How to create a sphere inside an ndarray? [duplicate] - python

This question already has answers here:
How to generate a sphere in 3D Numpy array
(5 answers)
Closed 3 months ago.
I have a ndarray of size 32x32x32. I want to create a sphere inside the array with the center at (x,y) and a radius of 4 pixels. The value of the sphere is 1 while value of the array is 0. How can this be done in python?
This is the code to generate the array:
import numpy as np
A = np.zeros((32,32,32))
print (A)

Very good question. You can try the following code. In the below mentioned code AA is the matrix that you want. =)
import numpy as np
from copy import deepcopy
''' size : size of original 3D numpy matrix A.
radius : radius of circle inside A which will be filled with ones.
'''
size, radius = 5, 2
''' A : numpy.ndarray of shape size*size*size. '''
A = np.zeros((size,size, size))
''' AA : copy of A (you don't want the original copy of A to be overwritten.) '''
AA = deepcopy(A)
''' (x0, y0, z0) : coordinates of center of circle inside A. '''
x0, y0, z0 = int(np.floor(A.shape[0]/2)), \
int(np.floor(A.shape[1]/2)), int(np.floor(A.shape[2]/2))
for x in range(x0-radius, x0+radius+1):
for y in range(y0-radius, y0+radius+1):
for z in range(z0-radius, z0+radius+1):
''' deb: measures how far a coordinate in A is far from the center.
deb>=0: inside the sphere.
deb<0: outside the sphere.'''
deb = radius - abs(x0-x) - abs(y0-y) - abs(z0-z)
if (deb)>=0: AA[x,y,z] = 1
Following is an example of the output for size=5 and radius=2 (a sphere of radius 2 pixels inside a numpy array of shape 5*5*5):
[[[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 1. 1. 1. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 1. 0. 0.]
[0. 1. 1. 1. 0.]
[1. 1. 1. 1. 1.]
[0. 1. 1. 1. 0.]
[0. 0. 1. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 1. 1. 1. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]]
I haven't printed the output for the size and radius that you had asked for (size=32 and radius=4), as the output will be very long.

Since array indexes only have a certain level of specificity (i.e. you can only subdivide down to the width, in this case 32), there's no one perfect way to represent a sphere in an array. Instead, we can treat each array index as a space of cubic area, where the [x][y][z] indices of the index represent the the cubic area's center coordinates. To create the sphere, we evaluate whether the sphere's presence in that area of space meets certain criteria.
We start with the equation for a sphere. From Wikipedia:
In analytic geometry, a sphere with center (x0, y0, z0) and radius r is the locus of all points (x, y, z) such that
(x - x0)^2 + (y - y0)^2 + (z - z0)^2 <= r^2.
For an array of dimensions N, the center will have the coordinate (N - 1) / 2 for all dimensions. (because for an even-numbered dimension, the center should be between the two center points, and for an odd-numbered dimension, the center should be an integer.) The magnitude of the radius can vary depending on where you decide the boundaries of the sphere relative to our imagined cubic array representation; re-reading the question, I notice you already gave the desired radius: 4.
There are two evaluation criteria I can think of:
Simple approach
In this approach, we will simply use a test of whether the array index's cubic area's center lies within the circle equation.
You can see Siddharth Satpathy's answer for some code using this approach.
Sophisticated approach
Ideally for me, the equation would decide whether an index lies within the sphere by assessing whether the proportion of sphere for that cubic area is greater than 50%. However, this approach unfortunately goes beyond my current working mathematical knowledge.
In regards to a discussion I had in the comments, neither approach is better than the other since they represent different perspectives: I personally imagine the array as being actually representative of the cubic area for each index, while others may imagine the indexes being the center points of these cubic areas.

Nothing above worked for me so there is my attempt:
def create_bin_sphere(arr_size, center, r):
coords = np.ogrid[:arr_size[0], :arr_size[1], :arr_size[2]]
distance = np.sqrt((coords[0] - center[0])**2 + (coords[1]-center[1])**2 + (coords[2]-center[2])**2)
return 1*(distance <= r)
where:
arr_size is a tuple with numpy array shape
center is a tuple with sphere center coords
r is a radius of the sphere
Example:
arr_size = (30,30,30)
sphere_center = (15,15,15)
r=10
sphere = create_bin_sphere(arr_size,sphere_center, r)
#Plot the result
fig =plt.figure(figsize=(6,6))
ax = fig.gca(projection='3d')
ax.voxels(sphere, facecolors=colors, edgecolor='k')
plt.show()
Vizualization rezult

