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Speed up computation for Distance Transform on Image in Python

I would like to find the find the distance transform of a binary image in the fastest way possible without using the scipy package distance_trnsform_edt(). The image is 256 by 256. The reason I don't want to use scipy is because using it is difficult in tensorflow. Evry time I want to use this package I need to start a new session and this takes a lot of time. So I would like to make a custom function that only utilizes numpy.
My approach is as follows: Find the coordinated for all the ones and all the zeros in the image. Find the euclidian distance between each of the zero pixels (a) and the one pixels (b) and then the value at each (a) position is the minimum distance to a (b) pixel. I do this for each 0 pixel. The resultant image has the same dimensions as the original binary map. My attempt at doing this is shown below.
I tried to do this as fast as possible using no loops and only vectorization. But my function still can't work as fast as the scipy package can. When I timed the code it looks like the assignment to the variable "a" is taking the longest time. But I do not know if there is a way to speed this up.
If anyone has any other suggestions for different algorithms to solve this problem of distance transforms or can direct me to other implementations in python, it would be very appreciated.
def get_dst_transform_img(og): #og is a numpy array of original image
ones_loc = np.where(og == 1)
ones = np.asarray(ones_loc).T # coords of all ones in og
zeros_loc = np.where(og == 0)
zeros = np.asarray(zeros_loc).T # coords of all zeros in og
a = -2 * np.dot(zeros, ones.T)
b = np.sum(np.square(ones), axis=1)
c = np.sum(np.square(zeros), axis=1)[:,np.newaxis]
dists = a + b + c
dists = np.sqrt(dists.min(axis=1)) # min dist of each zero pixel to one pixel
x = og.shape[0]
y = og.shape[1]
dist_transform = np.zeros((x,y))
dist_transform[zeros[:,0], zeros[:,1]] = dists
plt.figure()
plt.imshow(dist_transform)

The implementation in the OP is a brute-force approach to the distance transform. This algorithm is O(n2), as it computes the distance from each background pixel to each foreground pixel. Furthermore, because of the way it is vectorized, it requires a lot of memory. On my computer it couldn't compute the distance transform of a 256x256 image without thrashing. Many other algorithms are described in the literature, below I'll discuss two O(n) algorithms.
Note: Typically, the distance transform is computed for object pixels (value 1) to the nearest background pixel (value 0). The code in the OP does the reverse, and so the code I've pasted below follows OP's convention, not the more common convention.
The easiest to implement, IMO, is the chamfer distance algorithm. This is a recursive algorithm that does two passes over the image: one left to right and top to bottom, and one right to left and bottom to top. In each pass, the distance computed for previous pixels is propagated. This algorithm can be implemented using integer distances or floating-point distances between neighbors. The latter yields smaller errors, of course. But in both cases the errors can be reduced significantly by increasing the number of neighbors queried in this propagation. The algorithm is older, but G. Borgefors analyzed it and proposed suitable neighbor distances (G. Borgefors, Distance Transformations in Digital Images, Computer Vision, Graphics, and Image Processing 34:344-371, 1986).
Here is an implementation using 3-4 distance (distance to edge-connected neighbors is 3, distance to vertex-connected neighbors is 4):
def chamfer_distance(img):
w, h = img.shape
dt = np.zeros((w,h), np.uint32)
# Forward pass
x = 0
y = 0
if img[x,y] == 0:
dt[x,y] = 65535 # some large value
for x in range(1, w):
if img[x,y] == 0:
dt[x,y] = 3 + dt[x-1,y]
for y in range(1, h):
x = 0
if img[x,y] == 0:
dt[x,y] = min(3 + dt[x,y-1], 4 + dt[x+1,y-1])
for x in range(1, w-1):
if img[x,y] == 0:
dt[x,y] = min(4 + dt[x-1,y-1], 3 + dt[x,y-1], 4 + dt[x+1,y-1], 3 + dt[x-1,y])
x = w-1
if img[x,y] == 0:
dt[x,y] = min(4 + dt[x-1,y-1], 3 + dt[x,y-1], 3 + dt[x-1,y])
# Backward pass
for x in range(w-2, -1, -1):
y = h-1
if img[x,y] == 0:
dt[x,y] = min(dt[x,y], 3 + dt[x+1,y])
for y in range(h-2, -1, -1):
x = w-1
if img[x,y] == 0:
dt[x,y] = min(dt[x,y], 3 + dt[x,y+1], 4 + dt[x-1,y+1])
for x in range(1, w-1):
if img[x,y] == 0:
dt[x,y] = min(dt[x,y], 4 + dt[x+1,y+1], 3 + dt[x,y+1], 4 + dt[x-1,y+1], 3 + dt[x+1,y])
x = 0
if img[x,y] == 0:
dt[x,y] = min(dt[x,y], 4 + dt[x+1,y+1], 3 + dt[x,y+1], 3 + dt[x+1,y])
return dt
Note that a lot of the complication here is to avoid indexing out of bounds, but still computing distances all the way to the edges of the image. If we simply skip the pixels around the border of the image, the code becomes much simpler.
Because it is a recursive algorithm, it is not possible to vectorize its implementation. The Python code will not be very efficient. But programmed in C or the like will yield a very fast algorithm that yields a fairly good approximation to the Euclidean distance.
OpenCV's cv.distanceTransform implements this algorithm.
Another very efficient algorithm computes the square of the distance transform. The square distance is separable (i.e. can be computed independently for each axis and added). This leads to an algorithm that is easy to parallelize. For each image row, the algorithm does a forward and a backward pass. For each column in the result, the algorithm then does another forward and backward pass. This process leads to an exact Euclidean distance transform.
This algorithm was first proposed by R. van den Boomgaard in his Ph.D. thesis in 1992. Unfortunately this went unnoticed. The algorithm was then again proposed by A. Meijster, J.B.T.M. Roerdink and W.H. Hesselink (A General Algorithm for Computing Distance Transforms in Linear Time, Mathematical Morphology and its Applications to Image and Signal Processing, pp 331-340, 2002), and again by P. Felzenszwalb and D. Huttenlocher (Distance transforms of sampled functions, Technical report, Cornell University, 2004).
This is the most efficient algorithm known, in part because it is the only one that can be easily and efficiently parallelized (computation on each image row, and later on each image column, is independent of other rows/columns).
Unfortunately I don't have any Python code for this one to share, but you can find implementations online. For example OpenCV's cv.distanceTransform implements this algorithm, and DIPlib's dip.EuclideanDistanceTransform does too.

