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Finding k nearest neighbors in 3d numpy array

So I'm trying to find the k nearest neighbors in a pyvista numpy array from an example mesh. With the neighbors received, I want to implement some region growing in my 3d model.
But unfortunaley I receive some weird output, which you can see in the following picture.
It seems like I'm missing something on the KDTree implementation. I was following the answer on a similar question: https://stackoverflow.com/a/2486341/9812286
import numpy as np
from sklearn.neighbors import KDTree
import pyvista as pv
from pyvista import examples
# Example dataset with normals
mesh = examples.load_random_hills()
smooth = mesh
NDIM = 3
X = smooth.points
point = X[5000]
tree = KDTree(X, leaf_size=X.shape[0]+1)
# ind = tree.query_radius([point], r=10) # indices of neighbors within distance 0.3
distances, ind = tree.query([point], k=1000)
p = pv.Plotter()
p.add_mesh(smooth)
ids = np.arange(smooth.n_points)[ind[0]]
top = smooth.extract_cells(ids)
random_color = np.random.random(3)
p.add_mesh(top, color=random_color)
p.show()

You're almost there :) The problem is that you are using the points in the mesh to build the tree, but then extracting cells. Of course these are unrelated in the sense that indices for points will give you nonsense when applied as indices of cells.
Either you have to extract_points:
import numpy as np
from sklearn.neighbors import KDTree
import pyvista as pv
from pyvista import examples
# Example dataset with normals
mesh = examples.load_random_hills()
smooth = mesh
NDIM = 3
X = smooth.points
point = X[5000]
tree = KDTree(X, leaf_size=X.shape[0]+1)
# ind = tree.query_radius([point], r=10) # indices of neighbors within distance 0.3
distances, ind = tree.query([point], k=1000)
p = pv.Plotter()
p.add_mesh(smooth)
ids = np.arange(smooth.n_points)[ind[0]]
top = smooth.extract_points(ids) # changed here!
random_color = np.random.random(3)
p.add_mesh(top, color=random_color)
p.show()
Or you have to work with cell centers to begin with:
import numpy as np
from sklearn.neighbors import KDTree
import pyvista as pv
from pyvista import examples
# Example dataset with normals
mesh = examples.load_random_hills()
smooth = mesh
NDIM = 3
X = smooth.cell_centers().points # changed here!
point = X[5000]
tree = KDTree(X, leaf_size=X.shape[0]+1)
# ind = tree.query_radius([point], r=10) # indices of neighbors within distance 0.3
distances, ind = tree.query([point], k=1000)
p = pv.Plotter()
p.add_mesh(smooth)
ids = np.arange(smooth.n_points)[ind[0]]
top = smooth.extract_cells(ids)
random_color = np.random.random(3)
p.add_mesh(top, color=random_color)
p.show()
As you can see, the two results differ, since index 5000 (which we used for the reference point) means something else when indexing points or when indexing cells.

How to extract first component of FFT from 4D Image

I have 4D( 2D + slices along z axis + time frames) gray-scale image for the heart beating on different moments.
I do like to take Fourier Transform along time axis(for each slice separately), and analyze the fundamental Harmonic (also called H1 component, where H stands for Hilbert Space) so I can determine pixel regions corresponding to ROI which show strongest response to cardiac frequency.
I'm using python for this purpose, and I tried to do that with the following code, but I'm not sure that this is the correct way to do it, because I don't know how to determine the cut-frequency to keep only the fundamental Harmonic.
This link to the image which I'm dealing with
import nibabel as nib
import numpy as np
import matplotlib.pyplot as plt
img = nib.load('patient057_4d.nii.gz')
f = np.fft.fft2(img)
# Move the DC component of the FFT output to the center of the spectrum
fshift = np.fft.fftshift(f)
fshift_orig = fshift.copy()
# logarithmic transformation
magnitude_spectrum = 20*np.log(np.abs(fshift))
# Create mask
rows, cols = img.shape
crow, ccol = int(rows/2), int(cols/2)
# Use mask to remove low frequency components
dist1 = 20
dist2 = 10
fshift[crow-dist1:crow+dist1, ccol-dist1:ccol+dist1] = 0
#fshift[crow-dist2:crow+dist2, ccol-dist2:ccol+dist2] = fshift_orig[crow-dist2:crow+dist2, ccol-dist2:ccol+dist2]
# logarithmic transformation
magnitude_spectrum1 = 20*np.log(np.abs(fshift))
f_ishift = np.fft.ifftshift(fshift)
# inverse Fourier transform
img_back = np.fft.ifft2(f_ishift)
# get rid of imaginary part by abs
img_back = np.abs(img_back)
plt.figure(num = 'Im_Back')
plt.imshow(abs(fshift[:,:,2,2]).astype('uint8'),cmap='gray')
plt.show()

