Short version:
Is it possible to create a new scipy.spatial.Delaunay object with a subset of the triangles (2D data) from an existing object?
The goal would be to use the find_simplex method on the new object with filtered out simplices.
Similar but not quite the same
matplotlib contour/contourf of **concave** non-gridded data
How to deal with the (undesired) triangles that form between the edges of my geometry when using Triangulation in matplotlib
Long version:
I am looking at lat-lon data that I regrid with scipy.interpolate.griddata like in the pseudo-code below:
import numpy as np
from scipy.interpolate import griddata
from scipy.spatial import Delaunay
from scipy.interpolate.interpnd import _ndim_coords_from_arrays
#lat shape (a,b): 2D array of latitude values
#lon shape (a,b): 2D array of longitude values
#x shape (a,b): 2D array of variable of interest at lat and lon
# lat-lon data
nonan = ~np.isnan(lat)
flat_lat = lat[nonan]
flat_lon = lon[nonan]
flat_x = x[nonan]
# regular lat-lon grid for regridding
lon_ar = np.arange(loni,lonf,resolution)
lat_ar = np.arange(lati,latf,resolution)
lon_grid, lat_grid = np.meshgrid(lon_ar,lat_ar)
# regrid
x_grid = griddata((flat_lon,flat_lat),flat_x,(lon_grid,lat_grid), method='nearest')
# filter out extrapolated values
cloud_points = _ndim_coords_from_arrays((flat_lon,flat_lat))
regrid_points = _ndim_coords_from_arrays((lon_grid.ravel(),lat_grid.ravel()))
tri = Delaunay(cloud_points)
outside_hull = tri.find_simplex(regrid_points) < 0
x_grid[outside_hull.reshape(x_grid.shape)] = np.nan
# filter out large triangles ??
# it would be easy if I could "subset" tri into a new scipy.spatial.Delaunay object
# new_tri = ??
# outside_hull = new_tri.find_simplex(regrid_points) < 0
The problem is that the convex hull has low quality (very large, shown in blue in example below) triangles that I would like to filter out as they don't represent the data well. I know how to filter them out in input points, but not in the regridded output. Here is the filter function:
def filter_large_triangles(
points: np.ndarray, tri: Optional[Delaunay] = None, coeff: float = 2.0
):
"""
Filter out triangles that have an edge > coeff * median(edge)
Inputs:
tri: scipy.spatial.Delaunay object
coeff: triangles with an edge > coeff * median(edge) will be filtered out
Outputs:
valid_slice: boolean array that selects "normal" triangles
"""
if tri is None:
tri = Delaunay(points)
edge_lengths = np.zeros(tri.vertices.shape)
seen = {}
# loop over triangles
for i, vertex in enumerate(tri.vertices):
# loop over edges
for j in range(3):
id0 = vertex[j]
id1 = vertex[(j + 1) % 3]
# avoid calculating twice for non-border edges
if (id0,id1) in seen:
edge_lengths[i, j] = seen[(id0,id1)]
else:
edge_lengths[i, j] = np.linalg.norm(points[id1] - points[id0])
seen[(id0,id1)] = edge_lengths[i, j]
median_edge = np.median(edge_lengths.flatten())
valid_slice = np.all(edge_lengths < coeff * median_edge, axis=1)
return valid_slice
The bad triangles are shown in blue below:
import matplotlib.pyplot as plt
no_large_triangles = filter_large_triangles(cloud_points,tri)
fig,ax = plt.subplot()
ax.triplot(points[:,0],points[:,1],tri.simplices,c='blue')
ax.triplot(points[:,0],points[:,1],tri.simplices[no_large_triangles],c='green')
plt.show()
Is it possible to create a new scipy.spatial.Delaunay object with only the no_large_triangles simplices? The goal would be to use the find_simplex method on that new object to easily filter out points.
As an alternative how could I find the indices of points in regrid_points that fall inside the blue triangles? (tri.simplices[~no_large_triangles])
So it is possible to modify the Delaunay object for the purpose of using find_simplex on a subset of simplices, but it seems only with the bruteforce algorithm.