Calculating the intersection area between two rectangles with axes not aligned

I want to calculate the intersection over union IoU between two rectangles with axes not aligned, but with an angle of the axes smaller than 30 degrees. An approximate value is also seeked.
One possible solution is to check if the angle between the two rectangles is less than 30 degree and than rotate them parallel to aligne the axis. From here it is easy to calculate the IoU.
Another possibility is to use monte carlo methods for the intersection ( generate a point, find if the point is under some line of one rectangle and above some line of the other), but this seems expensive because I need to use this calculation a large number of times.
I was hopping that there is something better out there; maybe a geometry library, or maybe an algorithm from the computer vision folks.
I am trying to learn grasping positions using deep neural networks. My algorithem should predict a bounding box (rectangle) for an object in an rgb image. For any image I have also the ground truth (another rectangle) bounding box. From this two rectangles I need the IoU.
Any idea?

Since you're working in Python, I think the Shapely package would serve your needs.

There is quite effective algorithm for calculation of intersection between two convex polygons, described in O'Rourke book "Computational Geometry in C".
C code is available at the book page (convconv).
Algorithm traverse edges of the first polygon, checking orientations of the second polygon vertices in order to detect intersections. When two consequent vertices lie on the different sides of the edge, intersection occurs (there is a lot of trick cases). Algorithm outline is here

