I have two 2D numpy arrays (of the same dimensions) that I am plotting using matplotlib. The first array I've plotted as a color map in gray-scale. The second one represents an aperture, but it is an irregular shape (some of the pixels get outlined, and it is a set of horizontal and vertical lines that form the outline). I am not sure how to ask it to plot this second array. The array is composed of three numbers (0, 1, and 3), and I only need the pixels of one value (3) to be outlined, but I need the outline to encompass the region of these pixels, not the pixels individually. I need the interior of all the pixels to remain transparent so that I can see the gray-scale color map through it.
Does anyone know how to accomplish this?
That is an interesting question, if I understood it correctly. In order to make sure what you mean, you would like to draw a line with some color around all contiguous areas where the pixel value is 3.
I do not think there is a ready-made function for that, but let's not let that stop us. We will need to create our own function.
We can start by creating a boolean map of the area which needs to be outlined:
import numpy as np
import matplotlib.pyplot as plt
# our image with the numbers 1-3 is in array maskimg
# create a boolean image map which has trues only where maskimg[x,y] == 3
mapimg = (maskimg == 3)
# a vertical line segment is needed, when the pixels next to each other horizontally
# belong to diffferent groups (one is part of the mask, the other isn't)
# after this ver_seg has two arrays, one for row coordinates, the other for column coordinates
ver_seg = np.where(mapimg[:,1:] != mapimg[:,:-1])
# the same is repeated for horizontal segments
hor_seg = np.where(mapimg[1:,:] != mapimg[:-1,:])
# if we have a horizontal segment at 7,2, it means that it must be drawn between pixels
# (2,7) and (2,8), i.e. from (2,8)..(3,8)
# in order to draw a discountinuous line, we add Nones in between segments
l = []
for p in zip(*hor_seg):
l.append((p[1], p[0]+1))
l.append((p[1]+1, p[0]+1))
l.append((np.nan,np.nan))
# and the same for vertical segments
for p in zip(*ver_seg):
l.append((p[1]+1, p[0]))
l.append((p[1]+1, p[0]+1))
l.append((np.nan, np.nan))
# now we transform the list into a numpy array of Nx2 shape
segments = np.array(l)
# now we need to know something about the image which is shown
# at this point let's assume it has extents (x0, y0)..(x1,y1) on the axis
# drawn with origin='lower'
# with this information we can rescale our points
segments[:,0] = x0 + (x1-x0) * segments[:,0] / mapimg.shape[1]
segments[:,1] = y0 + (y1-y0) * segments[:,1] / mapimg.shape[0]
# and now there isn't anything else to do than plot it
plt.plot(segments[:,0], segments[:,1], color=(1,0,0,.5), linewidth=3)
Let us test this by generating some data and showing it:
image = np.cumsum(np.random.random((20,20))-.5, axis=1)
maskimg = np.zeros(image.shape, dtype='int')
maskimg[image > 0] = 3
x0 = -1.5
x1 = 1.5
y0 = 2.3
y1 = 3.8
plt.figure()
plt.imshow(maskimg, origin='lower', extent=[x0,x1,y0,y1], cmap=plt.cm.gray, interpolation='nearest')
plt.axis('tight')
After that we run the procedure on the top, and get:
The code can be made much denser, if needed, but now comments take a lot of space. With large images it might be wise to optimize the image segment creation by finding continuous paths. That will reduce the number of points to plot by a factor of up to three. However, doing that requires a bit different code, which is not as clear as this one. (If there will appear comments asking for that and an appropriate number of upvotes, I'll add it :)
Related
Assume that you have a digital image. You crop that image into 6 pieces vertically. After, you shuffled these pieces and rearrenge these pieces(sub images) randomly. So, you obtain a image like below.
I want to reconstruct the original image like a puzzle. First, we need to calcualte right points on X axis which sub images merged. For example, in the 100th pixel(X-axis), two subsample merged. I have to parse these points to regain sub-images. How can I do that? The image splitted vertically, Is sobel filter on X axis help me to find sharp transition? Any suggestion?
This is an interesting directed graph problem. In each slice the last column has a distance error to the first column. After you build the graph you just need to find the head, and walk along the minimum cost path.