The solution was to take Fourier transform 3D for each slice seperately, then to chose only the 2nd component of the Transform to transform it back to the spatial space, and that's it.
The benefit of this is to detect if something is moving along the third axis(time in my case).
for sl in range(img.shape[2]):
#-----Fourier--H1-----------------------------------------
# ff1[:, :, 1] H1 compnent 1, if 0 then DC
ff1 = FFT.fftn(img[:,:,sl,:])
fh = np.absolute(FFT.ifftn(ff1[:, :, 1]))
#-----Fourier--H1-----------------------------------------

Counterclockwise sorting of x, y data

I have a set of points in a text file: random_shape.dat.
The initial order of points in the file is random. I would like to sort these points in a counter-clockwise order as follows (the red dots are the xy data):
I tried to achieve that by using the polar coordinates: I calculate the polar angle of each point (x,y) then sort by the ascending angles, as follows:
"""
Script: format_file.py
Description: This script will format the xy data file accordingly to be used with a program expecting CCW order of data points, By soting the points in Counterclockwise order
Example: python format_file.py random_shape.dat
"""
import sys
import numpy as np
# Read the file name
filename = sys.argv[1]
# Get the header name from the first line of the file (without the newline character)
with open(filename, 'r') as f:
header = f.readline().rstrip('\n')
angles = []
# Read the data from the file
x, y = np.loadtxt(filename, skiprows=1, unpack=True)
for xi, yi in zip(x, y):
angle = np.arctan2(yi, xi)
if angle < 0:
angle += 2*np.pi # map the angle to 0,2pi interval
angles.append(angle)
# create a numpy array
angles = np.array(angles)
# Get the arguments of sorted 'angles' array
angles_argsort = np.argsort(angles)
# Sort x and y
new_x = x[angles_argsort]
new_y = y[angles_argsort]
print("Length of new x:", len(new_x))
print("Length of new y:", len(new_y))
with open(filename.split('.')[0] + '_formatted.dat', 'w') as f:
print(header, file=f)
for xi, yi in zip(new_x, new_y):
print(xi, yi, file=f)
print("Done!")
By running the script:
python format_file.py random_shape.dat
Unfortunately I don't get the expected results in random_shape_formated.dat! The points are not sorted in the desired order.
Any help is appreciated.
EDIT: The expected resutls:
Create a new file named: filename_formatted.dat that contains the sorted data according to the image above (The first line contains the starting point, the next lines contain the points as shown by the blue arrows in counterclockwise direction in the image).
EDIT 2: The xy data added here instead of using github gist:
random_shape
0.4919261070361315 0.0861956168831175
0.4860816807027076 -0.06601587301587264
0.5023029456281289 -0.18238249845392662
0.5194784026079869 0.24347943722943777
0.5395164357511545 -0.3140611471861465
0.5570497147514262 0.36010146103896146
0.6074231036252226 -0.4142604617604615
0.6397066014669927 0.48590810704447085
0.7048302091822873 -0.5173701298701294
0.7499157837544145 0.5698170011806378
0.8000108666123336 -0.6199254449254443
0.8601249660418364 0.6500974025974031
0.9002010323281716 -0.7196585989767801
0.9703341483292582 0.7299242424242429
1.0104102146155935 -0.7931355765446666
1.0805433306166803 0.8102046438410078
1.1206193969030154 -0.865251869342778
1.1907525129041021 0.8909386068476981
1.2308285791904374 -0.9360074773711129
1.300961695191524 0.971219008264463
1.3410377614778592 -1.0076702085792988
1.4111708774789458 1.051499409681228
1.451246943765281 -1.0788793781975592
1.5213800597663678 1.1317798110979933
1.561456126052703 -1.1509956709956706
1.6315892420537896 1.2120602125147582
1.671665308340125 -1.221751279024005
1.7417984243412115 1.2923406139315234
1.7818744906275468 -1.2943211334120424