You can consider a number of numerical approaches, practically "rendering" the rectangles into some "canvas"/canvases, and traverse the pixels for making your statistics. The size of the canvas should be the size of the bounding box for the entire scene, practically you can find that via picking the minimum and maximum coordinates occurring for each axis.
1) "most CG" approach: really get a rendering library, render one rectangle with red, other rectangle with transparent blue. Then visit each pixel and if it has a non-0 red component, it belongs to the first rectangle, if it has a non-0 blue component, it belongs to the second rectangle. And if it has both, it belongs to the intersection too. This approach is cheap for coding, but requires both writing and reading the canvas even in the rendering phase, which is slower than just writing. This might be even done on GPU too, though I am not sure if setup costs and getting back the result do not weight out the benefit for such a simple scene.
2) another CG-approach would be rendering into 2 arrays, preferably some 1-byte-per-pixel variant, for the sake of speed (you may have to go back in time a bit in order to find such dedicated rendering libraries). This way the renderer only writes, into one array per rectangle, and you read from two when creating the statistics
3) as writing and reading pixels take time, you can do some shortcut, but it needs more coding: convex shapes can be rendered via collecting the minimum and maximum coordinates per scanline, and just filling between the two. If you do it yourself, you can spare the filling part and also the read-and-check-every-pixel step at the end. Build such min-max list for both rectangles, and then you "just" have to check their relation/order for each scanline, to recognize overlaps
And then there is the mathematical way: this is not really useful, see EDIT below while it is unlikely that you would find some sane algorithm for calculating intersection area, specifically for the case of rectangles, if you find such algorithm for triangles, which is more probable, that would be enough. Both rectangles can be split into two triangles, 1A+1B and 2A+2B respectively, and then you just have to run such algorithm 4 times: 1A-2A, 1A-2B, 1B-2A, 1B-2B, sum the results and that is the area of your intersection.
EDIT: for the maths approach (though this also comes from graphics), I think https://en.wikipedia.org/wiki/Sutherland%E2%80%93Hodgman_algorithm can be applied here (as both rectangles are convex polygons, A-B and B-A should produce the same result) for finding the intersection polygon, and then the remaining task is to calculate the area of that polygon (here and now I think it is going to be convex, and then it is really easy).

I ended up using Sutherland-Hodgman algorithm implemented as this functions:
def clip(subjectPolygon, clipPolygon):
def inside(p):
return(cp2[0]-cp1[0])*(p[1]-cp1[1]) > (cp2[1]-cp1[1])*(p[0]-cp1[0])
def computeIntersection():
dc = [ cp1[0] - cp2[0], cp1[1] - cp2[1] ]
dp = [ s[0] - e[0], s[1] - e[1] ]
n1 = cp1[0] * cp2[1] - cp1[1] * cp2[0]
n2 = s[0] * e[1] - s[1] * e[0]
n3 = 1.0 / (dc[0] * dp[1] - dc[1] * dp[0])
return [(n1*dp[0] - n2*dc[0]) * n3, (n1*dp[1] - n2*dc[1]) * n3]
outputList = subjectPolygon
cp1 = clipPolygon[-1]
for clipVertex in clipPolygon:
cp2 = clipVertex
inputList = outputList
outputList = []
s = inputList[-1]
for subjectVertex in inputList:
e = subjectVertex
if inside(e):
if not inside(s):
outputList.append(computeIntersection())
outputList.append(e)
elif inside(s):
outputList.append(computeIntersection())
s = e
cp1 = cp2
return(outputList)
def PolygonArea(corners):
n = len(corners) # of corners
area = 0.0
for i in range(n):
j = (i + 1) % n
area += corners[i][0] * corners[j][1]
area -= corners[j][0] * corners[i][1]
area = abs(area) / 2.0
return area
intersection = clip(rec1, rec2)
intersection_area = PolygonArea(intersection)
iou = intersection_area/(PolygonArea(rec1)+PolygonArea(rec2)-intersection_area)
Another slower method (don't know what algorithm) could be:
from shapely.geometry import Polygon
p1 = Polygon(rec1)
p2 = Polygon(rec2)
inter_sec_area = p1.intersection(rec2).area
iou = inter_sec_area/(p1.area + p2.area - inter_sec_area)
It is worth mentioning that in just one case with bigger polygons (not my case) the shapely module had an area twice greater than the first method. I didn't test both methods intensively.

This might help
What about using Pythagorean theorem ? Since you have two rectangles, when they intersect, you will have one or more triangles, each with one angle of 90°.
Since you also know the angle between the two rectangles (20° in my example), and the coordinates of each rectangle, you can use the the appropriate function (cos/sin/tan) to know the length of all the edges of the triangles.
I hope this can help