Eventhough it's a little late - I recently faced the same problem and solved it somewhat like the solution supposed by Mstaino. Also, Mstainos solution doesn't work for asymetricly sized array's, because shapes will not match when calculating the distance.
So here's my approach in 3D which produces a sphere in the center of the array. :
# define array size and sphere radius
size = [size_x, size_y, size_z]
radius = sphere_radius
# compute center index of array
center = [int(size[0]/2), int(size[1]/2), int(size[2]/2)]
# create index grid for array
ind_0, ind_1, ind_2 = np.indices((size[0], size[1], size[2]))
# calculate "distance" of indices to center index
distance = ((ind_0 - center[0])**2 + (ind_1 - center[1])**2 + (ind_2 - center[2])**2)**.5
# create output
output = np.ones(shape = (size[0], size[1], size[2])) * (distance <= radius)

Using a combination of indexing, distance calculation and masking (all with numpy):
import numpy as np
center = (31/2, 31/2, 31/2) # if it is centered
size = (32, 32, 32)
max_dist = 4
distance = np.linalg.norm(np.subtract(np.indices(size).T, np.asarray(center)), axis=2)
#print(distance)
mask = np.ones(size) * (distance < max_dist)
print(mask)
np.indices creates an index in the form [[[(i, j, k)]]], np.substract calculates the vector difference to your center, and np.linalg.norm calculates the vector norm. The rest is just using a mask operation on the distance array.
Does that work?
EDIT: an example with (3,3,3) for clarity purposes
center = (1, 1, 1)
size = (3, 3, 3)
distance = np.linalg.norm(np.subtract(np.indices(size).T,np.asarray(center)), axis=len(center))
mask = np.ones(size) * (distance<=1)
print(mask)
>>[[[0. 0. 0.]
[0. 1. 0.]
[0. 0. 0.]]
[[0. 1. 0.]
[1. 1. 1.]
[0. 1. 0.]]
[[0. 0. 0.]
[0. 1. 0.]
[0. 0. 0.]]]

Related

Identity Matrix - only integer scalar arrays can be converted to a scalar index

I'm doing Identity Matrix, but it comes TypeError: only integer scalar arrays can be converted to a scalar index, and IDK how to fix it, plz help me!
Z = np.array([
[0,2,0,4,4],
[0,0,3,0,0],
[0,0,0,1,0],
[0,2,0,0,0],
[0,0,0,0,0]
])
I = np.eye(Z)
I = np.identity(Z)
Both np.eye and np.identify come to the same error.
The fucntion np.identity() takes an integer argument, not np.array() object as argument. So if you want to create an identity matrix of size nxn you need to calculate the length of Z:
import numpy as np
Z = np.array([
[0,2,0,4,4],
[0,0,3,0,0],
[0,0,0,1,0],
[0,2,0,0,0],
[0,0,0,0,0]
])
I = np.identity(len(Z))
print(I)
Output:
[[1. 0. 0. 0. 0.]
[0. 1. 0. 0. 0.]
[0. 0. 1. 0. 0.]
[0. 0. 0. 1. 0.]
[0. 0. 0. 0. 1.]]

How to efficiently concatenate Numpy Array based on position conditioning?

The objective is to concatenate a Numpy Array according to a set of position. However, I am curious whether the concatenate and step as shown in the code below can be optimized further without the need of for loop and if-else statement?
tot_length=0.2 implementation
steps=0.1
start_val=0
repeat_perm=3
list_no =np.arange(start_val, tot_length, steps)
x, y, z = np.meshgrid(*[list_no for _ in range(3)], sparse=True)
ix = np.array(((x>=y) & (y>=z)).nonzero()).T
final_opt=list_no[ix]
final_opt[:,[0, 1]] = final_opt[:,[1, 0]]
all_result=itertools.product(range(0,ix.shape[1]), repeat=repeat_perm)
for num, num_pair in enumerate(all_result, start=1):
for num_x, num_pair_x in enumerate ( num_pair, start=0 ):
if (num == 1) &(num_x==0) :
cont_arry = final_opt [num_pair_x, :]
else:
cont_arry= np.concatenate((cont_arry, final_opt [num_pair_x, :]), axis=0)
final_arr =np.reshape(cont_arry, (-1, 9))
print(final_arr)
Output of size (27, 9), but only partial are shown below
[[0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.1 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.1 0.1 0. ]
[0. 0. 0. 0.1 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.1 0. 0. 0.1 0. 0. ]
[0. 0. 0. 0.1 0. 0. 0.1 0.1 0. ]
[0.1 0.1 0. 0.1 0.1 0. 0.1 0.1 0. ]]
Just some heads up,the cont_arry will be vectorised multiply with a 1D array of similar length with the cont_arry. Knowing this, is there a way to avoid from storing the result of concatenation on memory or what not to minimise potential memory issue since in actual application, the worst possible parameter setting is as below
tot_length=200
steps=0.1
start_val=0
repeat_perm=1200
I think your concatenate loop can be replaced with:
alist = []
for num, num_pair in enumerate(all_result, start=1):
for num_x, num_pair_x in enumerate ( num_pair, start=0 ):
alist.append( final_opt [num_pair_x, :]))
arr = np.array(alist)
# arr = np.concatenate(alist, axis=0)
# arr = np.vstack(alist)
There may be some details in this that I didn't catch. I haven't tried to test it. List append is much faster than concatenate, especially when done repeatedly.
concatenate is most efficient when give a whole list of arrays to join.
Better yet, don't iterate at all; instead make use of whole-array math and indexing. But I haven't tried to master your code, so won't suggest how to do that.