I've sketched a script you coult start from start from:
import cv2
import numpy as np
import matplotlib.pyplot as plt
cut_thr = 0.19 # magic number , but kind of arbitrary as if you add a cut, you just make your graph bigger
im = cv2.imread(r'example.png').astype(np.float32)/255 #read image
im = cv2.cvtColor(im,cv2.COLOR_BGR2RGB)
dx=np.abs(np.diff(im,axis=1)) #x difference
dx = np.max(dx,axis=2) #max on all color channels
dx=np.median(dx,axis=0) #max on y axis
plt.plot(dx)
cuts = np.r_[0,np.where(dx>cut_thr)[0]+1,im.shape[1]] #inclusive borders
cuts = [im[:,cuts[i]:cuts[i+1]] for i in range(len(cuts)-1)]
n = len(cuts)
fig,ax = plt.subplots(1,n)
for a,c in zip(ax,cuts):
a.imshow(c, aspect='auto')
a.axis('off')
d = np.ones((n,n))*np.nan # directed connectivity
for y in range(n):
for x in range(y+1,n):
d[y][x]=np.median(np.abs(cuts[y][:,-1]-cuts[x][:,0]))
d[x][y]=np.median(np.abs(cuts[x][:,-1]-cuts[y][:,0]))
src = np.arange(n)
dst=np.nanargmin(d,axis=1) # the dest of source is the one with the lowest error
indx=np.where(d==np.nanmin(d))[0][-1] #head, where to begin
im = cuts[indx]
for i in range(n-1):
indx=dst[src[indx]]
im = np.concatenate([im,cuts[indx]],axis=1)
plt.figure()
plt.imshow(im, aspect='equal')
I need to write a python function or class with the following Input/Output
Input :
The position of the X-rays source (still not sure why it's needed)
The position of the board (still not sure why it's needed)
A three dimensional CT-Scan
Output :
A 2D X-ray Scan (simulate an X-Ray Scan which is a scan that goes through the whole body)
A few important remarks to what I'm trying to achieve:
You don’t need additional information from the real world or any advanced knowledge.
You can add any input parameter that you see fit.
If your method produces artifacts, you are excepted to fix them.
Please explain every step of your method.
What I've done until now: (.py file added)
I've read the .dicom files, which are located in "Case2" folder.
These .dicom files can be downloaded from my Google Drive:
https://drive.google.com/file/d/1lHoMJgj_8Dt62JaR2mMlK9FDnfkesH5F/view?usp=sharing
I've sorted the files by their position.
Finally, I've created a 3D array, and added all the images to that array in order to plot the results (you can see them in the added image) - which are slice of the CT Scans. (reference: https://pydicom.github.io/pydicom/stable/auto_examples/image_processing/reslice.html#sphx-glr-auto-examples-image-processing-reslice-py)
Here's the full code:
import pydicom as dicom
import os
import matplotlib.pyplot as plt
import sys
import glob
import numpy as np
path = "./Case2"
ct_images = os.listdir(path)
slices = [dicom.read_file(path + '/' + s, force=True) for s in ct_images]
slices[0].ImagePositionPatient[2]
slices = sorted(slices, key = lambda x: x.ImagePositionPatient[2])
#print(slices)
# Read a dicom file with a ctx manager
with dicom.dcmread(path + '/' + ct_images[0]) as ds:
# plt.imshow(ds.pixel_array, cmap=plt.cm.bone)
print(ds)
#plt.show()
fig = plt.figure()
for num, each_slice in enumerate(slices[:12]):
y= fig.add_subplot(3,4,num+1)
#print(each_slice)
y.imshow(each_slice.pixel_array)
plt.show()
for i in range(len(ct_images)):
with dicom.dcmread(path + '/' + ct_images[i], force=True) as ds:
plt.imshow(ds.pixel_array, cmap=plt.cm.bone)
plt.show()
# pixel aspects, assuming all slices are the same
ps = slices[0].PixelSpacing
ss = slices[0].SliceThickness
ax_aspect = ps[1]/ps[0]
sag_aspect = ps[1]/ss
cor_aspect = ss/ps[0]
# create 3D array
img_shape = list(slices[0].pixel_array.shape)
img_shape.append(len(slices))
img3d = np.zeros(img_shape)
# fill 3D array with the images from the files
for i, s in enumerate(slices):
img2d = s.pixel_array
img3d[:, :, i] = img2d
# plot 3 orthogonal slices
a1 = plt.subplot(2, 2, 1)
plt.imshow(img3d[:, :, img_shape[2]//2])
a1.set_aspect(ax_aspect)
a2 = plt.subplot(2, 2, 2)
plt.imshow(img3d[:, img_shape[1]//2, :])
a2.set_aspect(sag_aspect)
a3 = plt.subplot(2, 2, 3)
plt.imshow(img3d[img_shape[0]//2, :, :].T)
a3.set_aspect(cor_aspect)
plt.show()
The result isn't what I wanted because:
These are slice of the CT scans. I need to simulate an X-Ray Scan which is a scan that goes through the whole body.