Optimizations for Polygon Intersection Finding

I have ~1 billion 2D polygons, each with between 5 and 2000 vertices, laid out on a geographical map. I want to generate cutout masks of these polygons in the bounded range of a quadrangle which is a square portion of the map. I have a naively-coded solution which collects all the polygons that overlap the quadrangle bounds, then creates a mask and draws each of the polygons which it has found to be overlapping with the quadrangle bounds.
relevant code:
def generate_alteration_layers_by_extent(alterations, poly_locators, extents, output_res=1000):
left, right, upper, lower = extents
query_rectangle = pysal.cg.shapes.Rectangle(left, lower, right, upper)
num_layers = len(shapes_by_alterations)
boxed_alteration_layers = []
for i, (alteration, poly_locator) in enumerate(zip(alterations, poly_locators)):
print "computing overlapping polygons"
overlapping_polys = poly_locator.overlapping(query_rectangle)
print "drawing polygons onto mask"
img = Image.new('L', (output_res, output_res), 0)
polys = [np.array(map(list, poly.vertices)) for poly in overlapping_polys]
for poly in polys:
# normalize to the dimensions of img
poly[:,0] -= left; poly[:,0] *= (right-left); poly[:,0] *= output_res
poly[:,1] -= lower; poly[:,1] *= (upper-lower); poly[:,1] *= output_res
ImageDraw.Draw(img).polygon(poly, outline=1, fill=1)
mask = np.array(img)
boxed_alteration_layers.append(mask)
return np.dstack(boxed_alteration_layers)
The problem is that this computation takes too long -- I have run the program for three hours for one mask on a large-memory machine and it has still not finished. The most expensive operation is computing overlapping polygons. I think that the code can by looking over a smaller domain of polygons, excluding searching those that will definitely not overlap.
Note that the poly_locator variables are psyal.cg.locators.PolygonLocators. Any advice would be appreciated. I'm not hell-bent on using Python and think that C++ might have the functionality that I need.

I strongly recommend the BOOST.polygon library by Lucanus Simonson. It handles all sorts of polygon interactions, with a surprising reduction in computational complexity. Simonson's innovative calculus to reduce the dimensionality of the 2-D problems is really neat; it left my jaw hanging open for the last 20 minutes or so of his initial proposal.

How to find rotation angle of a stabilized video frame on Matlab

Consider I have the following a stabilized video frame where stabilization is done by only rotation and translation (no scaling):
As seen in the image, Right-hand side of the image is symmetric of the previous pixels, i.e the black region after rotation is filled with symmetry. I added a red line to indicate it more clearly.
I'd like to find the rotation angle which I will use later on. I could have done this via SURF or SIFT features, however, in real case scenario, I won't have the original frame.
I probably can find the angle by brute force but I wonder if there is any better and more elegant solution. Note that, the intensity value of the symmetric part is not precisely the same as the original part. I've checked some values, for example, upper right pixel of V character on the keyboard is [51 49 47] in original part but [50 50 47] in symmetric copy which means corresponding pixels are not guaranteed to be the same RGB value.
I'll implement this on Matlab or python and the video stabilization is done using ffmpeg.
EDIT: I only have stabilized video, don't have access to original video or files produced by ffmpeg.
Any help/suggestion is appreciated,

A pixel (probably) lies on the searched symmetry line if
Its (first/second/thrid/...) left and right point are equal (=> dG, Figure 1 left)
Its (first/second/thrid/...) left (or right) value is different (=> dGs, Figure 1 middle)
So, the points of interest are characterised by high values for |dGs| - |dG| (=> dGs_dG, Figure 1 right)
As can be seen on the right image of Figure 1, a lot of false positives still exist. Therefore, the Hough transform (Figure 2 left) will be used to detect all the points corresponding to the strongest line (Figure 2 right). The green line is indeed the searched line.
Tuning
Changing n: Higher values will discard more false positives, but also excludes n border pixels. This can be avoided by using a lower n for the border pixels.
Changing thresholds: A higher threshold on dGs_dG will discard more false positives. Discarding high values of dG may also be interesting to discard edge locations in the original image.
A priori knowledge of symmetry line: using the definition of the hough transform, you can discard all lines passing through the center part of the image.
The matlab code used to generate the images is:
I = imread('bnuqb.png');
G = int16(rgb2gray(I));
n = 3; % use the first, second and third left/right point
dG = int16(zeros(size(G) - [0 2*n+2]));
dGs = int16(zeros(size(G) - [0 2*n+2]));
for i=0:n
dG = dG + abs(G(:, 1+n-i:end-2-n-i) - G(:, 3+n+i:end-n+i));
dGs = dGs + abs(G(:, 1+n-i:end-2-n-i) - G(:, 2+n:end-n-1));
end
dGs_dG = dGs - dG;
dGs_dG(dGs_dG < 0) = 0;
figure
subplot(1,3,1);
imshow(dG, [])
subplot(1,3,2);
imshow(dGs, [])
subplot(1,3,3);
imshow(dGs_dG, [])
BW = dGs_dG > 0;
[H,theta,rho] = hough(BW);
P = houghpeaks(H,1);
lines = houghlines(BW,theta,rho,P,'FillGap',50000,'MinLength',7);
figure
subplot(1,2,1);
imshow(H, [])
hold on
plot(P(:, 2),P(:, 1),'r.');
subplot(1,2,2);
imshow(I(:, n+2:end-n-1, :))
hold on
max_len = 0;
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1),xy(:,2),'g');
end