Vectorize extracting sub-multidimensional array from multidimensional array with list of indices

I have this multidimensional array of shape (500000,3,2,3),let's call it data. The data is basically 500000 sets of 3 points,each of the 3 points seperated into its x and y coordinates (hence the 2). The last 3 in the shape represents different rotations of the 3 points. Now, I've got this 1d array of 500000 numbers between 0 and 2 that tell me which of the rotations I want to keep, let's call it rot_index. I would like to construct a multidimensional array of shape (500000,3,2) that only keeps the correctly rotated data points. Any ideas on how to extract the data with the correct index from the original data array? I tried something like this, but it didn't work
data[:,:,:,rot_index]
Edit:
here is some example data (giving 10 sets of points instead of 500000)
data =
[[[[0.70846822 0.98552876 0.66736535]
[0. 0. 0. ]]
[[0.66736535 0.70846822 0.98552876]
[1.54545219 2.39798549 2.33974762]]
[[0.98552876 0.66736535 0.70846822]
[3.88519982 3.94343768 4.73773311]]]
[[[0.8132551 1.18845796 1.53004225]
[0. 0. 0. ]]
[[1.18845796 1.53004225 0.8132551 ]
[1.43211754 2.58720625 2.26386152]]
[[1.53004225 0.8132551 1.18845796]
[4.01932379 4.85106777 3.69597906]]]
[[[0.66123513 0.93651048 0.83170562]
[0. 0. 0. ]]
[[0.93651048 0.83170562 0.66123513]
[2.09747072 2.38383457 1.80188002]]
[[0.83170562 0.66123513 0.93651048]
[4.48130529 4.18571459 3.89935074]]]
[[[1.31047414 0.67740955 1.42020073]
[0. 0. 0. ]]
[[0.67740955 1.42020073 1.31047414]
[1.66061575 1.97600777 2.64656179]]
[[1.42020073 1.31047414 0.67740955]
[3.63662352 4.62256956 4.30717753]]]
[[[1.4085555 1.64177102 0.27708893]
[0. 0. 0. ]]
[[0.27708893 1.4085555 1.64177102]
[0.62154257 3.04315813 2.61848461]]
[[1.64177102 0.27708893 1.4085555 ]
[3.24002718 3.6647007 5.66164274]]]
[[[0.48080385 0.85910831 0.52342904]
[0. 0. 0. ]]
[[0.52342904 0.48080385 0.85910831]
[1.08970318 2.57102289 2.62245924]]
[[0.85910831 0.52342904 0.48080385]
[3.71216242 3.66072607 5.19348213]]]
[[[1.13610207 1.51237019 0.47256909]
[0. 0. 0. ]]
[[1.51237019 0.47256909 1.13610207]
[2.92304081 2.59328103 0.76686347]]
[[0.47256909 1.13610207 1.51237019]
[5.51632184 3.3601445 3.68990428]]]
[[[1.08397801 1.16506242 0.84703646]
[0. 0. 0. ]]
[[1.16506242 0.84703646 1.08397801]
[2.37250664 2.04419242 1.86648625]]
[[0.84703646 1.08397801 1.16506242]
[4.41669906 3.91067866 4.23899289]]]
[[[0.98734317 1.11177984 0.90283297]
[0. 0. 0. ]]
[[1.11177984 0.90283297 0.98734317]
[2.25981006 2.13666143 1.88671382]]
[[0.90283297 0.98734317 1.11177984]
[4.39647149 4.02337525 4.14652387]]]
[[[1.94118244 1.14738719 1.98251535]
[0. 0. 0. ]]
[[1.14738719 1.98251535 1.94118244]
[1.83291888 1.90183408 2.54843234]]
[[1.98251535 1.94118244 1.14738719]
[3.73475296 4.45026642 4.38135123]]]]
And here is a list of the indices I want to keep:
rot_index = np.array([1 2 1 1 1 1 1 2 1 1])
So just as an example, if you consider
data[0,:,:,0] = [[0.70846822 0.]
[0.66736535 1.54545219]
[0.98552876 3.88519982]]
data[0,:,:,1] = [[0.98552876 0.]
[0.70846822 2.39798549]
[0.66736535 3.94343768]]
data[0,:,:,2] = [[0.66736535 0.]
[0.98552876 2.33974762]
[0.70846822 4.73773311]]
These are 3 different "rotations" of the same sample, and if we look at the first element of rot_index, it is a 1. So I only want to keep
data[0,:,:,1] = [[0.98552876 0.]
[0.70846822 2.39798549]
[0.66736535 3.94343768]]
Using numpy advanced indexing, and under that, the specific subtopic of combining advanced and basic indexing this should work (where data_array is a numpy ndarray having your data):
result = data_array[range(500000),...,rot_index]
For your sample data, this produces:
[[[0.98552876 0. ]
[0.70846822 2.39798549]
[0.66736535 3.94343768]]
[[1.53004225 0. ]
[0.8132551 2.26386152]
[1.18845796 3.69597906]]
[[0.93651048 0. ]
[0.83170562 2.38383457]
[0.66123513 4.18571459]]
[[0.67740955 0. ]
[1.42020073 1.97600777]
[1.31047414 4.62256956]]
[[1.64177102 0. ]
[1.4085555 3.04315813]
[0.27708893 3.6647007 ]]
[[0.85910831 0. ]
[0.48080385 2.57102289]
[0.52342904 3.66072607]]
[[1.51237019 0. ]
[0.47256909 2.59328103]
[1.13610207 3.3601445 ]]
[[0.84703646 0. ]
[1.08397801 1.86648625]
[1.16506242 4.23899289]]
[[1.11177984 0. ]
[0.90283297 2.13666143]
[0.98734317 4.02337525]]
[[1.14738719 0. ]
[1.98251535 1.90183408]
[1.94118244 4.45026642]]]