Would love your help to simulate an X-Ray scan that goes through the body.
I've read that it could be done in the following way: "A normal 2D X-ray image is a sum projection through the volume. Send parallel rays through the volume and add up the densities." Which I'm not sure how it's accomplished in code.
References that may help: https://pydicom.github.io/pydicom/stable/index.html
EDIT: as further answers noted, this solution yields a parallel projection, not a perspective projection.
From what I understand of the definition of "A normal 2D X-ray image", this can be done by summing each density for each pixel, for each slice of a projection in a given direction.
With your 3D volume, this means performing a sum over a given axis, which can be done with ndarray.sum(axis) in numpy.
# plot 3 orthogonal slices
a1 = plt.subplot(2, 2, 1)
plt.imshow(img3d.sum(2), cmap=plt.cm.bone)
a1.set_aspect(ax_aspect)
a2 = plt.subplot(2, 2, 2)
plt.imshow(img3d.sum(1), cmap=plt.cm.bone)
a2.set_aspect(sag_aspect)
a3 = plt.subplot(2, 2, 3)
plt.imshow(img3d.sum(0).T, cmap=plt.cm.bone)
a3.set_aspect(cor_aspect)
plt.show()
This yields the following result:
Which, to me, looks like a X-ray image.
EDIT : the result is a bit too "bright", so you may want to apply gamma correction. With matplotlib, import matplotlib.colors as colors and add a colors.PowerNorm(gamma_value) as the norm parameter in plt.imshow:
plt.imshow(img3d.sum(0).T, norm=colors.PowerNorm(gamma=3), cmap=plt.cm.bone)
Result:
The way I understand the task you are expected to write a ray-tracer that follows the X-rays from the source (that's why you need its position) to the projection plane (That's why you need its position).
Sum up the values as you go and do a mapping to the allowed grey-values in the end.
Take a look at line drawing algorithms to see how you can do this.
It is really no black magic, I have done this kind of stuff more than 30 years ago. Damn, I'm old...
What you want is a perspective projection instead of a parallel projection. In order to obtain this, you need to know which values to sum for each point on the projection plane. There are multiple considerations to keep in mind:
We are talking about voxels, so you need to a method to determine whether a certain point in space belongs to a certain voxel in your volume.
A line between two points is straight, but because voxels are a discrete representation of space different methods of determining the above can lead to different (mostly minor) results. This difference will ultimately also lead to slightly different images depending on the alogrithms used. This is expected.
Let's say you have a CT scan volume comprising of 256 512x512 pixel slices. This gives you a volume of 512x512x256 voxels. For each of these voxels you need to know what their positions in x,y,z coordinates are. You can do this as follows:
- Use the ImagePositionPatient attribute to find out the x,y,z coordinate of the upper left hand corner pixel in mm for a given slice.
- Use the PixelSpacing attribute to calculate the x,y,z coordinates of the other pixels in your slice. Repeat for all slices
edit: i just found a counterexample against below method, the rest is still helpful. will update
Now to find out for a given point (Xa, Ya, Za) what voxel values need to be summed if the source is at (Xb, Yb, Zb):
Find the voxel that belongs to (Xa,Ya, Za). Keep pixel/voxel data.
Calculate (you can do this with NumPy) the distance between voxel(Xa, Ya, Za) and (Xb, Yb, Zb). There is an optimalization possible here :)
For all directly surrounding voxels (that will be a number of 3x3x3-1 voxels) also calculate this distance. Can also be optimized :)
Take the voxel with the shortest distance as the starting point for a next iteration of the above. Add pixel/voxel data.