Peak detection in a 2D array

I'm helping a veterinary clinic measuring pressure under a dogs paw. I use Python for my data analysis and now I'm stuck trying to divide the paws into (anatomical) subregions.
I made a 2D array of each paw, that consists of the maximal values for each sensor that has been loaded by the paw over time. Here's an example of one paw, where I used Excel to draw the areas I want to 'detect'. These are 2 by 2 boxes around the sensor with local maxima's, that together have the largest sum.
So I tried some experimenting and decide to simply look for the maximums of each column and row (can't look in one direction due to the shape of the paw). This seems to 'detect' the location of the separate toes fairly well, but it also marks neighboring sensors.
So what would be the best way to tell Python which of these maximums are the ones I want?
Note: The 2x2 squares can't overlap, since they have to be separate toes!
Also I took 2x2 as a convenience, any more advanced solution is welcome, but I'm simply a human movement scientist, so I'm neither a real programmer or a mathematician, so please keep it 'simple'.
Here's a version that can be loaded with np.loadtxt
Results
So I tried #jextee's solution (see the results below). As you can see, it works very on the front paws, but it works less well for the hind legs.
More specifically, it can't recognize the small peak that's the fourth toe. This is obviously inherent to the fact that the loop looks top down towards the lowest value, without taking into account where this is.
Would anyone know how to tweak #jextee's algorithm, so that it might be able to find the 4th toe too?
Since I haven't processed any other trials yet, I can't supply any other samples. But the data I gave before were the averages of each paw. This file is an array with the maximal data of 9 paws in the order they made contact with the plate.
This image shows how they were spatially spread out over the plate.
Update:
I have set up a blog for anyone interested and I have setup a OneDrive with all the raw measurements. So to anyone requesting more data: more power to you!
New update:
So after the help I got with my questions regarding paw detection and paw sorting, I was finally able to check the toe detection for every paw! Turns out, it doesn't work so well in anything but paws sized like the one in my own example. Off course in hindsight, it's my own fault for choosing the 2x2 so arbitrarily.
Here's a nice example of where it goes wrong: a nail is being recognized as a toe and the 'heel' is so wide, it gets recognized twice!
The paw is too large, so taking a 2x2 size with no overlap, causes some toes to be detected twice. The other way around, in small dogs it often fails to find a 5th toe, which I suspect is being caused by the 2x2 area being too large.
After trying the current solution on all my measurements I came to the staggering conclusion that for nearly all my small dogs it didn't find a 5th toe and that in over 50% of the impacts for the large dogs it would find more!
So clearly I need to change it. My own guess was changing the size of the neighborhood to something smaller for small dogs and larger for large dogs. But generate_binary_structure wouldn't let me change the size of the array.
Therefore, I'm hoping that anyone else has a better suggestion for locating the toes, perhaps having the toe area scale with the paw size?

I detected the peaks using a local maximum filter. Here is the result on your first dataset of 4 paws:
I also ran it on the second dataset of 9 paws and it worked as well.
Here is how you do it:
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
#for some reason I had to reshape. Numpy ignored the shape header.
paws_data = np.loadtxt("paws.txt").reshape(4,11,14)
#getting a list of images
paws = [p.squeeze() for p in np.vsplit(paws_data,4)]
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 (xor operation)
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()
All you need to do after is use scipy.ndimage.measurements.label on the mask to label all distinct objects. Then you'll be able to play with them individually.
Note that the method works well because the background is not noisy. If it were, you would detect a bunch of other unwanted peaks in the background. Another important factor is the size of the neighborhood. You will need to adjust it if the peak size changes (the should remain roughly proportional).

