Efficiently apply function to spheric neighbourhood in numpy array - python

I have a 3D numpy array of float values in Python.
I need to retrieve all the elements in a sphere of radius r starting from
a center point P(x, y, z). Then, I want to apply to the sphere points a function that
updates their values and needs the distance to the center point to do this. I do these steps a lot of times and for
large radius values, so I would like to have a solution that is as efficient
as possible.
My current solution checks only the points in the bounding box of the sphere,
as indicated here: Using a QuadTree to get all points within a bounding circle.
A sketch of the code looks like this:
# P(x, y, z): center of the sphere
for k1 in range(x - r, x + r + 1):
for k2 in range(y - r, y + r + 1):
for k3 in range(z - r, z + r + 1):
# Sphere center - current point distance
dist = np.sum((np.array([k1, k2, k3]) - np.array([x, y, z])) ** 2)
if (dist <= r * r):
# computeUpdatedValue(distance, radius): function that computes the new value of the matrix in the current point
newValue = computeUpdatedValue(dist, r)
# Update the matrix
mat[k1, k2, k3] = newValue
However, I thought that applying a mask to retrive the points and, then,
update them based on distance in a vectorized manner is more efficient.
I have seen how to apply a circular kernel
(How to apply a disc shaped mask to a numpy array?),
but I do no know how to efficiently apply the function (depending on the indices) on each of the mask's elements.

EDIT: If your array is very big compared to the region you are updating, the solution below will take much more memory than necessary. You can apply the same idea but only to the region where the sphere may fall:
def updateSphereBetter(mat, center, radius):
# Find beginning and end of region of interest
center = np.asarray(center)
start = np.minimum(np.maximum(center - radius, 0), mat.shape)
end = np.minimum(np.maximum(center + radius + 1, 0), mat.shape)
# Slice region of interest
mat_sub = mat[tuple(slice(s, e) for s, e in zip(start, end))]
# Center coordinates relative to the region of interest
center_rel = center - start
# Same as before but with mat_sub and center_rel
ind = np.indices(mat_sub.shape)
ind = np.moveaxis(ind, 0, -1)
dist_squared = np.sum(np.square(ind - center_rel), axis=-1)
mask = dist_squared <= radius * radius
mat_sub[mask] = computeUpdatedValue(dist_squared[mask], radius)
Note that since mat_sub is a view of mat, updating it updates the original array, so this produces the same result as before, but with less resources.
Here is a little proof of concept. I defined computeUpdatedValue so that it shows the distance from the center, and then plotted a few "sections" of an example:
import numpy as np
import matplotlib.pyplot as plt
def updateSphere(mat, center, radius):
# Make array of all index coordinates
ind = np.indices(mat.shape)
# Compute the squared distances to each point
ind = np.moveaxis(ind, 0, -1)
dist_squared = np.sum(np.square(ind - center), axis=-1)
# Make a mask for squared distances within squared radius
mask = dist_squared <= radius * radius
# Update masked values
mat[mask] = computeUpdatedValue(dist_squared[mask], radius)
def computeUpdatedValue(dist_squared, radius):
# 1 at the center of the sphere and 0 at the surface
return np.clip(1 - np.sqrt(dist_squared) / radius, 0, 1)
mat = np.zeros((100, 60, 80))
updateSphere(mat, [50, 20, 40], 20)
plt.subplot(131)
plt.imshow(mat[:, :, 30], vmin=0, vmax=1)
plt.subplot(132)
plt.imshow(mat[:, :, 40], vmin=0, vmax=1)
plt.subplot(133)
plt.imshow(mat[:, :, 55], vmin=0, vmax=1)
Output:

Related

Binning 2D data with circles instead of rectangles - from pandas df

I have a dataframe of x, y data and need to bin it into circles. Ie a grid of circles of certain size and spacing centered on some point. So for example some data would be left out after this sampling/binning. How is this possible?
I have tried np.histogram2d and creating masks/broadcasting. The mask was too slow, and I don't seem able to broadcast into a circle. Only to tell if the point is within said grid of circles via this answer: Binning 2D data into overlapping circles in x,y.
If there is a way to input edges or something into histogram2d and make the edges circular please let me know. Cheers
The only way this can be done is by looping over your points and grid of circles like so:
def inside_circle(x, y, x0, y0, r):
return (x - x0)*(x - x0) + (y - y0)*(y - y0) < r*r
x_bins = np.linspace(-9, 9, 30)
y_bins = np.linspace(-9, 9, 30)
h = df['upmm']
w = df['anode_entrance']
histo = np.zeros((32,32))
for i in range(0, len(h)):
for j in range(0, len(x_bins)):
for k in range(0, len(y_bins)):
if inside_circle(h[i], w[i], x_bins[j], y_bins[k], 0.01):
histo[j][k] = histo[j][k] + 1
plt.imshow(histo, cmap='hot', interpolation='nearest')
plt.show()

