This might seem a bit strange, but I really feel like there should be a relatively straightforward solution to it. Basically I've got an image in the form of a 3D numpy array (x, y, color). I was following along with this tutorial for a slightly different product area, and found that these methods did not extend well.
As a result, I'm making a modified edge detection algorithm for my use case. As of now this is just some basic signal processing on top of a 1d array. This works great if I only want to sample in the x and y directions, as I can just use the existing rows and columns of the array.
However, to determine orientation of these edges, I would like to be able to sample any arbitrary vector across the image below is an image to help illustrate:
I tried hacking together something that would just append pixels as it crossed them, but it was inefficient, inelegant, and non-ideal in a number of ways. I feel like there must be some relatively elegant way of doing this.
Any ideas? The size of the sample across the vector doesn't really matter to me if that makes things any easier.
I would make an equation for the line you want to cut along, then make a mask around it and keep all pixels that come within some width of it. For example, say you want a cut along i = 2*j + 34, where i and j are measured in pixels:
h, w = im.shape[:2]
width = 2 # width of slice in pixels, too narrow and it will have gaps
i, j = np.ogrid[:h, :w]
mask = np.abs(2*j + 34 - i) < width
im[mask]
Note that im[mask] will be a 2d array, since it should still have the colors. It will be ordered so that the uppermost pixels are first, and the bottom pixels are last, opposite of that shown in your arrow, unless of course you have origin=lower in your plotting :) And if several pixels are selected in each row (if width > 1), then they'll go left to right, so the shape for a slice like your drawing would be a tiny sequence of z's, and for the other direction, backwards z's (s's?).
Keep in mind that for an array there doesn't exist a diagonal slice without some weird zigzag (or alternatively, interpolation) no matter how elegant your implementation is. You could rotate the image (by some algorithm) and take a horizontal slice.
Using the equation
x2 = x1 + length * cos(θ)
y2 = y1 + length * sin(θ)
where
θ = angle * 3.14 / 180.0
You can iterate through the pixel using angle and length like
int angle =45; //angle of iteration
int length = 0; //Alternately you can skip the pixel by giving value other than 0
Point P1(starX,startY); //Your starting point.
Point P2;//??
while(1){
length++;
P2.x = (int)round(P1.x + length * cos(angle * CV_PI / 180.0));
P2.y = (int)round(P1.y + length * sin(angle * CV_PI / 180.0));
if(P2_exceed_boundary()) break;
do_Whatever_with_P2();
}
Related
I have a problem where in a grid of x*y size I am provided a single dot, and I need to find the nearest neighbour. In practice, I am trying to find the closest dot to the cursor in pygame that crosses a color distance threshold that is calculated as following:
sqrt(((rgb1[0]-rgb2[0])**2)+((rgb1[1]-rgb2[1])**2)+((rgb1[2]-rgb2[2])**2))
So far I have a function that calculates the different resolutions for the grid and reduces it by a factor of two while always maintaining the darkest pixel. It looks as following:
from PIL import Image
from typing import Dict
import numpy as np
#we input a pillow image object and retrieve a dictionary with every grid version of the 3 dimensional array:
def calculate_resolutions(image: Image) -> Dict[int, np.ndarray]:
resolutions = {}
#we start with the highest resolution image, the size of which we initially divide by 1, then 2, then 4 etc.:
divisor = 1
#reduce the grid by 5 iterations
resolution_iterations = 5
for i in range(resolution_iterations):
pixel_lookup = image.load() #convert image to PixelValues object, which allows for pixellookup via [x,y] index
#calculate the resolution of the new grid, round upwards:
resolution = (int((image.size[0] - 1) // divisor + 1), int((image.size[1] - 1) // divisor + 1))
#generate 3d array with new grid resolution, fill in values that are darker than white:
new_grid = np.full((resolution[0],resolution[1],3),np.array([255,255,255]))
for x in range(image.size[0]):
for y in range(image.size[1]):
if not x%divisor and not y%divisor:
darkest_pixel = (255,255,255)
x_range = divisor if x+divisor<image.size[0] else (0 if image.size[0]-x<0 else image.size[0]-x)
y_range = divisor if y+divisor<image.size[1] else (0 if image.size[1]-y<0 else image.size[1]-y)