1.8520076066286335 1.3726210153482883
1.8920836729149686 -1.3596340023612745
1.9622167889160553 1.4533549783549786
2.0022928552023904 -1.4086186540731989
2.072425971203477 1.5331818181818184
2.1125020374898122 -1.451707005116095
2.182635153490899 1.6134622195985833
2.2227112197772345 -1.4884454939000387
2.292844335778321 1.6937426210153486
2.3329204020646563 -1.5192876820149541
2.403053518065743 1.774476584022039
2.443129584352078 -1.5433264462809912
2.513262700353165 1.8547569854388037
2.5533387666395 -1.561015348288075
2.6234718826405867 1.9345838252656438
2.663547948926922 -1.5719008264462806
2.7336810649280086 1.9858362849271942
2.7737571312143436 -1.5750757575757568
2.8438902472154304 2.009421487603306
2.883966313501766 -1.5687258953168035
2.954099429502852 2.023481896890988
2.9941754957891877 -1.5564797323888229
3.0643086117902745 2.0243890200708385
3.1043846780766096 -1.536523022432113
3.1745177940776963 2.0085143644234558
3.2145938603640314 -1.5088557654466737
3.284726976365118 1.9749508067689887
3.324803042651453 -1.472570838252656
3.39493615865254 1.919162731208186
3.435012224938875 -1.4285753640299088
3.5051453409399618 1.8343467138921687
3.545221407226297 -1.3786835891381335
3.6053355066557997 1.7260966810966811
3.655430589513719 -1.3197205824478546
3.6854876392284703 1.6130086580086582
3.765639771801141 -1.2544077134986225
3.750611246943765 1.5024152236652237
3.805715838087476 1.3785173160173163
3.850244800627849 1.2787337662337666
3.875848954088563 -1.1827449822904361
3.919007794704616 1.1336638361638363
3.9860581363759846 -1.1074537583628485
3.9860581363759846 1.0004485329485333
4.058012891753723 0.876878197560016
4.096267318663407 -1.0303482880755608
4.15638141809291 0.7443374218374221
4.206476500950829 -0.9514285714285711
4.256571583808748 0.6491902794175526
4.3166856832382505 -0.8738695395513574
4.36678076609617 0.593855765446675
4.426894865525672 -0.7981247540338443
4.476989948383592 0.5802489177489183
4.537104047813094 -0.72918339236521
4.587199130671014 0.5902272727272733
4.647313230100516 -0.667045454545454
4.697408312958435 0.6246979535615904
4.757522412387939 -0.6148858717040526
4.807617495245857 0.6754968516332154
4.8677315946753605 -0.5754260133805582
4.917826677533279 0.7163173947264858
4.977940776962782 -0.5500265643447455
5.028035859820701 0.7448917748917752
5.088149959250204 -0.5373268398268394
5.138245042108123 0.7702912239275879
5.198359141537626 -0.5445838252656432
5.2484542243955445 0.7897943722943728
5.308568323825048 -0.5618191656828015
5.358663406682967 0.8052154663518301
5.41877750611247 -0.5844972451790631
5.468872588970389 0.8156473829201105
5.5289866883998915 -0.6067217630853987
5.579081771257811 0.8197294372294377
5.639195870687313 -0.6248642266824076
5.689290953545233 0.8197294372294377
5.749405052974735 -0.6398317591499403
5.799500135832655 0.8142866981503349
5.859614235262157 -0.6493565525383702
5.909709318120076 0.8006798504525783
5.969823417549579 -0.6570670995670991
6.019918500407498 0.7811767020857934
6.080032599837001 -0.6570670995670991
6.13012768269492 0.7562308146399057
6.190241782124423 -0.653438606847697
6.240336864982342 0.7217601338055886
6.300450964411845 -0.6420995670995664
6.350546047269764 0.6777646595828419
6.410660146699267 -0.6225964187327819
6.4607552295571855 0.6242443919716649
6.520869328986689 -0.5922077922077915
6.570964411844607 0.5548494687131056
6.631078511274111 -0.5495730027548205
6.681173594132029 0.4686727666273125
6.7412876935615325 -0.4860743801652889
6.781363759847868 0.3679316979316982
6.84147785927737 -0.39541245791245716
6.861515892420538 0.25880333951762546
6.926639500135833 -0.28237987012986965
6.917336127605076 0.14262677798392165
6.946677533279001 0.05098957832291173
6.967431210462995 -0.13605442176870675
6.965045730326905 -0.03674603174603108