Repeat until out of bounds of you CT volume.
In order to obtain a projection repeat these steps for all points on your projection plane and visualize the result. Good luck with your assignment! :)
I have a square 2D array data that I would like to add to a larger 2D array frame at some given set of non-integer coordinates coords. The idea is that data will be interpolated onto frame with it's center at the new coordinates.
Some toy data:
# A gaussian to add to the frame
x, y = np.meshgrid(np.linspace(-1,1,10), np.linspace(-1,1,10))
data = 50*np.exp(-np.sqrt(x**2+y**2)**2)
# The frame to add the gaussian to
frame = np.random.normal(size=(100,50))
# The desired (x,y) location of the gaussian center on the new frame
coords = 23.4, 22.6
Here's the idea. I want to add this:
to this:
to get this:
If the coordinates were integers (indexes), of course I could simply add them like this:
frame[23:33,22:32] += data
But I want to be able to specify non-integer coordinates so that data is regridded and added to frame.
I've looked into PIL.Image methods but my use case is just for 2D data, not images. Is there a way to do this with just scipy? Can this be done with interp2d or a similar function? Any guidance would be greatly appreciated!
Scipy's shift function from scipy.ndimage.interpolation is what you are looking for, as long as the grid spacings between data and frame overlap. If not, look to the other answer. The shift function can take floating point numbers as input and will do a spline interpolation. First, I put the data into an array as large as frame, then shift it, and then add it. Make sure to reverse the coordinate list, as x is the rightmost dimension in numpy arrays. One of the nice features of shift is that it sets to zero those values that go out of bounds.
import numpy as np
import matplotlib.pyplot as plt
from scipy.ndimage.interpolation import shift
# A gaussian to add to the frame.
x, y = np.meshgrid(np.linspace(-1,1,10), np.linspace(-1,1,10))
data = 50*np.exp(-np.sqrt(x**2+y**2)**2)
# The frame to add the gaussian to
frame = np.random.normal(size=(100,50))
x_frame = np.arange(50)
y_frame = np.arange(100)
# The desired (x,y) location of the gaussian center on the new frame.
coords = np.array([23.4, 22.6])
# First, create a frame as large as the frame.
data_large = np.zeros(frame.shape)
data_large[:data.shape[0], :data.shape[1]] = data[:,:]
# Subtract half the distance as the bottom left is at 0,0 instead of the center.
# The shift of 4.5 is because data is 10 points wide.
# Reverse the coords array as x is the last coordinate.
coords_shift = -4.5
data_large = shift(data_large, coords[::-1] + coords_shift)
frame += data_large
# Plot the result and add lines to indicate to coordinates
plt.figure()
plt.pcolormesh(x_frame, y_frame, frame, cmap=plt.cm.jet)
plt.axhline(coords[1], color='w')
plt.axvline(coords[0], color='w')
plt.colorbar()
plt.gca().invert_yaxis()
plt.show()
The script gives you the following figure, which has the desired coordinates indicated with white dotted lines.
One possible solution is to use scipy.interpolate.RectBivariateSpline. In the code below, x_0 and y_0 are the coordinates of a feature from data (i.e., the position of the center of the Gaussian in your example) that need to be mapped to the coordinates given by coords. There are a couple of advantages to this approach:
If you need to "place" the same object into multiple locations in the output frame, the spline needs to be computed only once (but evaluated multiple times).
In case you actually need to compute integrated flux of the model over a pixel, you can use the integral method of scipy.interpolate.RectBivariateSpline.