for x_ in range(x,x+x_range):
for y_ in range(y,y+y_range):
if pixel_lookup[x_,y_][0]+pixel_lookup[x_,y_][1]+pixel_lookup[x_,y_][2] < darkest_pixel[0]+darkest_pixel[1]+darkest_pixel[2]:
darkest_pixel = pixel_lookup[x_,y_]
if darkest_pixel != (255,255,255):
new_grid[int(x/divisor)][int(y/divisor)] = np.array(darkest_pixel)
resolutions[i] = new_grid
divisor = divisor*2
return resolutions
This is the most performance efficient solution I was able to come up with. If this function is run on a grid that continually changes, like a video with x fps, it will be very performance intensive. I also considered using a kd-tree algorithm that simply adds and removes any dots that happen to change on the grid, but when it comes to finding individual nearest neighbours on a static grid this solution has the potential to be more resource efficient. I am open to any kinds of suggestions in terms of how this function could be improved in terms of performance.
Now, I am in a position where for example, I try to find the nearest neighbour of the current cursor position in a 100x100 grid. The resulting reduced grids are 50^2, 25^2, 13^2, and 7^2. In a situation where a part of the grid looks as following:
And I am on the aggregation step where a part of the grid consisting of six large squares, the black one being the current cursor position and the orange dots being dots where the color distance threshold is crossed, I would not know which diagonally located closest neighbour I would want to pick to search next. In this case, going one aggregation step down shows that the lower left would be the right choice. Depending on how many grid layers I have this could result in a very large error in terms of the nearest neighbour search. Is there a good way how I can solve this problem? If there are multiple squares that show they have a relevant location, do I have to search them all in the next step to be sure? And if that is the case, the further away I get the more I would need to make use of math functions such as the pythagorean theorem to assert whether the two positive squares I find are overlapping in terms of distance and could potentially contain the closest neighbour, which would start to be performance intensive again if the function is called frequently. Would it still make sense to pursue this solution over a regular kd tree? For now the grid size is still fairly small (~800-600) but if the grid gets larger the performance may start suffering again. Is there a good scalable solution to this problem that could be applied here?
I am trying to pack hard-spheres in a unit cubical box, such that these spheres cannot overlap on each other. This is being done in Python.
I am given some packing fraction f, and the number of spheres in the system is N.
So, I say that the diameter of each sphere will be
d = (p*6/(math.pi*N)**)1/3).
My box has periodic boundary conditions - which means that there is a recurring image of my box in all direction. If there is a particle who is at the edge of the box and has a portion of it going beyond the wall, it will stick out at the other side.
My attempt:
Create a numpy N-by-3 array box which holds the position vector of each particle [x,y,z]
The first particle is fine as it is.
The next particle in the array is checked with all the previous particles. If the distance between them is more than d, move on to the next particle. If they overlap, randomly change the position vector of the particle in question. If the new position does not overlap with the previous atoms, accept it.
Repeat steps 2-3 for the next particle.
I am trying to populate my box with these hard spheres, in the following manner:
for i in range(1,N):
mybool=True
print("particles in box: " + str(i))
while (mybool): #the deal with this while loop is that if we place a bad particle, we need to change its position, and restart the process of checking
for j in range(0,i):
displacement=box[j,:]-box[i,:]
for k in range(3):
if abs(displacement[k])>L/2:
displacement[k] -= L*np.sign(displacement[k])
distance = np.linalg.norm(displacement,2) #check distance between ith particle and the trailing j particles
if distance<diameter:
box[i,:] = np.random.uniform(0,1,(1,3)) #change the position of the ith particle randomly, restart the process
break
if j==i-1 and distance>diameter:
mybool = False
break
The problem with this code is that if p=0.45, it is taking a really, really long time to converge. Is there a better method to solve this problem, more efficiently?