Insert Numpy Array into Array with extending of the embedding array

First of all, I work with byte array (>= 400x400x1000) bytes.
I wrote a small function which can insert a multidimensional array (or a fraction of) into another one by indicating an offset. This works if the embedded array is smaller than the embedding array (case A). Otherwise the embedded array is truncated (case B).
case A) Inserting a 3x3 into a 5x5 matrix with offset 1,1 would look like this.
[[ 0. 0. 0. 0. 0.]
[ 0. 1. 1. 1. 0.]
[ 0. 1. 1. 1. 0.]
[ 0. 1. 1. 1. 0.]
[ 0. 0. 0. 0. 0.]]
case B) If the offsets are exceeding the dimensions of the embedding matrix, the smaller array is truncated. E.g. a (-1,-1) offset would result in this.
[[ 1. 1. 0. 0. 0.]
[ 1. 1. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]]
case C) Now, instead of truncating the embedded array, I want to extend the embedding array (by zeroes) if the embedded array is either bigger than the embedding array or the offsets enforce it (e.g. case B). Is there a smart way with numpy or scipy to solve this?
[[ 1. 1. 1. 0. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0.]]
Actually I work with 3D array, but for simplicity I wrote an example for 2D arrays. Current source:
import numpy as np
import nibabel as nib
def addAtPos(mat_bigger, mat_smaller, xyz_coor):
size_sm_x, size_sm_y = np.shape(mat_smaller)
size_gr_x, size_gr_y = np.shape(mat_bigger)
start_gr_x, start_gr_y = xyz_coor
start_sm_x, start_sm_y = 0,0
end_x, end_y = (start_gr_x + size_sm_x), (start_gr_y + size_sm_y)
print(size_sm_x, size_sm_y)
print(size_gr_x, size_gr_y)
print(end_x, end_y)
if start_gr_x < 0:
start_sm_x = -start_gr_x
start_gr_x = 0
if start_gr_y < 0:
start_sm_y = -start_gr_y
start_gr_y = 0
if end_x > size_gr_x:
size_sm_x = size_sm_x - (end_x - size_gr_x)
end_x = size_gr_x
if end_y > size_gr_y:
size_sm_y = size_sm_y - (end_y - size_gr_y)
end_y = size_gr_y
# copy all or a chunk (if offset is small/big enough) of the smaller matrix into the bigger matrix
mat_bigger[start_gr_x:end_x, start_gr_y:end_y] = mat_smaller[start_sm_x:size_sm_x, start_sm_y:size_sm_y]
return mat_bigger
a_gr = np.zeros([5,5])
a_sm = np.ones([3,3])
a_res = addAtPos(a_gr, a_sm, [-2,1])
#print (a_gr)
print (a_res)
Actually there is an easier way to do it.
For your first example of a 3x3 array embedded to a 5x5 one you can do it with something like:
A = np.array([[1,1,1], [1,1,1], [1,1,1]])
(N, M) = A.shape
B = np.zeros(shape=(N + 2, M + 2))
B[1:-1:, 1:-1] = A
By playing with slicing you can select a subset of A and insert it anywhere within a continuous subset of B.
Hope it helps! ;-)

Does this function compute convolution correctly?