Draw a circle in a numpy array given index and radius without external libraries

I need to draw a circle in a 2D numpy array given [i,j] as indexes of the array, and r as the radius of the circle. Each time a condition is met at index [i,j], a circle should be drawn with that as the center point, increasing all values inside the circle by +1. I want to avoid the for-loops at the end where I draw the circle (where I use p,q to index) because I have to draw possibly millions of circles. Is there a way without for loops? I also don't want to import another library for just a single task.
Here is my current implementation:
for i in range(array_shape[0]):
for j in range(array_shape[1]):
if (condition): # Draw circle if condition is fulfilled
# Create a square of pixels with side lengths equal to radius of circle
x_square_min = i-r
x_square_max = i+r+1
y_square_min = j-r
y_square_max = j+r+1
# Clamp this square to the edges of the array so circles near edges don't wrap around
if x_square_min < 0:
x_square_min = 0
if y_square_min < 0:
y_square_min = 0
if x_square_max > array_shape[0]:
x_square_max = array_shape[0]
if y_square_max > array_shape[1]:
y_square_max = array_shape[1]
# Now loop over the box and draw circle inside of it
for p in range(x_square_min , x_square_max):
for q in range(y_square_min , y_square_max):
if (p - i) ** 2 + (q - j) ** 2 <= r ** 2:
new_array[p,q] += 1 # Incrementing because need to have possibility of
# overlapping circles
If you're using the same radius for every single circle, you can simplify things significantly by only calculating the circle coordinates once and then adding the center coordinates to the circle points when needed. Here's the code:
# The main array of values is called array.
shape = array.shape
row_indices = np.arange(0, shape[0], 1)
col_indices = np.arange(0, shape[1], 1)
# Returns xy coordinates for a circle with a given radius, centered at (0,0).
def points_in_circle(radius):
a = np.arange(radius + 1)
for x, y in zip(*np.where(a[:,np.newaxis]**2 + a**2 <= radius**2)):
yield from set(((x, y), (x, -y), (-x, y), (-x, -y),))
# Set the radius value before running code.
radius = RADIUS
circle_r = np.array(list(points_in_circle(radius)))
# Note that I'm using x as the row number and y as the column number.
# Center of circle is at (x_center, y_center). shape_0 and shape_1 refer to the main array
# so we can get rid of coordinates outside the bounds of array.
def add_center_to_circle(circle_points, x_center, y_center, shape_0, shape_1):
circle = np.copy(circle_points)
circle[:, 0] += x_center
circle[:, 1] += y_center
# Get rid of rows where coordinates are below 0 (can't be indexed)
bad_rows = np.array(np.where(circle < 0)).T[:, 0]
circle = np.delete(circle, bad_rows, axis=0)
# Get rid of rows that are outside the upper bounds of the array.
circle = circle[circle[:, 0] < shape_0, :]
circle = circle[circle[:, 1] < shape_1, :]
return circle
for x in row_indices:
for y in col_indices:
# You need to set CONDITION before running the code.
if CONDITION:
# Because circle_r is the same for all circles, it doesn't need to be recalculated all the time. All you need to do is add x and y to circle_r each time CONDITION is met.
circle_coords = add_center_to_circle(circle_r, x, y, shape[0], shape[1])
array[tuple(circle_coords.T)] += 1
When I set radius = 10, array = np.random.rand(1200).reshape(40, 30) and replaced if CONDITION with if (x == 20 and y == 20) or (x == 25 and y == 20), I got this, which seems to be what you want:
Let me know if you have any questions.
Adding each circle can be vectorized. This solution iterates over the coordinates where the condition is met. On a 2-core colab instance ~60k circles with radius 30 can be added per second.
import numpy as np
np.random.seed(42)
arr = np.random.rand(400,300)
r = 30
xx, yy = np.mgrid[-r:r+1, -r:r+1]
circle = xx**2 + yy**2 <= r**2
condition = np.where(arr > .999) # np.where(arr > .5) to benchmark 60k circles
for x,y in zip(*condition):
# valid indices of the array
i = slice(max(x-r,0), min(x+r+1, arr.shape[0]))
j = slice(max(y-r,0), min(y+r+1, arr.shape[1]))
# visible slice of the circle
ci = slice(abs(min(x-r, 0)), circle.shape[0] - abs(min(arr.shape[0]-(x+r+1), 0)))
cj = slice(abs(min(y-r, 0)), circle.shape[1] - abs(min(arr.shape[1]-(y+r+1), 0)))
arr[i, j] += circle[ci, cj]
Visualizing np.array arr
import matplotlib.pyplot as plt
plt.figure(figsize=(8,8))
plt.imshow(arr)
plt.show()