Resample using spline interpolation:
from scipy.interpolate import RectBivariateSpline
x = np.arange(data.shape[1], dtype=np.float)
y = np.arange(data.shape[0], dtype=np.float)
kx = 3; ky = 3; # spline degree
spline = RectBivariateSpline(
x, y, data.T, kx=kx, ky=ky, s=0
)
# Define coordinates of a feature in the data array.
# This can be the center of the Gaussian:
x_0 = (data.shape[1] - 1.0) / 2.0
y_0 = (data.shape[0] - 1.0) / 2.0
# create output grid, shifted as necessary:
yg, xg = np.indices(frame.shape, dtype=np.float64)
xg += x_0 - coords[0] # see below how to account for pixel scale change
yg += y_0 - coords[1] # see below how to account for pixel scale change
# resample and fill extrapolated points with 0:
resampled_data = spline.ev(xg, yg)
extrapol = (((xg < -0.5) | (xg >= data.shape[1] - 0.5)) |
((yg < -0.5) | (yg >= data.shape[0] - 0.5)))
resampled_data[extrapol] = 0
Now plot the frame and resampled data:
plt.figure(figsize=(14, 14));
plt.imshow(frame+resampled_data, cmap=plt.cm.jet,
origin='upper', interpolation='none', aspect='equal')
plt.show()
If you also want to allow for scale changes, then replace code for computing xg and yg above with:
coords = 20, 80 # change coords to easily identifiable (in plot) values
zoom_x = 2 # example scale change along X axis
zoom_y = 3 # example scale change along Y axis
yg, xg = np.indices(frame.shape, dtype=np.float64)
xg = (xg - coords[0]) / zoom_x + x_0
yg = (yg - coords[1]) / zoom_y + y_0
Most likely this is what you actually want based on your example. Specifically, the coordinates of pixels in data are "spaced" by 0.222(2) distance units. Therefore it actually seems that for your particular example (whether accidental or intentional), you have a zoom factor of 0.222(2). In that case your data image would shrink to almost 2 pixels in the output frame.
Comparison to #Chiel answer
In the image below, I compare the results from my method (left), #Chiel's method (center) and difference (right panel):
Fundamentally, the two methods are quite similar and possibly even use the same algorithm (I did not look at the code for shift but based on the description - it also uses splines). From comparison image it is visible that the biggest differences are at the edges and, for unknown to me reasons, shift seems to truncate the shifted image slightly too soon.
I think the biggest difference is that my method allows for pixel scale changes and it also allows re-use of the same interpolator to place the original image at different locations in the output frame. #Chiel's method is somewhat simpler but (what I did not like about it is that) it requires creation of a larger array (data_large) into which the original image is placed in the corner.
While the other answers have gone into detail, but here's my lazy solution:
xc,yc = 23.4, 22.6
x, y = np.meshgrid(np.linspace(-1,1,10)-xc%1, np.linspace(-1,1,10)-yc%1)
data = 50*np.exp(-np.sqrt(x**2+y**2)**2)
frame = np.random.normal(size=(100,50))
frame[23:33,22:32] += data
And it's the way you liked it. As you mentioned, the coordinates of both are the same, so the origin of data is somewhere between the indices. Now just simply shift it by the amount you want it to be off a grid point (remainder to one) in the second line and you're good to go (you might need to flip the sign, but I think this is correct).
I have a coarse skymap made up of 128 points, of which I would like to make a smooth healpix map (see attached Figure, LHS). Figures referenced in the text:
I load my data, then make new longitude and latitude arrays of the appropriate pixel length for the final map (with e.g. nside=32).
My input data are:
lats = pi/2 + ths # theta from 0, pi, size 8
lons = phs # phi from 0, 2pi, size 16
data = sky_data[0] # shape (8,16)
New lon/lat array size based on number of pixels from nside:
nside = 32
pixIdx = hp.nside2npix(nside) # number of pixels I can get from this nside
pixIdx = np.arange(pixIdx) # pixel index numbers
I then find the new data values for those pixels by interpolation, and then convert back from angles to pixels.
# new lon/lat
new_lats = hp.pix2ang(nside, pixIdx)[0] # thetas I need to populate with interpolated theta values
new_lons = hp.pix2ang(nside, pixIdx)[1] # phis, same
# interpolation
lut = RectSphereBivariateSpline(lats, lons, data, pole_values=4e-14)
data_interp = lut.ev(new_lats.ravel(), new_lons.ravel()) #interpolate the data
pix = hp.ang2pix(nside, new_lats, new_lons) # convert latitudes and longitudes back to pixels
Then, I construct a healpy map with the interpolated values:
healpix_map = np.zeros(hp.nside2npix(nside), dtype=np.double) # create empty map
healpix_map[pix] = data_interp # assign pixels to new interpolated values
testmap = hp.mollview(healpix_map)
The result of the map is the upper RHS of the attached Figure.