Solution
Data file: paw.txt. Source code:
from scipy import *
from operator import itemgetter
n = 5 # how many fingers are we looking for
d = loadtxt("paw.txt")
width, height = d.shape
# Create an array where every element is a sum of 2x2 squares.
fourSums = d[:-1,:-1] + d[1:,:-1] + d[1:,1:] + d[:-1,1:]
# Find positions of the fingers.
# Pair each sum with its position number (from 0 to width*height-1),
pairs = zip(arange(width*height), fourSums.flatten())
# Sort by descending sum value, filter overlapping squares
def drop_overlapping(pairs):
no_overlaps = []
def does_not_overlap(p1, p2):
i1, i2 = p1[0], p2[0]
r1, col1 = i1 / (width-1), i1 % (width-1)
r2, col2 = i2 / (width-1), i2 % (width-1)
return (max(abs(r1-r2),abs(col1-col2)) >= 2)
for p in pairs:
if all(map(lambda prev: does_not_overlap(p,prev), no_overlaps)):
no_overlaps.append(p)
return no_overlaps
pairs2 = drop_overlapping(sorted(pairs, key=itemgetter(1), reverse=True))
# Take the first n with the heighest values
positions = pairs2[:n]
# Print results
print d, "\n"
for i, val in positions:
row = i / (width-1)
column = i % (width-1)
print "sum = %f # %d,%d (%d)" % (val, row, column, i)
print d[row:row+2,column:column+2], "\n"
Output without overlapping squares. It seems that the same areas are selected as in your example.
Some comments
The tricky part is to calculate sums of all 2x2 squares. I assumed you need all of them, so there might be some overlapping. I used slices to cut the first/last columns and rows from the original 2D array, and then overlapping them all together and calculating sums.
To understand it better, imaging a 3x3 array:
>>> a = arange(9).reshape(3,3) ; a
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
Then you can take its slices:
>>> a[:-1,:-1]
array([[0, 1],
[3, 4]])
>>> a[1:,:-1]
array([[3, 4],
[6, 7]])
>>> a[:-1,1:]
array([[1, 2],
[4, 5]])
>>> a[1:,1:]
array([[4, 5],
[7, 8]])
Now imagine you stack them one above the other and sum elements at the same positions. These sums will be exactly the same sums over the 2x2 squares with the top-left corner in the same position:
>>> sums = a[:-1,:-1] + a[1:,:-1] + a[:-1,1:] + a[1:,1:]; sums
array([[ 8, 12],
[20, 24]])
When you have the sums over 2x2 squares, you can use max to find the maximum, or sort, or sorted to find the peaks.
To remember positions of the peaks I couple every value (the sum) with its ordinal position in a flattened array (see zip). Then I calculate row/column position again when I print the results.
Notes
I allowed for the 2x2 squares to overlap. Edited version filters out some of them such that only non-overlapping squares appear in the results.
Choosing fingers (an idea)
Another problem is how to choose what is likely to be fingers out of all the peaks. I have an idea which may or may not work. I don't have time to implement it right now, so just pseudo-code.
I noticed that if the front fingers stay on almost a perfect circle, the rear finger should be inside of that circle. Also, the front fingers are more or less equally spaced. We may try to use these heuristic properties to detect the fingers.
Pseudo code:
select the top N finger candidates (not too many, 10 or 12)
consider all possible combinations of 5 out of N (use itertools.combinations)
for each combination of 5 fingers:
for each finger out of 5:
fit the best circle to the remaining 4
=> position of the center, radius
check if the selected finger is inside of the circle
check if the remaining four are evenly spread
(for example, consider angles from the center of the circle)
assign some cost (penalty) to this selection of 4 peaks + a rear finger
(consider, probably weighted:
circle fitting error,
if the rear finger is inside,
variance in the spreading of the front fingers,
total intensity of 5 peaks)
choose a combination of 4 peaks + a rear peak with the lowest penalty
This is a brute-force approach. If N is relatively small, then I think it is doable. For N=12, there are C_12^5 = 792 combinations, times 5 ways to select a rear finger, so 3960 cases to evaluate for every paw.