# filter out extrapolated values
cloud_points = _ndim_coords_from_arrays((flat_lon,flat_lat))
regrid_points = _ndim_coords_from_arrays((lon_grid.ravel(),lat_grid.ravel()))
tri = Delaunay(cloud_points)
outside_hull = tri.find_simplex(regrid_points) < 0
# filter out large triangles
large_triangles = ~filter_large_triangles(cloud_points,tri)
large_triangle_ids = np.where(large_triangles)[0]
subset_tri = tri # this doesn't preserve tri, effectively just a renaming
# the _find_simplex_bruteforce method only needs the simplices and neighbors
subset_tri.nsimplex = large_triangle_ids.size
subset_tri.simplices = tri.simplices[large_triangles]
subset_tri.neighbors = tri.neighbors[large_triangles]
# update neighbors
for i,triangle in enumerate(subset_tri.neighbors):
for j,neighbor_id in enumerate(triangle):
if neighbor_id in large_triangle_ids:
# reindex the neighbors to match the size of the subset
subset_tri.neighbors[i,j] = np.where(large_triangle_ids==neighbor_id)[0]
elif neighbor_id>=0 and (neighbor_id not in large_triangle_ids):
# that neighbor was a "normal" triangle that should not exist in the subset
subset_tri.neighbors[i,j] = -1
inside_large_triangles = subset_tri.find_simplex(regrid_points,bruteforce=True) >= 0
invalid_slice = np.logical_or(outside_hull,inside_large_triangles)
x_grid[invalid_slice.reshape(x_grid.shape)] = np.nan
Showing that the new Delaunay object has only the subset of large triangles
import matplotlib.pyplot as plt
fig,ax = plt.subplot()
ax.triplot(cloud_points[:,0],cloud_points[:,1],subset_tri.simplices,color='red')
plt.show()
Plotting x_grid with pcolormesh before the filtering for large triangles (zoomed in the blue circle above):
After the filtering:
I have a random distribution of points in 2D space like so:
from sklearn import datasets
import pandas as pd
import numpy as np
arr, labels = datasets.make_moons()
arr, labels = datasets.make_blobs(n_samples=1000, centers=3)
pd.DataFrame(arr, columns=['x', 'y']).plot.scatter('x', 'y', s=1)
I'm trying to assign each of these points to the nearest unoccupied slot on an imaginary hex grid to ensure the points don't overlap. The code I'm using to accomplish this goal produces the plot below. The general idea is to create the hex bins, then iterate over each point and find the minimal radius that allows the algorithm to assign that point to an unoccupied hex bin:
from scipy.spatial.distance import euclidean
def get_bounds(arr):
'''Given a 2D array return the y_min, y_max, x_min, x_max'''
return [
np.min(arr[:,1]),
np.max(arr[:,1]),
np.min(arr[:,0]),
np.max(arr[:,0]),
]
def create_mesh(arr, h=100, w=100):
'''arr is a 2d array; h=number of vertical divisions; w=number of horizontal divisions'''
print(' * creating mesh with size', h, w)
bounds = get_bounds(arr)
# create array of valid positions
y_vals = np.arange(bounds[0], bounds[1], (bounds[1]-bounds[0])/h)
x_vals = np.arange(bounds[2], bounds[3], (bounds[3]-bounds[2])/w)
# create the dense mesh
data = np.tile(
[[0, 1], [1, 0]],
np.array([
int(np.ceil(len(y_vals) / 2)),
int(np.ceil(len(x_vals) / 2)),
]))
# ensure each axis has an even number of slots
if len(y_vals) % 2 != 0 or len(x_vals) % 2 != 0:
data = data[0:len(y_vals), 0:len(x_vals)]
return pd.DataFrame(data, index=y_vals, columns=x_vals)
def align_points_to_grid(arr, h=100, w=100, verbose=False):
'''arr is a 2d array; h=number of vertical divisions; w=number of horizontal divisions'''
h = w = len(arr)/10
grid = create_mesh(arr, h=h, w=w)
# fill the mesh
print(' * filling mesh')
df = pd.DataFrame(arr, columns=['x', 'y'])
bounds = get_bounds(arr)
# store the number of points slotted
c = 0
for site, point in df[['x', 'y']].iterrows():
# skip points not in original points domain
if point.y < bounds[0] or point.y > bounds[1] or \
point.x < bounds[2] or point.x > bounds[3]:
raise Exception('Input point is out of bounds', point.x, point.y, bounds)
# assign this point to the nearest open slot
r_y = (bounds[1]-bounds[0])/h
r_x = (bounds[3]-bounds[2])/w
slotted = False
while not slotted:
bottom = grid.index.searchsorted(point.y - r_y)
top = grid.index.searchsorted(point.y + r_y, side='right')
left = grid.columns.searchsorted(point.x - r_x)
right = grid.columns.searchsorted(point.x + r_x, side='right')
close_grid_points = grid.iloc[bottom:top, left:right]
# store the position in this point's radius that minimizes distortion
best_dist = np.inf
grid_loc = [np.nan, np.nan]
for x, col in close_grid_points.iterrows():
for y, val in col.items():
if val != 1: continue
dist = euclidean(point, (x,y))
if dist < best_dist:
best_dist = dist
grid_loc = [x,y]
# no open slots were found so increase the search radius
if np.isnan(grid_loc[0]):
r_y *= 2
r_x *= 2
# success! report the slotted position to the user
else:
# assign a value other than 1 to mark this slot as filled
grid.loc[grid_loc[0], grid_loc[1]] = 2
df.loc[site, ['x', 'y']] = grid_loc
slotted = True
c += 1
if verbose:
print(' * completed', c, 'of', len(arr), 'assignments')
return df
# plot sample data
df = align_points_to_grid(arr, verbose=False)
df.plot.scatter('x', 'y', s=1)
I'm satisfied with the result of this algorithm, but not with the performance.
Is there a faster solution to this kind of hexbin assignment problem in Python? I feel others with more exposure to the Linear Assignment Problem or the Hungarian Algorithm might have valuable insight into this problem. Any suggestions would be hugely helpful!
It turned out assigning each point to the first available grid spot within its current radius was sufficiently performant.
For others who end up here, my lab wrapped this functionality into a little Python package for convenience. You can pip install pointgrid and then:
from pointgrid import align_points_to_grid
from sklearn import datasets
# create fake data
arr, labels = datasets.make_blobs(n_samples=1000, centers=5)
# get updated point positions
updated = align_points_to_grid(arr)
updated will be a numpy array with the same shape as the input array arr.
Suppose I have a 3D numpy array of nonzero values and "background" = 0. As an example I will take a sphere of random values:
array = np.random.randint(1, 5, size = (100,100,100))
z,y,x = np.ogrid[-50:50, -50:50, -50:50]
mask = x**2 + y**2 + z**2<= 20**2
array[np.invert(mask)] = 0
First, I would like to find the "border voxels" (all nonzero values that have a zero within their 3x3x3 neigbourhood). Second, I would like to replace all border voxels with the mean of their nonzero neighbours. So far I tried to use scipy's generic filter in the following way:
Function to apply at each element:
def borderCheck(values):
#check if the footprint center is on a nonzero value
if values[13] != 0:
#replace border voxels with the mean of nonzero neighbours
if 0 in values:
return np.sum(values)/np.count_nonzero(values)
else:
return values[13]
else:
return 0
Generic filter:
from scipy import ndimage
result = ndimage.generic_filter(array, borderCheck, footprint = np.ones((3,3,3)))
Is this a proper way to handle this problem? I feel that I am trying to reinvent the wheel here and that there must be a shorter, nicer way to achieve the result. Are there any other suitable (numpy, scipy ) functions that I can use?
EDIT
I messed one thing up: I would like to replace all border voxels with the mean of their nonzero AND non-border neighbours. For this, I tried to clean up the neighbours from ali_m's code (2D case):
#for each neighbour voxel, check whether it also appears in the border/edges
non_border_neighbours = []
for each in neighbours:
non_border_neighbours.append([i for i in each if nonzero_idx[i] not in edge_idx])
Now I can't figure out why non_border_neighbours comes back empty?