I think what you are looking for is either the hexagonal closed-packed (HCP or sometime called face-centered cubic, FCC) lattice or the cubic closed-packed one (CCP). See e.g. Wikipedia on Close-packing of equal spheres.
Since your space has periodic conditions, I believe it doesn't matter which one you chose (hcp or ccp), and they both achieve the same density of ~74.04%, which was proved by Gauss to be the highest density by lattice packing.
Update:
For the follow-up question on how to generate efficiently one such lattice, let's take as an example the HCP lattice. First, let's create a bunch of (i, j, k) indices [(0,0,0), (1,0,0), (2,0,0), ..., (0,1,0), ...]. Then, get xyz coordinates from those indices and return a DataFrame with them:
def hcp(n):
dim = 3
k, j, i = [v.flatten()
for v in np.meshgrid(*([range(n)] * dim), indexing='ij')]
df = pd.DataFrame({
'x': 2 * i + (j + k) % 2,
'y': np.sqrt(3) * (j + 1/3 * (k % 2)),
'z': 2 * np.sqrt(6) / 3 * k,
})
return df
We can plot the result as scatter3d using plotly for interactive exploration:
import plotly.graph_objects as go
df = hcp(12)
fig = go.Figure(data=go.Scatter3d(
x=df.x, y=df.y, z=df.z, mode='markers',
marker=dict(size=df.x*0 + 30, symbol="circle", color=-df.z, opacity=1),
))
fig.show()
Note: plotly's scatter3d is not a very good rendering of spheres: the marker sizes are constant (so when you zoom in and out, the "spheres" will appear to change relative size), and there is no shading, limited z-ordering faithfulness, etc., but it's convenient to interact with the plot.
Resize and clip to the unit box:
Here, a strict clipping (each sphere needs to be completely inside the unit box). Your "periodic boundary condition" is something you will need to address separately (see further below for ideas).
def hcp_unitbox(r):
n = int(np.ceil(1 / (np.sqrt(3) * r)))
df = hcp(n) * r
df += r
df = df[(df <= 1 - r).all(axis=1)]
return df
With this, you find that a radius of 0.06 gives you 608 fully enclosed spheres:
hcp_unitbox(.06).shape # (608, 3)
Where you would go next:
You may dig deeper into the effect of your so-called "periodic boundary conditions", and perhaps play with some rotations (and small translations).
To do so, you may try to generate an HCP-lattice that is large enough that any rotation will still fully enclose your unit cube. For example:
r = 0.2 # example
n = int(np.ceil(2 / r))
df = hcp(n) * r - 1
Then rotate it (by any amount) and translate it (by up to 1 radius in any direction) as you wish for your research, and clip. The "periodic boundary conditions", as you call them, present a bit of extra challenge, as the clipping becomes trickier. First, clip any sphere whose center is outside your box. Then select spheres close enough to the boundaries, or even partition the regions of interest into overlapping regions along the walls of your cube, then check for collisions among the spheres (as per your periodic boundary conditions) that fall in each such region.
I want to implement affine transformation by not using library functions.