This is an image registration problem. The general strategy is:
Have a known example, or some kind of prior on the data.
Fit your data to the example, or fit the example to your data.
It helps if your data is roughly aligned in the first place.
Here's a rough and ready approach, "the dumbest thing that could possibly work":
Start with five toe coordinates in roughly the place you expect.
With each one, iteratively climb to the top of the hill. i.e. given current position, move to maximum neighbouring pixel, if its value is greater than current pixel. Stop when your toe coordinates have stopped moving.
To counteract the orientation problem, you could have 8 or so initial settings for the basic directions (North, North East, etc). Run each one individually and throw away any results where two or more toes end up at the same pixel. I'll think about this some more, but this kind of thing is still being researched in image processing - there are no right answers!
Slightly more complex idea: (weighted) K-means clustering. It's not that bad.
Start with five toe coordinates, but now these are "cluster centres".
Then iterate until convergence:
Assign each pixel to the closest cluster (just make a list for each cluster).
Calculate the center of mass of each cluster. For each cluster, this is: Sum(coordinate * intensity value)/Sum(coordinate)
Move each cluster to the new centre of mass.
This method will almost certainly give much better results, and you get the mass of each cluster which may help in identifying the toes.
(Again, you've specified the number of clusters up front. With clustering you have to specify the density one way or another: Either choose the number of clusters, appropriate in this case, or choose a cluster radius and see how many you end up with. An example of the latter is mean-shift.)
Sorry about the lack of implementation details or other specifics. I would code this up but I've got a deadline. If nothing else has worked by next week let me know and I'll give it a shot.

Using persistent homology to analyze your data set I get the following result (click to enlarge):
This is the 2D-version of the peak detection method described in this SO answer. The above figure simply shows 0-dimensional persistent homology classes sorted by persistence.
I did upscale the original dataset by a factor of 2 using scipy.misc.imresize(). However, note that I did consider the four paws as one dataset; splitting it into four would make the problem easier.
Methodology.
The idea behind this quite simple: Consider the function graph of the function that assigns each pixel its level. It looks like this:
Now consider a water level at height 255 that continuously descents to lower levels. At local maxima islands pop up (birth). At saddle points two islands merge; we consider the lower island to be merged to the higher island (death). The so-called persistence diagram (of the 0-th dimensional homology classes, our islands) depicts death- over birth-values of all islands:
The persistence of an island is then the difference between the birth- and death-level; the vertical distance of a dot to the grey main diagonal. The figure labels the islands by decreasing persistence.
The very first picture shows the locations of births of the islands. This method not only gives the local maxima but also quantifies their "significance" by the above mentioned persistence. One would then filter out all islands with a too low persistence. However, in your example every island (i.e., every local maximum) is a peak you look for.
Python code can be found here.

This problem has been studied in some depth by physicists. There is a good implementation in ROOT. Look at the TSpectrum classes (especially TSpectrum2 for your case) and the documentation for them.
References:
M.Morhac et al.: Background elimination methods for multidimensional coincidence gamma-ray spectra. Nuclear Instruments and Methods in Physics Research A 401 (1997) 113-132.
M.Morhac et al.: Efficient one- and two-dimensional Gold deconvolution and its application to gamma-ray spectra decomposition. Nuclear Instruments and Methods in Physics Research A 401 (1997) 385-408.
M.Morhac et al.: Identification of peaks in multidimensional coincidence gamma-ray spectra. Nuclear Instruments and Methods in Research Physics A 443(2000), 108-125.
...and for those who don't have access to a subscription to NIM:
Spectrum.doc
SpectrumDec.ps.gz
SpectrumSrc.ps.gz
SpectrumBck.ps.gz

I'm sure you have enough to go on by now, but I can't help but suggest using the k-means clustering method. k-means is an unsupervised clustering algorithm which will take you data (in any number of dimensions - I happen to do this in 3D) and arrange it into k clusters with distinct boundaries. It's nice here because you know exactly how many toes these canines (should) have.
Additionally, it's implemented in Scipy which is really nice (http://docs.scipy.org/doc/scipy/reference/cluster.vq.html).
Here's an example of what it can do to spatially resolve 3D clusters:
What you want to do is a bit different (2D and includes pressure values), but I still think you could give it a shot.

Here is an idea: you calculate the (discrete) Laplacian of the image. I would expect it to be (negative and) large at maxima, in a way that is more dramatic than in the original images. Thus, maxima could be easier to find.
Here is another idea: if you know the typical size of the high-pressure spots, you can first smooth your image by convoluting it with a Gaussian of the same size. This may give you simpler images to process.