(Forgive the use of jet -- viridis doesn't have a "white" zero, so using that colormap adds a blue background.)
The map doesn't look right: you can see from the coarse map in the Figure that there should be a "hotspot" on the lower RHS, but here it appears in the upper left.
As a sanity check, I used matplotlib to make a scatter plot of the interpolated points in a mollview projection, Figure 2, where I removed the edges of the markers to make it look like a map ;)
ax = plt.subplot(111, projection='astro mollweide')
ax.grid()
colors = data_interp
sky=plt.scatter(new_lons, new_lats-pi/2, c = colors, edgecolors='none', cmap ='jet')
plt.colorbar(sky, orientation = 'horizontal')
You can see that this map, lower RHS of attached Figure, produces exactly what I expect! So the coordinates are ok, and I am completely confused.
Has anyone encountered this before? What can I do? I'd like to use the healpy functions on this and future maps, so just using matplotlib isn't an option.
Thanks!
I figured it out -- I had to add pi/2 to my thetas for the interpolation to work, so in the end need to apply the following transformation for the image to render correctly:
newnew_lats = pi - new_lats
newnew_lons = pi + new_lons
There still seems to be a bit of an issue with the interpolation, although the seem is not so visible now. I may try a different one to compare.
I'm no expert in healpix (actually I've never used it before - I'm a particle physicist), but as far as I can tell it's just a matter of conventions: in a Mollweide projection, healpy places the north pole (positive latitude) at the bottom of the map, for some reason. I'm not sure why it would do that, or whether this is intentional behavior, but it seems pretty clear that's what is happening. If I mask out everything below the equator, i.e. keep only the positive-latitude points
mask = new_lats - pi/2 > 0
pix = hp.ang2pix(nside, new_lats[mask], new_lons[mask])
healpix_map = np.zeros(hp.nside2npix(nside), dtype=np.double)
healpix_map[pix] = data_interp[mask]
testmap = hp.mollview(healpix_map)
it comes up with a plot with no data above the center line:
At least it's easy enough to fix. mollview admits a rot parameter that will effectively rotate the sphere around the viewing axis before projecting it, and a flip parameter which can be set to 'astro' (default) or 'geo' to set whether east is shown at the left or right. A little experimentation shows that you get the coordinate system you want with
hp.mollview(healpix_map, rot=(180, 0, 180), flip='geo')
In the tuple, the first two elements are longitude and latitude of the point to set in the center of the plot, and the third element is the rotation. All are in degrees. With no mask it gives this:
which I believe is just what you're looking for.
I'm trying to get python to return, as close as possible, the center of the most obvious clustering in an image like the one below:
In my previous question I asked how to get the global maximum and the local maximums of a 2d array, and the answers given worked perfectly. The issue is that the center estimation I can get by averaging the global maximum obtained with different bin sizes is always slightly off than the one I would set by eye, because I'm only accounting for the biggest bin instead of a group of biggest bins (like one does by eye).
I tried adapting the answer to this question to my problem, but it turns out my image is too noisy for that algorithm to work. Here's my code implementing that answer:
import numpy as np
from scipy.ndimage.filters import maximum_filter
from scipy.ndimage.morphology import generate_binary_structure, binary_erosion
import matplotlib.pyplot as pp
from os import getcwd
from os.path import join, realpath, dirname
# Save path to dir where this code exists.
mypath = realpath(join(getcwd(), dirname(__file__)))
myfile = 'data_file.dat'
x, y = np.loadtxt(join(mypath,myfile), usecols=(1, 2), unpack=True)
xmin, xmax = min(x), max(x)
ymin, ymax = min(y), max(y)
rang = [[xmin, xmax], [ymin, ymax]]
paws = []
for d_b in range(25, 110, 25):
# Number of bins in x,y given the bin width 'd_b'
binsxy = [int((xmax - xmin) / d_b), int((ymax - ymin) / d_b)]
H, xedges, yedges = np.histogram2d(x, y, range=rang, bins=binsxy)
paws.append(H)
def detect_peaks(image):
"""
Takes an image and detect the peaks usingthe local maximum filter.