I find that an easy way to sort points with x,y-coordinates like that is to sort them dependent on the angle between the line from the points and the center of mass of the whole polygon and the horizontal line which is called alpha in the example. The coordinates of the center of mass (x0 and y0) can easily be calculated by averaging the x,y coordinates of all points. Then you calculate the angle using numpy.arccos for instance. When y-y0 is larger than 0 you take the angle directly, otherwise you subtract the angle from 360° (2𝜋). I have used numpy.where for the calculation of the angle and then numpy.argsort to produce a mask for indexing the initial x,y-values. The following function sort_xy sorts all x and y coordinates with respect to this angle. If you want to start from any other point you could add an offset angle for that. In your case that would be zero though.
def sort_xy(x, y):
x0 = np.mean(x)
y0 = np.mean(y)
r = np.sqrt((x-x0)**2 + (y-y0)**2)
angles = np.where((y-y0) > 0, np.arccos((x-x0)/r), 2*np.pi-np.arccos((x-x0)/r))
mask = np.argsort(angles)
x_sorted = x[mask]
y_sorted = y[mask]
return x_sorted, y_sorted
Plotting x, y before sorting using matplotlib.pyplot.plot (points are obvisously not sorted):
Plotting x, y using matplotlib.pyplot.plot after sorting with this method:

If it is certain that the curve does not cross the same X coordinate (i.e. any vertical line) more than twice, then you could visit the points in X-sorted order and append a point to one of two tracks you follow: to the one whose last end point is the closest to the new one. One of these tracks will represent the "upper" part of the curve, and the other, the "lower" one.
The logic would be as follows:
dist2 = lambda a,b: (a[0]-b[0])*(a[0]-b[0]) + (a[1]-b[1])*(a[1]-b[1])
z = list(zip(x, y)) # get the list of coordinate pairs
z.sort() # sort by x coordinate
cw = z[0:1] # first point in clockwise direction
ccw = z[1:2] # first point in counter clockwise direction
# reverse the above assignment depending on how first 2 points relate
if z[1][1] > z[0][1]:
cw = z[1:2]
ccw = z[0:1]
for p in z[2:]:
# append to the list to which the next point is closest
if dist2(cw[-1], p) < dist2(ccw[-1], p):
cw.append(p)
else:
ccw.append(p)
cw.reverse()
result = cw + ccw
This would also work for a curve with steep fluctuations in the Y-coordinate, for which an angle-look-around from some central point would fail, like here:
No assumption is made about the range of the X nor of the Y coordinate: like for instance, the curve does not necessarily have to cross the X axis (Y = 0) for this to work.

Counter-clock-wise order depends on the choice of a pivot point. From your question, one good choice of the pivot point is the center of mass.
Something like this:
# Find the Center of Mass: data is a numpy array of shape (Npoints, 2)
mean = np.mean(data, axis=0)
# Compute angles
angles = np.arctan2((data-mean)[:, 1], (data-mean)[:, 0])
# Transform angles from [-pi,pi] -> [0, 2*pi]
angles[angles < 0] = angles[angles < 0] + 2 * np.pi
# Sort
sorting_indices = np.argsort(angles)
sorted_data = data[sorting_indices]

Not really a python question I think, but still I think you could try sorting by - sign(y) * x doing something like:
def counter_clockwise_sort(points):
return sorted(points, key=lambda point: point['x'] * (-1 if point['y'] >= 0 else 1))
should work fine, assuming you read your points properly into a list of dicts of format {'x': 0.12312, 'y': 0.912}
EDIT: This will work as long as you cross the X axis only twice, like in your example.