I have an image named "transformed" and I want to apply inverse transformation to obtain "img_org" image. Right now, I am using my own basic GetBilinearPixel function to set the intensity value. But, the image is not transforming properly.This is what I came up with. :
This is image("transformed.png"):
This is image("img_org.png"):
But My goal is to produce this image:
You can see the transformation matrix here:
pts1 = np.float32( [[693,349] , [605,331] , [445,59]] )
pts2 = np.float32 ( [[1379,895] , [1213,970] ,[684,428]] )
Mat = cv2.getAffineTransform(pts2,pts1)
B=Mat
code:
img_org=np.zeros(shape=(780,1050))
img_size=np.zeros(shape=(780,1050))
def GetBilinearPixel(imArr, posX, posY):
return imArr[posX][posY]
for i in range(1,img.shape[0]-1):
for j in range(1,img.shape[1]-1):
pos=np.array([[i],[j],[1]],np.float32)
#print pos
pos=np.matmul(B,pos)
r=int(pos[0][0])
c=int(pos[1][0])
#print r,c
if(c<=1024 and r<=768 and c>=0 and r>=0):
img_size[r][c]=img_size[r][c]+1
img_org[r][c] += GetBilinearPixel(img, i, j)
for i in range(0,img_org.shape[0]):
for j in range(0,img_org.shape[1]):
if(img_size[i][j]>0):
img_org[i][j] = img_org[i][j]/img_size[i][j]
Is my logic wrong? I know that i have applied very inefficient algorithm.
Is there any insight that i am missing?
Or can you give me any other algorithm which will work fine.
(Request) . I don't want to use warpAffine function.
So I vectorized the code and this method works---I can't find the exact issue with your implementation, but maybe this will shed some light (plus the speed is way faster).
The setup to vectorize is to create a linear (homogeneous) array containing every point in the image. We want an array that looks like
x0 x1 ... xN x0 x1 ... xN ..... x0 x1 ... xN
y0 y0 ... y0 y1 y1 ... y1 ..... yM yM ... yM
1 1 ... 1 1 1 ... 1 ..... 1 1 ... 1
So that every point (xi, yi, 1) is included. Then transforming is just a single matrix multiplication with your transformation matrix and this array.
To simplify matters (partially because your image naming conventions confused me), I'll say the original starting image is the "destination" or dst because we want to transform back to the "source" or src image. Bearing that in mind, creating this linear homogenous array could look something like this:
dst = cv2.imread('img.jpg', 0)
h, w = dst.shape[:2]
dst_y, dst_x = np.indices((h, w)) # similar to meshgrid/mgrid
dst_lin_homg_pts = np.stack((dst_x.ravel(), dst_y.ravel(), np.ones(dst_y.size)))
Then, to transform the points, just create the transformation matrix and multiply. I'll round the transformed pixel locations because I'm using them as an index and not bothering with interpolation:
src_pts = np.float32([[693, 349], [605, 331], [445, 59]])
dst_pts = np.float32([[1379, 895], [1213, 970], [684, 428]])
transf = cv2.getAffineTransform(dst_pts, src_pts)
src_lin_pts = np.round(transf.dot(dst_lin_homg_pts)).astype(int)
Now this transformation will send some pixels to negative indices, and if we index with those, it'll wrap around the image---probably not what we want to do. Of course in the OpenCV implementation, it just cuts those pixels off completely. But we can just shift all the transformed pixels so that all of the locations are positive and we don't cut off any (you can of course do whatever you want in this regard):
min_x, min_y = np.amin(src_lin_pts, axis=1)
src_lin_pts -= np.array([[min_x], [min_y]])
Then we'll need to create the source image src which the transform maps into. I'll create it with a gray background so we can see the extent of the black from the dst image.
trans_max_x, trans_max_y = np.amax(src_lin_pts, axis=1)
src = np.ones((trans_max_y+1, trans_max_x+1), dtype=np.uint8)*127
Now all we have to do is place some corresponding pixels from the destination image into the source image. Since I didn't cut off any of the pixels and there's the same number of pixels in both linear points array, I can just assign the transformed pixels the color they had in the original image.
src[src_lin_pts[1], src_lin_pts[0]] = dst.ravel()
Now, of course, this isn't interpolating on the image. But there's no built-ins in OpenCV for interpolation (there is backend C functions for other methods to use but not that you can access in Python AFAIK). But, you have the important parts---where the destination image gets mapped to, and the original image, so you can use any number of libraries to interpolate onto that grid. Or just implement a linear interpolation yourself as it's not too difficult. You'll probably want to un-round the warped pixel locations of course before then.
cv2.imshow('src', src)
cv2.waitKey()
Edit: Also this same method will work for warpPerspective too, although your resulting matrix multiplication will give a three-rowed (homogeneous) vector, and you'll need to divide the first two rows by the third row to set them back into Cartesian world. Other than that, everything else stays the same.