Returns a boolean mask of the peaks (i.e. 1 when
the pixel's value is the neighborhood maximum, 0 otherwise)
"""
# define an 8-connected neighborhood
neighborhood = generate_binary_structure(2,2)
#apply the local maximum filter; all pixel of maximal value
#in their neighborhood are set to 1
local_max = maximum_filter(image, footprint=neighborhood)==image
#local_max is a mask that contains the peaks we are
#looking for, but also the background.
#In order to isolate the peaks we must remove the background from the mask.
#we create the mask of the background
background = (image==0)
#a little technicality: we must erode the background in order to
#successfully subtract it form local_max, otherwise a line will
#appear along the background border (artifact of the local maximum filter)
eroded_background = binary_erosion(background, structure=neighborhood, border_value=1)
#we obtain the final mask, containing only peaks,
#by removing the background from the local_max mask
detected_peaks = local_max - eroded_background
return detected_peaks
#applying the detection and plotting results
for i, paw in enumerate(paws):
detected_peaks = detect_peaks(paw)
pp.subplot(4,2,(2*i+1))
pp.imshow(paw)
pp.subplot(4,2,(2*i+2) )
pp.imshow(detected_peaks)
pp.show()
and here's the result of that (varying the bin size):
Clearly my background is too noisy for that algorithm to work, so the question is: how can I make that algorithm less sensitive? If an alternative solution exists then please let me know.
EDIT
Following Bi Rico advise I attempted smoothing my 2d array before passing it on to the local maximum finder, like so:
H, xedges, yedges = np.histogram2d(x, y, range=rang, bins=binsxy)
H1 = gaussian_filter(H, 2, mode='nearest')
paws.append(H1)
These were the results with a sigma of 2, 4 and 8:
EDIT 2
A mode ='constant' seems to work much better than nearest. It converges to the right center with a sigma=2 for the largest bin size:
So, how do I get the coordinates of the maximum that shows in the last image?
Answering the last part of your question, always you have points in an image, you can find their coordinates by searching, in some order, the local maximums of the image. In case your data is not a point source, you can apply a mask to each peak in order to avoid the peak neighborhood from being a maximum while performing a future search. I propose the following code:
import matplotlib.image as mpimg
import matplotlib.pyplot as plt
import numpy as np
import copy
def get_std(image):
return np.std(image)
def get_max(image,sigma,alpha=20,size=10):
i_out = []
j_out = []
image_temp = copy.deepcopy(image)
while True:
k = np.argmax(image_temp)
j,i = np.unravel_index(k, image_temp.shape)
if(image_temp[j,i] >= alpha*sigma):
i_out.append(i)
j_out.append(j)
x = np.arange(i-size, i+size)
y = np.arange(j-size, j+size)
xv,yv = np.meshgrid(x,y)
image_temp[yv.clip(0,image_temp.shape[0]-1),
xv.clip(0,image_temp.shape[1]-1) ] = 0
print xv
else:
break
return i_out,j_out
#reading the image
image = mpimg.imread('ggd4.jpg')
#computing the standard deviation of the image
sigma = get_std(image)
#getting the peaks
i,j = get_max(image[:,:,0],sigma, alpha=10, size=10)
#let's see the results
plt.imshow(image, origin='lower')
plt.plot(i,j,'ro', markersize=10, alpha=0.5)
plt.show()
The image ggd4 for the test can be downloaded from:
http://www.ipac.caltech.edu/2mass/gallery/spr99/ggd4.jpg
The first part is to get some information about the noise in the image. I did it by computing the standard deviation of the full image (actually is better to select an small rectangle without signal). This is telling us how much noise is present in the image.
The idea to get the peaks is to ask for successive maximums, which are above of certain threshold (let's say, 3, 4, 5, 10, or 20 times the noise). This is what the function get_max is actually doing. It performs the search of maximums until one of them is below the threshold imposed by the noise. In order to avoid finding the same maximum many times it is necessary to remove the peaks from the image. In the general way, the shape of the mask to do so depends strongly on the problem that one want to solve. for the case of stars, it should be good to remove the star by using a Gaussian function, or something similar. I have chosen for simplicity a square function, and the size of the function (in pixels) is the variable "size".
I think that from this example, anybody can improve the code by adding more general things.