If:
the shape is arbitrarily complex and
the point spacing is ~random
then I think this is a really hard problem.
For what it's worth, I have faced a similar problem in the past, and I used a traveling salesman solver. In particular, I used the LKH solver. I see there is a Python repo for solving the problem, LKH-TSP. Once you have an order to the points, I don't think it will be too hard to decide on a clockwise vs clockwise ordering.

If we want to answer your specific problem, we need to pick a pivot point.
Since you want to sort according to the starting point you picked, I would take a pivot in the middle (x=4,y=0 will do).
Since we're sorting counterclockwise, we'll take arctan2(-(y-pivot_y),-(x-center_x)) (we're flipping the x axis).
We get the following, with a gradient colored scatter to prove correctness (fyi I removed the first line of the dat file after downloading):
import numpy as np
import matplotlib.pyplot as plt
points = np.loadtxt('points.dat')
#oneliner for ordering points (transform, adjust for 0 to 2pi, argsort, index at points)
ordered_points = points[np.argsort(np.apply_along_axis(lambda x: np.arctan2(-x[1],-x[0]+4) + np.pi*2, axis=1,arr=points)),:]
#color coding 0-1 as str for gray colormap in matplotlib
plt.scatter(ordered_points[:,0], ordered_points[:,1],c=[str(x) for x in np.arange(len(ordered_points)) / len(ordered_points)],cmap='gray')
Result (in the colormap 1 is white and 0 is black), they're numbered in the 0-1 range by order:

For points with comparable distances between their neighbouring pts, we can use KDTree to get two closest pts for each pt. Then draw lines connecting those to give us a closed shape contour. Then, we will make use of OpenCV's findContours to get contour traced always in counter-clockwise manner. Now, since OpenCV works on images, we need to sample data from the provided float format to uint8 image format. Given, comparable distances between two pts, that should be pretty safe. Also, OpenCV handles it well to make sure it traces even sharp corners in curvatures, i.e. smooth or not-smooth data would work just fine. And, there's no pivot requirement, etc. As such all kinds of shapes would be good to work with.
Here'e the implementation -
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial.distance import pdist
from scipy.spatial import cKDTree
import cv2
from scipy.ndimage.morphology import binary_fill_holes
def counter_clockwise_order(a, DEBUG_PLOT=False):
b = a-a.min(0)
d = pdist(b).min()
c = np.round(2*b/d).astype(int)
img = np.zeros(c.max(0)[::-1]+1, dtype=np.uint8)
d1,d2 = cKDTree(c).query(c,k=3)
b = c[d2]
p1,p2,p3 = b[:,0],b[:,1],b[:,2]
for i in range(len(b)):
cv2.line(img,tuple(p1[i]),tuple(p2[i]),255,1)
cv2.line(img,tuple(p1[i]),tuple(p3[i]),255,1)
img = (binary_fill_holes(img==255)*255).astype(np.uint8)
if int(cv2.__version__.split('.')[0])>=3:
_,contours,hierarchy = cv2.findContours(img.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
else:
contours,hierarchy = cv2.findContours(img.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
cont = contours[0][:,0]
f1,f2 = cKDTree(cont).query(c,k=1)
ordered_points = a[f2.argsort()[::-1]]
if DEBUG_PLOT==1:
NPOINTS = len(ordered_points)
for i in range(NPOINTS):
plt.plot(ordered_points[i:i+2,0],ordered_points[i:i+2,1],alpha=float(i)/(NPOINTS-1),color='k')
plt.show()
return ordered_points
Sample run -
# Load data in a 2D array with 2 columns
a = np.loadtxt('random_shape.csv',delimiter=' ')
ordered_a = counter_clockwise_order(a, DEBUG_PLOT=1)
Output -

How to reshape a networkx graph in Python?