I need to write a basic function that computes a 2D convolution between a matrix and a kernel.
I have recently got into Python, so I'm sorry for my mistakes.
My dissertation teacher said that I should write one by myself so I can handle it better and to be able to modify it for future improvements.
I have found an example of this function on a website, but I don't understand how the returned values are obtained.
This is the code (from http://docs.cython.org/src/tutorial/numpy.html )
from __future__ import division
import numpy as np
def naive_convolve(f, g):
# f is an image and is indexed by (v, w)
# g is a filter kernel and is indexed by (s, t),
# it needs odd dimensions
# h is the output image and is indexed by (x, y),
# it is not cropped
if g.shape[0] % 2 != 1 or g.shape[1] % 2 != 1:
raise ValueError("Only odd dimensions on filter supported")
# smid and tmid are number of pixels between the center pixel
# and the edge, ie for a 5x5 filter they will be 2.
#
# The output size is calculated by adding smid, tmid to each
# side of the dimensions of the input image.
vmax = f.shape[0]
wmax = f.shape[1]
smax = g.shape[0]
tmax = g.shape[1]
smid = smax // 2
tmid = tmax // 2
xmax = vmax + 2*smid
ymax = wmax + 2*tmid
# Allocate result image.
h = np.zeros([xmax, ymax], dtype=f.dtype)
# Do convolution
for x in range(xmax):
for y in range(ymax):
# Calculate pixel value for h at (x,y). Sum one component
# for each pixel (s, t) of the filter g.
s_from = max(smid - x, -smid)
s_to = min((xmax - x) - smid, smid + 1)
t_from = max(tmid - y, -tmid)
t_to = min((ymax - y) - tmid, tmid + 1)
value = 0
for s in range(s_from, s_to):
for t in range(t_from, t_to):
v = x - smid + s
w = y - tmid + t
value += g[smid - s, tmid - t] * f[v, w]
h[x, y] = value
return h
I don't know if this function does the weighted sum from input and filter, because I see no sum here.
I applied this with
kernel = np.array([(1, 1, -1), (1, 0, -1), (1, -1, -1)])
file = np.ones((5,5))
naive_convolve(file, kernel)
I got this matrix:
[[ 1. 2. 1. 1. 1. 0. -1.]
[ 2. 3. 1. 1. 1. -1. -2.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 2. 1. -1. -1. -1. -3. -2.]
[ 1. 0. -1. -1. -1. -2. -1.]]
I tried to do a manual calculation (on paper) for the first full iteration of the function and I got 'h[0,0] = 0', because of the matrix product: 'filter[0, 0] * matrix[0, 0]', but the function returns 1. I am very confused with this.
If anyone can help me understand what is going on here, I would be very grateful. Thanks! :)
Yes, that function computes the convolution correctly. You can check this using scipy.signal.convolve2d
import numpy as np
from scipy.signal import convolve2d
kernel = np.array([(1, 1, -1), (1, 0, -1), (1, -1, -1)])
file = np.ones((5,5))
x = convolve2d(file, kernel)
print x
Which gives:
[[ 1. 2. 1. 1. 1. 0. -1.]
[ 2. 3. 1. 1. 1. -1. -2.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 3. 3. 0. 0. 0. -3. -3.]
[ 2. 1. -1. -1. -1. -3. -2.]
[ 1. 0. -1. -1. -1. -2. -1.]]
It's impossible to know how to explain all this to you since I don't know where to start, and I don't know how all the other explanations aren't working for you. I think, though, that you are doing all of this as a learning exercise so you can figure this out for yourself. From what I've seen on SO, asking big questions on SO is not a substitute for working it through yourself.
Your specific question of why does
h[0,0] = 0
in your calculation not match this matrix is a good one. In fact, both are correct. The reason for mismatch is that the output of the convolution doesn't have the mathematical indices specified, but instead they are implied. The center, which is mathematically indicated by the indices [0,0] corresponds to x[3,3] in the matrix above.

Categories