I'm trying to build a basic heatmap based on points. Each point has a heat radius, and therefore is represented by a circle.
Problem is that the circle needs to be converted in a list of pixels colored based on the distance from the circle's center.
Finding it hard to find an optimal solution for many points, what I have for now is something similar to this:
for pixels in pixels:
if (pixel.x - circle.x)**2 + (pixel.y - circle.y)**2 <= circle.radius:
pixel.set_color(circle.color)
Edit:
data I have:
pixel at the center of the circle
circle radius (integer)
Any tips?
Instead of doing it pixel-by-pixel, use a higher level interface with anti-aliasing, like the aggdraw module and its ellipse(xy, pen, brush) function.
Loop over the number of color steps you want (lets say, radius/2) and use 255/number_of_steps*current_step as the alpha value for the fill color.
For plotting it is usually recommended to use the matplotlib library (e.g. using imshow for heatmaps). Of course matplotlib also supports color gradients.
However, I don't really understand what you are trying to accomplish. If you just want to draw a bunch of colored circles then pretty much any graphics library will do (e.g. using the ellipse function in PIL).
It sounds like you want to color the pixel according to their distance from the center, but your own example code suggests that the color is constant?
If you are handling your pixels by yourself and your point is to increase performances, you can just focus on the square [x - radius; x + radius] * [y - radius; y + radius] since the points of your circle live here. That will save you a lot of useless iterations, if of course you CAN focus on this region (i.e. your pixels are not just an array without index per line and column).
You can even be sure that the pixels in the square [x - radius*sqrt(2)/2; x + radius*sqrt(2)/2] * [y - radius*sqrt(2)/2; y + radius*sqrt(2)/2] must be colored, with basic trigonometry (maximum square inside the circle).
So you could do:
import math
half_sqrt = math.sqrt(2) / 2
x_max = x + half_sqrt
y_max = y + half_sqrt
for (i in range(x, x + radius + 1):
for (j in range(y, y + radius + 1):
if (x <= x_max and y <= y_max):
colorize_4_parts(i, j)
else:
pixel = get_pixel(i, j)
if (pixel.x - circle.x)**2 + (pixel.y - circle.y)**2 <= circle.radius:
# Apply same colors as above, could be a function
colorize_4_parts(i, j)
def colorize_4_parts(i, j):
# Hoping you have access to such a function get_pixel !
pixel_top_right = get_pixel(i, j)
pixel_top_right.set_color(circle.color)
pixel_top_left = get_pixel(2 * x - i, j)
pixel_top_leftt.set_color(circle.color)
pixel_bot_right = get_pixel(i, 2 * y - j)
pixel_bot_right.set_color(circle.color)
pixel_bot_left = get_pixel(2 * x - i, 2 * y - j)
pixel_bot_leftt.set_color(circle.color)
This is optimized to reduce costly computations to the minimum.
EDIT: function updated to be more efficient again: I had forgotten that we had a double symetry horizontal and vertical, so we can compute only for the top right corner !
This is a very common operation, and here's how people do it...
summary: Represent the point density on a grid, smooth this using a 2D convolution if needed (this gives your points to circles), and plot this as a heatmap using matplotlib.
In more detail: First, make a 2D grid for your heatmap, and add your data points to the grid, incrementing by the cells by 1 when a data point lands in the cell. Second, make another grid to represents the shape you want to give each point (usually people use a cylinder or gaussian or something like this). Third, convolve these two together, using, say scipy.signal.convolve2d. Finally, use matplotlib's imshow function to plot the convolution, and this will be your heatmap.
If you can't use the tools suggested in the standard approach then you might find work-arounds, but it has advantages. For example, the convolution will deal well with cases when the circles overlap.