So I created a really naive (probably inefficient) way of generating hasse diagrams.
Question:
I have 4 dimensions... p q r s .
I want to display it uniformly (tesseract) but I have no idea how to reshape it. How can one reshape a networkx graph in Python?
I've seen some examples of people using spring_layout() and draw_circular() but it doesn't shape in the way I'm looking for because they aren't uniform.
Is there a way to reshape my graph and make it uniform? (i.e. reshape my hasse diagram into a tesseract shape (preferably using nx.draw() )
Here's what mine currently look like:
Here's my code to generate the hasse diagram of N dimensions
#!/usr/bin/python
import networkx as nx
import matplotlib.pyplot as plt
import itertools
H = nx.DiGraph()
axis_labels = ['p','q','r','s']
D_len_node = {}
#Iterate through axis labels
for i in xrange(0,len(axis_labels)+1):
#Create edge from empty set
if i == 0:
for ax in axis_labels:
H.add_edge('O',ax)
else:
#Create all non-overlapping combinations
combinations = [c for c in itertools.combinations(axis_labels,i)]
D_len_node[i] = combinations
#Create edge from len(i-1) to len(i) #eg. pq >>> pqr, pq >>> pqs
if i > 1:
for node in D_len_node[i]:
for p_node in D_len_node[i-1]:
#if set.intersection(set(p_node),set(node)): Oops
if all(p in node for p in p_node) == True: #should be this!
H.add_edge(''.join(p_node),''.join(node))
#Show Plot
nx.draw(H,with_labels = True,node_shape = 'o')
plt.show()
I want to reshape it like this:
If anyone knows of an easier way to make Hasse Diagrams, please share some wisdom but that's not the main aim of this post.

This is a pragmatic, rather than purely mathematical answer.
I think you have two issues - one with layout, the other with your network.
1. Network
You have too many edges in your network for it to represent the unit tesseract. Caveat I'm not an expert on the maths here - just came to this from the plotting angle (matplotlib tag). Please explain if I'm wrong.
Your desired projection and, for instance, the wolfram mathworld page for a Hasse diagram for n=4 has only 4 edges connected all nodes, whereas you have 6 edges to the 2 and 7 edges to the 3 bit nodes. Your graph fully connects each "level", i.e. 4-D vectors with 0 1 values connect to all vectors with 1 1 value, which then connect to all vectors with 2 1 values and so on. This is most obvious in the projection based on the Wikipedia answer (2nd image below)
2. Projection
I couldn't find a pre-written algorithm or library to automatically project the 4D tesseract onto a 2D plane, but I did find a couple of examples, e.g. Wikipedia. From this, you can work out a co-ordinate set that would suit you and pass that into the nx.draw() call.
Here is an example - I've included two co-ordinate sets, one that looks like the projection you show above, one that matches this one from wikipedia.
import networkx as nx
import matplotlib.pyplot as plt
import itertools
H = nx.DiGraph()
axis_labels = ['p','q','r','s']
D_len_node = {}
#Iterate through axis labels
for i in xrange(0,len(axis_labels)+1):
#Create edge from empty set
if i == 0:
for ax in axis_labels:
H.add_edge('O',ax)
else:
#Create all non-overlapping combinations
combinations = [c for c in itertools.combinations(axis_labels,i)]
D_len_node[i] = combinations
#Create edge from len(i-1) to len(i) #eg. pq >>> pqr, pq >>> pqs
if i > 1:
for node in D_len_node[i]:
for p_node in D_len_node[i-1]:
if set.intersection(set(p_node),set(node)):
H.add_edge(''.join(p_node),''.join(node))
#This is manual two options to project tesseract onto 2D plane
# - many projections are available!!
wikipedia_projection_coords = [(0.5,0),(0.85,0.25),(0.625,0.25),(0.375,0.25),
(0.15,0.25),(1,0.5),(0.8,0.5),(0.6,0.5),
(0.4,0.5),(0.2,0.5),(0,0.5),(0.85,0.75),
(0.625,0.75),(0.375,0.75),(0.15,0.75),(0.5,1)]
#Build the "two cubes" type example projection co-ordinates
half_coords = [(0,0.15),(0,0.6),(0.3,0.15),(0.15,0),
(0.55,0.6),(0.3,0.6),(0.15,0.4),(0.55,1)]
#make the coords symmetric
example_projection_coords = half_coords + [(1-x,1-y) for (x,y) in half_coords][::-1]
print example_projection_coords
def powerset(s):
ch = itertools.chain.from_iterable(itertools.combinations(s, r) for r in range(len(s)+1))
return [''.join(t) for t in ch]
pos={}
for i,label in enumerate(powerset(axis_labels)):
if label == '':
label = 'O'
pos[label]= example_projection_coords[i]
#Show Plot
nx.draw(H,pos,with_labels = True,node_shape = 'o')
plt.show()
Note - unless you change what I've mentioned in 1. above, they still have your edge structure, so won't look exactly the same as the examples from the web. Here is what it looks like with your existing network generation code - you can see the extra edges if you compare it to your example (e.g. I don't this pr should be connected to pqs:
'Two cube' projection
Wikimedia example projection
Note
If you want to get into the maths of doing your own projections (and building up pos mathematically), you might look at this research paper.
EDIT:
Curiosity got the better of me and I had to search for a mathematical way to do this. I found this blog - the main result of which being the projection matrix:
This led me to develop this function for projecting each label, taking the label containing 'p' to mean the point has value 1 on the 'p' axis, i.e. we are dealing with the unit tesseract. Thus:
def construct_projection(label):
r1 = r2 = 0.5
theta = math.pi / 6
phi = math.pi / 3
x = int( 'p' in label) + r1 * math.cos(theta) * int('r' in label) - r2 * math.cos(phi) * int('s' in label)
y = int( 'q' in label) + r1 * math.sin(theta) * int('r' in label) + r2 * math.sin(phi) * int('s' in label)
return (x,y)
Gives a nice projection into a regular 2D octagon with all points distinct.
This will run in the above program, just replace
pos[label] = example_projection_coords[i]
with
pos[label] = construct_projection(label)
This gives the result:
play with r1,r2,theta and phi to your heart's content :)

