lons = (data_lons - (-90) ) / 0.1
>>> lattice_points_in_quadrilateral(idx_lats, idx_lons)
[array([2381, 2380, 2381, 2379, 2380, 2381, 2378, 2379, 2378]),
array([2385, 2386, 2386, 2387, 2387, 2387, 2388, 2388, 2389])]
Timing wise this approach is going to be, in my system, about 10x too slow for your needs:
In [8]: %timeit lattice_points_in_quadrilateral(idx_lats, idx_lons)
1000 loops, best of 3: 269 us per loop
So you are looking at over 20 sec. to process your 80,000 polygons.

How to properly set projection and modelview matrices in OpenGL using camera parameters

I have a set of points (3D) taken from a range scanner. Sample data can be found here: http://pastebin.com/RBfQLm56
I also have the following parameters for the scanner:
camera matrix
[3871.88184, 0, 950.736938;
0, 3871.88184, 976.1383059999999;
0, 0, 1]
distortion coeffs
[0.020208003; -1.41251862; -0.00355229038; -0.00438868301; 6.55825615]
camera to reference point (transform)
[0.0225656671, 0.0194614234, 0.9995559233, 1.2656986283;
-0.9994773883, -0.0227084301, 0.0230060289, 0.5798922567;
0.0231460759, -0.99955269, 0.0189388219, -0.2110195758;
0, 0, 0, 1]
I am trying to render these points properly using opengl but the rendering does not look right. What is the correct way to set openGL projection and modelview matrix? This is what I currently do -
znear = 0.00001
zfar = 100
K = array([[3871.88184, 0, 950.736938],[0, 3871.88184, 976.1383059999999],[0, 0, 1]])
Rt =array([[0.0225656671, 0.0194614234, 0.9995559233, 1.2656986283],[-0.9994773883, -0.0227084301, 0.0230060289, 0.5798922567],[0.0231460759, -0.99955269, 0.0189388219, -0.2110195758]])
ren.set_projection(K,zfar,znear)
ren.set_projection_from_camera(Rt)
The function being used are:
def set_projection(self,K,zfar,znear):
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
f_x = K[0,0]
f_y = K[1,1]
c_x = K[0,2]
c_y = K[1,2]
fovY = 1/(float(f_x)/height * 2);
aspectRatio = (float(width)/height) * (float(f_y)/f_x);
near = zfar
far = znear
frustum_height = near * fovY;
frustum_width = frustum_height * aspectRatio;
offset_x = (width/2 - c_x)/width * frustum_width * 2;
offset_y = (height/2 - c_y)/height * frustum_height * 2;
glFrustum(-frustum_width - offset_x, frustum_width - offset_x, -frustum_height - offset_y, frustum_height - offset_y, near, far);
def set_modelview_from_camera(self,Rt):
glMatrixMode(GL_MODELVIEW)
glLoadIdentity()
Rx = array([[1,0,0],[0,0,-1],[0,1,0]])
R = Rt[:,:3]
U,S,V = linalg.svd(R)
R = dot(U,V)
R[0,:]=-R[0,:]
t=Rt[:,3]
M=eye(4)
M[:3,:3]=dot(R,Rx)
M[:3,3]=t
M=M.T
m=M.flatten()
glLoadMatrixf(m)
Then I just render points (pasting snippet):
def renderLIDAR(self,filename):
glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT)
glPushMatrix();
glEnable(GL_DEPTH_TEST)
glClear(GL_DEPTH_BUFFER_BIT)
glPointSize(1.0)
f = open(filename,'r')
f.readline() #Contains number of particles
for line in f:
line = line.split(' ')
glBegin(GL_POINTS)
glColor3f (0.0,1.0,0.0);
x = float(line[0])
y = float(line[1])
z = float(line[2])
glVertex3f(x,y,z)
#print x,y,z
glEnd()
glPopMatrix();

The matrices you get back, most notably the last one in your question are what in OpenGL is the composition of projection and modelview, also called Modelviewprojection, i.e.
MVP = P · M
As long as you're not interested in performing illumination calculations, you can use just that in a vertex shader, i.e.
#version 330
uniform mat4 MVP;
in vec3 position;
void main()
{
gl_Position = MVP * vec4(position, 1);
}
BTW, OpenGL and probably the library you're using as well, are using column major order, i.e. the order of the elements in memory is
0 4 8 c
1 5 9 d
2 6 a e
3 7 b f
so what's written in source code must be thought as "transposed" (of course it is not). Since the matrix you wrote follows the same scheme you can just put it into the uniform as it is. The only question that remains are the boundaries of the NDC space used by the range scanner. But that could be taken care of with an additional matrix applied. OpenGL uses the range [-1, 1]^3 so the worst thing that can happen is, that if it's in the other popular NDC range [0, 1]^3, you'll see your geometry just squeezed into the upper left hand corner of your window, and maybe turned "inside out" if the Z axis goes into the other direction. Just try it, I'd say it already matches OpenGL.
Anyway, if you want to use it with illumination, you have to decompose it into a projection and a modelview part. Easier said than done, but a good starting point is to orthonormalize the upper left 3×3 submatrix, which yields the rotational part of the modelview 'M'. You then have to find a matrix P, that, when left multiplied with M yields the original matrix. That's an overdetermined set of linear equations, so a Gauss-Jordan scheme can do it. And if I'm not entirely mistaken, what you already got in form of that camera matrix is either the decomposed M or P (I'd go for M).
Once you got that you may want to get the translational part (the 4th column) into the modelview matrix as well.