I have created a Python file to generate a Mandelbrot set image. The original maths code was not mine, so I do not understand it - I only heavily modified it to make it about 250x faster (Threads rule!).
Anyway, I was wondering how I could modify the maths part of the code to make it render one specific bit. Here is the maths part:
for y in xrange(size[1]):
coords = (uleft[0] + (x/size[0]) * (xwidth),uleft[1] - (y/size[1]) * (ywidth))
z = complex(coords[0],coords[1])
o = complex(0,0)
dotcolor = 0 # default, convergent
for trials in xrange(n):
if abs(o) <= 2.0:
o = o**2 + z
else:
dotcolor = trials
break # diverged
im.putpixel((x,y),dotcolor)
And the size definitions:
size1 = 500
size2 = 500
n=64
box=((-2,1.25),(0.5,-1.25))
plus = size[1]+size[0]
uleft = box[0]
lright = box[1]
xwidth = lright[0] - uleft[0]
ywidth = uleft[1] - lright[1]
what do I need to modify to make it render a certain section of the set?
The line:
box=((-2,1.25),(0.5,-1.25))
is the bit that defines the area of coordinate space that is being rendered, so you just need to change this line. First coordinate pair is the top-left of the area, the second is the bottom right.
To get a new coordinate from the image should be quite straightforward. You've got two coordinate systems, your "image" system 100x100 pixels in size, origin at (0,0). And your "complex" plane coordinate system defined by "box". For X:
X_complex=X_complex_origin+(X_image/X_image_width)*X_complex_width
The key in understanding how to do this is to understand what the coords = line is doing:
coords = (uleft[0] + (x/size[0]) * (xwidth),uleft[1] - (y/size[1]) * (ywidth))
Effectively, the x and y values you are looping through which correspond to the coordinates of the on-screen pixel are being translated to the corresponding point on the complex plane being looked at. This means that (0,0) screen coordinate will translate to the upper left region being looked at (-2,1.25), and (1,0) will be the same, but moved 1/500 of the distance (assuming a 500 pixel width window) between the -2 and 0.5 x-coordinate.
That's exactly what that line is doing - I'll expand just the X-coordinate bit with more illustrative variable names to indicate this:
mandel_x = mandel_start_x + (screen_x / screen_width) * mandel_width
(The mandel_ variables refer to the coordinates on the complex plane, the screen_ variables refer to the on-screen coordinates of the pixel being plotted.)
If you want then to take a region of the screen to zoom into, you want to do exactly the same: take the screen coordinates of the upper-left and lower-right region, translate them to the complex-plane coordinates, and make those the new uleft and lright variables. ie to zoom in on the box delimited by on-screen coordinates (x1,y1)..(x2,y2), use:
new_uleft = (uleft[0] + (x1/size[0]) * (xwidth), uleft[1] - (y1/size[1]) * (ywidth))
new_lright = (uleft[0] + (x2/size[0]) * (xwidth), uleft[1] - (y2/size[1]) * (ywidth))
(Obviously you'll need to recalculate the size, xwidth, ywidth and other dependent variables based on the new coordinates)
In case you're curious, the maths behind the mandelbrot set isn't that complicated (just complex).
All it is doing is taking a particular coordinate, treating it as a complex number, and then repeatedly squaring it and adding the original number to it.
For some numbers, doing this will cause the result diverge, constantly growing towards infinity as you repeat the process. For others, it will always stay below a certain level (eg. obviously (0.0, 0.0) never gets any bigger under this process. The mandelbrot set (the black region) is those coordinates which don't diverge. Its been shown that if any number gets above the square root of 5, it will diverge - your code is just using 2.0 as its approximation to sqrt(5) (~2.236), but this won't make much noticeable difference.
Usually the regions that diverge get plotted with the number of iterations of the process that it takes for them to exceed this value (the trials variable in your code) which is what produces the coloured regions.