Isolate greatest/smallest labeled patches from numpy array

i have a large numpy array and labeled it with the connected component labeling in scipy. Now i want to create subsets of this array, where only the biggest or smallest labels in size are left.
Both extrema can of course occur several times.
import numpy
from scipy import ndimage
....
# Loaded in my image file here. To big to paste
....
s = ndimage.generate_binary_structure(2,2) # iterate structure
labeled_array, numpatches = ndimage.label(array,s) # labeling
# get the area (nr. of pixels) of each labeled patch
sizes = ndimage.sum(array,labeled_array,range(1,numpatches+1))
# To get the indices of all the min/max patches. Is this the correct label id?
map = numpy.where(sizes==sizes.max())
mip = numpy.where(sizes==sizes.min())
# This here doesn't work! Now i want to create a copy of the array and fill only those cells
# inside the largest, respecitively the smallest labeled patches with values
feature = numpy.zeros_like(array, dtype=int)
feature[labeled_array == map] = 1
Someone can give me hint how to move on?

Here is the full code:
import numpy
from scipy import ndimage
array = numpy.zeros((100, 100), dtype=np.uint8)
x = np.random.randint(0, 100, 2000)
y = np.random.randint(0, 100, 2000)
array[x, y] = 1
pl.imshow(array, cmap="gray", interpolation="nearest")
s = ndimage.generate_binary_structure(2,2) # iterate structure
labeled_array, numpatches = ndimage.label(array,s) # labeling
sizes = ndimage.sum(array,labeled_array,range(1,numpatches+1))
# To get the indices of all the min/max patches. Is this the correct label id?
map = numpy.where(sizes==sizes.max())[0] + 1
mip = numpy.where(sizes==sizes.min())[0] + 1
# inside the largest, respecitively the smallest labeled patches with values
max_index = np.zeros(numpatches + 1, np.uint8)
max_index[map] = 1
max_feature = max_index[labeled_array]
min_index = np.zeros(numpatches + 1, np.uint8)
min_index[mip] = 1
min_feature = min_index[labeled_array]
Notes:
numpy.where returns a tuple
the size of label 1 is sizes[0], so you need to add 1 to the result of numpy.where
To get a mask array with multiple labels, you can use labeled_array as the index of a label mask array.
The results:

first you need a labeled mask, given a mask with only 0(background) and 1(foreground):
labeled_mask, cc_num = ndimage.label(mask)
then find the largest connected component:
largest_cc_mask = (labeled_mask == (np.bincount(labeled_mask.flat)[1:].argmax() + 1))
you can deduce the smallest object finding by using argmin()..