How to use mgrid to interpolate between a rectangle and a circle

I am trying to create a 3D surface that has a 1/4 rectangle for the exterior and 1/4 circle for the interior. I had help before to create the 3D surface with an ellipse as an exterior but I cannot do this for a rectangle for some reason. I have done the math by hand which makes sense, but my code does not. I would greatly appreciate any help with this.
import numpy as np
import pyvista as pv
# parameters for the waveguide
# diameter of the inner circle
waveguide_throat = 30
# axes of the outer ellipse
ellipse_x = 250
ellipse_y = 170
# shape parameters for the z profile
depth_factor = 4
angle_factor = 40
# number of grid points in radial and angular direction
array_length = 100
phase_plug = 0
phase_plug_dia = 20
plug_offset = 5
dome_dia = 28
# theta is angle where x and y intersect
theta = np.arctan(ellipse_x / ellipse_y)
# chi is for x direction and lhi is for y direction
chi = np.linspace(0, theta, 100)
lhi = np.linspace(theta, np.pi/2, 100)
# mgrid to create structured grid
r, phi = np.mgrid[0:1:array_length*1j, 0:np.pi/2:array_length*1j]
# Rectangle exterior, circle interior
x = (ellipse_y * np.tan(chi)) * r + ((waveguide_throat / 2 * (1 - r)) * np.cos(phi))
y = (ellipse_x / np.tan(lhi)) * r + ((waveguide_throat / 2 * (1 - r)) * np.sin(phi))
# compute z profile
angle_factor = angle_factor / 10000
z = (ellipse_x / 2 * r / angle_factor) ** (1 / depth_factor)
plotter = pv.Plotter()
waveguide_mesh = pv.StructuredGrid(x, y, z)
plotter.add_mesh(waveguide_mesh)
plotter.show()
The linear interpolation you're trying to use is a general tool that should work (with one small caveat). So the issue is first with your rectangular edge.
Here's a sanity check which plots your interior and exterior lines:
# debugging: plot interior and exterior
exterior_points = np.array([
ellipse_y * np.tan(chi),
ellipse_x / np.tan(lhi),
np.zeros_like(chi)
]).T
phi_aux = np.linspace(0, np.pi/2, array_length)
interior_points = np.array([
waveguide_throat / 2 * np.cos(phi_aux),
waveguide_throat / 2 * np.sin(phi_aux),
np.zeros_like(phi_aux)
]).T
plotter = pv.Plotter()
plotter.add_mesh(pv.wrap(exterior_points))
plotter.add_mesh(pv.wrap(interior_points))
plotter.show()
The bottom left is your interior circle, looks good. The top right is what's supposed to be a rectangle, but isn't.
To see why your original surface looks the way it does, we have to note one more thing (this is the small caveat I mentioned): the orientation of your curves is also the opposite. This implies that you interpolate the "top" (in the screenshot) point of your interior curve with the "bottom" point of the exterior curve. This explains the weird fan shape.
So you need to fix the exterior curve, and make sure the orientation of the two edges is the same. Note that you can just create the two 1d arrays for the two edges, and then interpolate them. You don't have to come up with a symbolic formula that you plug into the interpolation step. If you have 1d arrays of the same shape x_interior, y_interior, x_exterior, y_exterior then you can then do x_exterior * r + x_interior * (1 - r) and the same for y. This means removing the mgrid call, only using an array r of shape (n, 1), and making use of array broadcasting to do the interpolation. This means doing r = np.linspace(0, 1, array_length)[:, None].