Furthermore, correct me if I am wrong but doesn't tree.query_ball_point with radius 1 adress only the 6 next neighbours (euclidean distance 1)? Should I set sqrt(3) (3D case) as radius to get the 26-neighbourhood?
I think it's best to start out with the 2D case first, since it can be visualized much more easily:
import numpy as np
from matplotlib import pyplot as plt
A = np.random.randint(1, 5, size=(100, 100)).astype(np.double)
y, x = np.ogrid[-50:50, -50:50]
mask = x**2 + y**2 <= 30**2
A[~mask] = 0
To find the edge pixels you could perform binary erosion on your mask, then XOR the result with your mask
# rank 2 structure with full connectivity
struct = ndimage.generate_binary_structure(2, 2)
erode = ndimage.binary_erosion(mask, struct)
edges = mask ^ erode
One approach to find the nearest non-zero neighbours of each edge pixel would be to use a scipy.spatial.cKDTree:
from scipy.spatial import cKDTree
# the indices of the non-zero locations and their corresponding values
nonzero_idx = np.vstack(np.where(mask)).T
nonzero_vals = A[mask]
# build a k-D tree
tree = cKDTree(nonzero_idx)
# use it to find the indices of all non-zero values that are at most 1 pixel
# away from each edge pixel
edge_idx = np.vstack(np.where(edges)).T
neighbours = tree.query_ball_point(edge_idx, r=1, p=np.inf)
# take the average value for each set of neighbours
new_vals = np.hstack(np.mean(nonzero_vals[n]) for n in neighbours)
# use these to replace the values of the edge pixels
A_new = A.astype(np.double, copy=True)
A_new[edges] = new_vals
Some visualisation:
fig, ax = plt.subplots(1, 3, figsize=(10, 4), sharex=True, sharey=True)
norm = plt.Normalize(0, A.max())
ax[0].imshow(A, norm=norm)
ax[0].set_title('Original', fontsize='x-large')
ax[1].imshow(edges)
ax[1].set_title('Edges', fontsize='x-large')
ax[2].imshow(A_new, norm=norm)
ax[2].set_title('Averaged', fontsize='x-large')
for aa in ax:
aa.set_axis_off()
ax[0].set_xlim(20, 50)
ax[0].set_ylim(50, 80)
fig.tight_layout()
plt.show()
This approach will also generalize to the 3D case:
B = np.random.randint(1, 5, size=(100, 100, 100)).astype(np.double)
z, y, x = np.ogrid[-50:50, -50:50, -50:50]
mask = x**2 + y**2 + z**2 <= 20**2
B[~mask] = 0
struct = ndimage.generate_binary_structure(3, 3)
erode = ndimage.binary_erosion(mask, struct)
edges = mask ^ erode
nonzero_idx = np.vstack(np.where(mask)).T
nonzero_vals = B[mask]
tree = cKDTree(nonzero_idx)
edge_idx = np.vstack(np.where(edges)).T
neighbours = tree.query_ball_point(edge_idx, r=1, p=np.inf)
new_vals = np.hstack(np.mean(nonzero_vals[n]) for n in neighbours)
B_new = B.astype(np.double, copy=True)
B_new[edges] = new_vals
Test against your version:
def borderCheck(values):
#check if the footprint center is on a nonzero value
if values[13] != 0:
#replace border voxels with the mean of nonzero neighbours
if 0 in values:
return np.sum(values)/np.count_nonzero(values)
else:
return values[13]
else:
return 0
result = ndimage.generic_filter(B, borderCheck, footprint=np.ones((3, 3, 3)))
print(np.allclose(B_new, result))
# True
I'm sure this isn't the most efficient way to do it, but it will still be significantly faster than using generic_filter.
Update
The performance could be further improved by reducing the number of points that are considered as candidate neighbours of the edge pixels/voxels:
# ...
# the edge pixels/voxels plus their immediate non-zero neighbours
erode2 = ndimage.binary_erosion(erode, struct)
candidate_neighbours = mask ^ erode2
nonzero_idx = np.vstack(np.where(candidate_neighbours)).T
nonzero_vals = B[candidate_neighbours]
# ...
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.
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.