EDIT:
The original image looks like:
While the image after identifying the luminous points looks like this:
Too much of a n00b on Stack Overflow to comment on Alejandro's answer elsewhere here. I would refine his code a bit to use a preallocated numpy array for output:
def get_max(image,sigma,alpha=3,size=10):
from copy import deepcopy
import numpy as np
# preallocate a lot of peak storage
k_arr = np.zeros((10000,2))
image_temp = deepcopy(image)
peak_ct=0
while True:
k = np.argmax(image_temp)
j,i = np.unravel_index(k, image_temp.shape)
if(image_temp[j,i] >= alpha*sigma):
k_arr[peak_ct]=[j,i]
# this is the part that masks already-found peaks.
x = np.arange(i-size, i+size)
y = np.arange(j-size, j+size)
xv,yv = np.meshgrid(x,y)
# the clip here handles edge cases where the peak is near the
# image edge
image_temp[yv.clip(0,image_temp.shape[0]-1),
xv.clip(0,image_temp.shape[1]-1) ] = 0
peak_ct+=1
else:
break
# trim the output for only what we've actually found
return k_arr[:peak_ct]
In profiling this and Alejandro's code using his example image, this code about 33% faster (0.03 sec for Alejandro's code, 0.02 sec for mine.) I expect on images with larger numbers of peaks, it would be even faster - appending the output to a list will get slower and slower for more peaks.
I think the first step needed here is to express the values in H in terms of the standard deviation of the field:
import numpy as np
H = H / np.std(H)
Now you can put a threshold on the values of this H. If the noise is assumed to be Gaussian, picking a threshold of 3 you can be quite sure (99.7%) that this pixel can be associated with a real peak and not noise. See here.
Now the further selection can start. It is not exactly clear to me what exactly you want to find. Do you want the exact location of peak values? Or do you want one location for a cluster of peaks which is in the middle of this cluster?
Anyway, starting from this point with all pixel values expressed in standard deviations of the field, you should be able to get what you want. If you want to find clusters you could perform a nearest neighbour search on the >3-sigma gridpoints and put a threshold on the distance. I.e. only connect them when they are close enough to each other. If several gridpoints are connected you can define this as a group/cluster and calculate some (sigma-weighted?) center of the cluster.
Hope my first contribution on Stackoverflow is useful for you!
The way I would do it:
1) normalize H between 0 and 1.
2) pick a threshold value, as tcaswell suggests. It could be between .9 and .99 for example
3) use masked arrays to keep only the x,y coordinates with H above threshold:
import numpy.ma as ma
x_masked=ma.masked_array(x, mask= H < thresold)
y_masked=ma.masked_array(y, mask= H < thresold)
4) now you can weight-average on the masked coordinates, with weight something like (H-threshold)^2, or any other power greater or equal to one, depending on your taste/tests.
Comment:
1) This is not robust with respect to the type of peaks you have, since you may have to adapt the thresold. This is the minor problem;
2) This DOES NOT work with two peaks as it is, and will give wrong results if the 2nd peak is above threshold.
Nonetheless, it will always give you an answer without crashing (with pros and cons of the thing..)
I'm adding this answer because it's the solution I ended up using. It's a combination of Bi Rico's comment here (May 30 at 18:54) and the answer given in this question: Find peak of 2d histogram.
As it turns out using the peak detection algorithm from this question Peak detection in a 2D array only complicates matters. After applying the Gaussian filter to the image all that needs to be done is to ask for the maximum bin (as Bi Rico pointed out) and then obtain the maximum in coordinates.
So instead of using the detect-peaks function as I did above, I simply add the following code after the Gaussian 2D histogram is obtained:
# Get 2D histogram.
H, xedges, yedges = np.histogram2d(x, y, range=rang, bins=binsxy)
# Get Gaussian filtered 2D histogram.
H1 = gaussian_filter(H, 2, mode='nearest')
# Get center of maximum in bin coordinates.
x_cent_bin, y_cent_bin = np.unravel_index(H1.argmax(), H1.shape)
# Get center in x,y coordinates.
x_cent_coor , y_cent_coord = np.average(xedges[x_cent_bin:x_cent_bin + 2]), np.average(yedges[y_cent_g:y_cent_g + 2])