So the question is how to define your rectangle. You need to have the same number of points on the rectangular curve than what you have on the circle (I would strongly recommend using the array_length parameter everywhere to ensure this!). Since you want to span the whole rectangle, I believe you have to choose an array index (i.e. a certain angle in the circular arc) which will map to the corner of the rectangle. Then it's a simple matter of varying only y for the points until that index, and x for the rest (or vice versa).
Here's what I mean: you know that the rectangle's corner is at angle theta in your code (although I think you have x and y mixed up if we assume the conventional relationship between "x", "y" and the tangent of the angle). Since theta goes from 0 to pi/2, and your phi values also go from 0 to pi/2, you should choose index (array_length * (2*theta/np.pi)).round().astype(int) - 1 (or something similar) that will map to the rectangle's corner. If you have a square, this gives you theta = pi/4, and consequently (array_length / 2).round().astype(int) - 1. For array_length = 3 this is index (2 - 1) == 1, which is the middle index for 3-length arrays. (The more points you have along the edge, the less it will matter if you commit an off-by-one error here.)
The only remaining complication then is that we have to explicitly broadcast the 1d z array to the common shape. And we can use the same math you used to get a rectangular edge that is equidistant in angles.
Your code fixed with this suggestion (note that I've added 1 to the corner index because I'm using it as a right-exclusive range index):
import numpy as np
import pyvista as pv
# parameters for the waveguide
# diameter of the inner circle
waveguide_throat = 30
# axes of the outer ellipse
ellipse_x = 250
ellipse_y = 170
# shape parameters for the z profile
depth_factor = 4
angle_factor = 40
# number of grid points in radial and angular direction
array_length = 100
# quarter circle interior line
phi = np.linspace(0, np.pi/2, array_length)
x_interior = waveguide_throat / 2 * np.cos(phi)
y_interior = waveguide_throat / 2 * np.sin(phi)
# theta is angle where x and y intersect
theta = np.arctan2(ellipse_y, ellipse_x)
# find array index which maps to the corner of the rectangle
corner_index = (array_length * (2*theta/np.pi)).round().astype(int)
# construct rectangular coordinates manually
x_exterior = np.zeros_like(x_interior)
y_exterior = x_exterior.copy()
phi_aux = np.linspace(0, theta, corner_index)
x_exterior[:corner_index] = ellipse_x
y_exterior[:corner_index] = ellipse_x * np.tan(phi_aux)
phi_aux = np.linspace(np.pi/2, theta, array_length - corner_index, endpoint=False)[::-1] # mind the reverse!
x_exterior[corner_index:] = ellipse_y / np.tan(phi_aux)
y_exterior[corner_index:] = ellipse_y
# interpolate between two curves
r = np.linspace(0, 1, array_length)[:, None] # shape (array_length, 1) for broadcasting
x = x_exterior * r + x_interior * (1 - r)
y = y_exterior * r + y_interior * (1 - r)
# debugging: plot interior and exterior
exterior_points = np.array([
x_exterior,
y_exterior,
np.zeros_like(x_exterior),
]).T
interior_points = np.array([
x_interior,
y_interior,
np.zeros_like(x_interior),
]).T
plotter = pv.Plotter()
plotter.add_mesh(pv.wrap(exterior_points))
plotter.add_mesh(pv.wrap(interior_points))
plotter.show()
# compute z profile
angle_factor = angle_factor / 10000
z = (ellipse_x / 2 * r / angle_factor) ** (1 / depth_factor)
# explicitly broadcast to the shape of x and y
z = np.broadcast_to(z, x.shape)
plotter = pv.Plotter()
waveguide_mesh = pv.StructuredGrid(x, y, z)
plotter.add_mesh(waveguide_mesh, style='wireframe')
plotter.show()
The curves look reasonable:
As does the interpolated surface:

Generate profiles through a 2D array at an angle without altering pixels

I'd like to plot two profiles through the highest intensity point in a 2D numpy array, which is an image of a blob (i.e. a line through the semi-major axis, and another line through the semi-minor axis). The blob is rotated at an angle theta counterclockwise from the standard x-axis and is asymmetric.
It is a 600x600 array with a max intensity of 1 (at only one pixel) that is located right at the center at (300, 300). The angle rotation from the x-axis (which then gives the location of the semi-major axis when rotated by that angle) is theta = 89.54 degrees. I do not want to use scipy.ndimage.rotate because it uses spline interpolation, and I do not want to change any of my pixel values. But I suppose a nearest-neighbor interpolation method would be okay.
I tried generating lines corresponding to the major and minor axes across the image, but the result was not right at all (the peak was far less than 1), so maybe I did something wrong. The code for this is below:
import numpy as np
import matplotlib.pyplot as plt
from scipy import ndimage
def profiles_at_angle(image, axis, theta):
theta = np.deg2rad(theta)
if axis == 'major':
x_0, y_0 = 0, 300-300*np.tan(theta)
x_1, y_1 = 599, 300+300*np.tan(theta)
elif axis=='minor':
x_0, y_0 = 300-300*np.tan(theta), 599
x_1, y_1 = 300+300*np.tan(theta), -599
num = 600
x, y = np.linspace(x_0, x_1, num), np.linspace(y_0, y_1, num)
z = ndimage.map_coordinates(image, np.vstack((x,y)))
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(image, cmap='gray')
axes[0].axis('image')
axes[1].plot(z)
plt.xlim(250,350)
plt.show()
profiles_at_angle(image, 'major', theta)
Did I do something obviously wrong in my code above? Or how else can I accomplish this? Thank you.
Edit: Here are some example images. Sorry for the bad quality; my browser crashed every time I tried uploading them anywhere so I had to take photos of the screen.
Figure 1: This is the result of my code above, which is clearly wrong since the peak should be at 1. I'm not sure what I did wrong though.
Figure 2: I made this plot below by just taking the profiles through the standard x and y axes, ignoring any rotation (this only looks good coincidentally because the real angle of rotation is so close to 90 degrees, so I was able to just switch the labels and get this). I want my result to look something like this, but taking the correction rotation angle into account.
Edit: It could be useful to run tests on this method using data very much like my own (it's a 2D Gaussian with nearly the same parameters):
image = np.random.random((600,600))
def generate(data_set):
xvec = np.arange(0, np.shape(data_set)[1], 1)
yvec = np.arange(0, np.shape(data_set)[0], 1)
X, Y = np.meshgrid(xvec, yvec)
return X, Y
def gaussian_func(xy, x0, y0, sigma_x, sigma_y, amp, theta, offset):
x, y = xy
a = (np.cos(theta))**2/(2*sigma_x**2) + (np.sin(theta))**2/(2*sigma_y**2)
b = -np.sin(2*theta)/(4*sigma_x**2) + np.sin(2*theta)/(4*sigma_y**2)
c = (np.sin(theta))**2/(2*sigma_x**2) + (np.cos(theta))**2/(2*sigma_y**2)
inner = a * (x-x0)**2
inner += 2*b*(x-x0)*(y-y0)
inner += c * (y-y0)**2
return (offset + amp * np.exp(-inner)).ravel()
xx, yy = generate(image)
image = gaussian_func((xx.ravel(), yy.ravel()), 300, 300, 5, 4, 1, 1.56, 0)
image = np.reshape(image, (600, 600))
This should do it for you. You just did not properly compute your lines.
theta = 65
peak = np.argwhere(image==1)[0]
x = np.linspace(peak[0]-100,peak[0]+100,1000)
y = lambda x: (x-peak[1])*np.tan(np.deg2rad(theta))+peak[0]
y_maj = np.linspace(y(peak[1]-100),y(peak[1]+100),1000)
y = lambda x: -(x-peak[1])/np.tan(np.deg2rad(theta))+peak[0]
y_min = np.linspace(y(peak[1]-100),y(peak[1]+100),1000)
del y
z_min = scipy.ndimage.map_coordinates(image, np.vstack((x,y_min)))
z_maj = scipy.ndimage.map_coordinates(image, np.vstack((x,y_maj)))
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(image)
axes[0].plot(x,y_maj)
axes[0].plot(x,y_min)
axes[0].axis('image')
axes[1].plot(z_min)
axes[1].plot(z_maj)
plt.show()

computing weight based on distance from centre of ellipse

I was using a circular image mask before and I was calculating a weight based on the distance from the centre of the circle as follows:
import numpy as np
def create_mask(image, radius, center=(0, 0)):
r, c, d = image.shape
x, y = np.ogrid[:r, :c]
distance = np.sqrt((x-center[0])**2 + (y-center[1])**2)
m = distance < radius
distance[m] = 1.0 - distance[m]/radius
array = np.zeros((r, c))
array[m] = distance[m]
return array
This was basically setting the height weight at the centre and the weight was dropping linearly towards the edges.
Now, I want to do something similar with an ellipse. Again, the ellipse can have very different radii along the two dimensions and I would like the weight to drop linearly with distance as well. However, regardless of the long or the short radii, I would like the weights to decay similarly towards the edges. I am guessing I need to include a weight based on both the radius to achieve this but was unable to figure it out.
I'm not sure about linear weights, but you can achieve a continuous array weights from 1 to 0 using (and I'm sure there's a more efficient way to do this)
ellipse = lambda x0, y0, r_x, r_y: lambda x, y: ((x - x0) / r_x)**2 + ((y - y0) / r_y)**2
def gen_ellipse(el, lower, upper, step):
coords = np.arange(lower, upper, step)
x, y = np.meshgrid(coords, coords)
mask = el(x, y)
mask[np.where(mask > 1)] = 0
return 1 - mask
For example:
> %pylab
> el = ellipse(0.0, 0.0, 0.3, 0.8)
> mask = gen_ellipse(el, -1.0, 1.0, 0.0025)
> imshow(mask, cmap=get_cmap('Greys'))
Where black is 1 and white is 0.

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