Just a couple of ideas off the top of my head:
take the gradient (derivative) of the scan, see if that eliminates the false calls
take the maximum of the local maxima
You might also want to take a look at OpenCV, it's got a fairly decent Python API and might have some functions you'd find useful.

thanks for the raw data. I'm on the train and this is as far as I've gotten (my stop is coming up). I massaged your txt file with regexps and have plopped it into a html page with some javascript for visualization. I'm sharing it here because some, like myself, might find it more readily hackable than python.
I think a good approach will be scale and rotation invariant, and my next step will be to investigate mixtures of gaussians. (each paw pad being the center of a gaussian).
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Physicist's solution:
Define 5 paw-markers identified by their positions X_i and init them with random positions.
Define some energy function combining some award for location of markers in paws' positions with some punishment for overlap of markers; let's say:
E(X_i;S)=-Sum_i(S(X_i))+alfa*Sum_ij (|X_i-Xj|<=2*sqrt(2)?1:0)
(S(X_i) is the mean force in 2x2 square around X_i, alfa is a parameter to be peaked experimentally)
Now time to do some Metropolis-Hastings magic:
1. Select random marker and move it by one pixel in random direction.
2. Calculate dE, the difference of energy this move caused.
3. Get an uniform random number from 0-1 and call it r.
4. If dE<0 or exp(-beta*dE)>r, accept the move and go to 1; if not, undo the move and go to 1.
This should be repeated until the markers will converge to paws. Beta controls the scanning to optimizing tradeoff, so it should be also optimized experimentally; it can be also constantly increased with the time of simulation (simulated annealing).

Just wanna tell you guys there is a nice option to find local maxima in images with python:
from skimage.feature import peak_local_max
or for skimage 0.8.0:
from skimage.feature.peak import peak_local_max
http://scikit-image.org/docs/0.8.0/api/skimage.feature.peak.html

It's probably worth to try with neural networks if you are able to create some training data... but this needs many samples annotated by hand.

Heres another approach that I used when doing something similar for a large telescope:
1) Search for the highest pixel.
Once you have that, search around that for the best fit for 2x2 (maybe maximizing the 2x2 sum), or do a 2d gaussian fit inside the sub region of say 4x4 centered on the highest pixel.
Then set those 2x2 pixels you have found to zero (or maybe 3x3) around the peak center
go back to 1) and repeat till the highest peak falls below a noise threshold, or you have all the toes you need

a rough outline...
you'd probably want to use a connected components algorithm to isolate each paw region. wiki has a decent description of this (with some code) here: http://en.wikipedia.org/wiki/Connected_Component_Labeling
you'll have to make a decision about whether to use 4 or 8 connectedness. personally, for most problems i prefer 6-connectedness. anyway, once you've separated out each "paw print" as a connected region, it should be easy enough to iterate through the region and find the maxima. once you've found the maxima, you could iteratively enlarge the region until you reach a predetermined threshold in order to identify it as a given "toe".
one subtle problem here is that as soon as you start using computer vision techniques to identify something as a right/left/front/rear paw and you start looking at individual toes, you have to start taking rotations, skews, and translations into account. this is accomplished through the analysis of so-called "moments". there are a few different moments to consider in vision applications:
central moments: translation invariant
normalized moments: scaling and translation invariant
hu moments: translation, scale, and rotation invariant
more information about moments can be found by searching "image moments" on wiki.

Perhaps you can use something like Gaussian Mixture Models. Here's a Python package for doing GMMs (just did a Google search)
http://www.ar.media.kyoto-u.ac.jp/members/david/softwares/em/

Interesting problem. The solution I would try is the following.
Apply a low pass filter, such as convolution with a 2D gaussian mask. This will give you a bunch of (probably, but not necessarily floating point) values.
Perform a 2D non-maximal suppression using the known approximate radius of each paw pad (or toe).
This should give you the maximal positions without having multiple candidates which are close together. Just to clarify, the radius of the mask in step 1 should also be similar to the radius used in step 2. This radius could be selectable, or the vet could explicitly measure it beforehand (it will vary with age/breed/etc).
Some of the solutions suggested (mean shift, neural nets, and so on) probably will work to some degree, but are overly complicated and probably not ideal.

It seems you can cheat a bit using jetxee's algorithm. He is finding the first three toes fine, and you should be able to guess where the fourth is based off that.

Well, here's some simple and not terribly efficient code, but for this size of a data set it is fine.
import numpy as np
grid = np.array([[0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0.4,0.4,0.4,0,0,0],
[0,0,0,0,0.4,1.4,1.4,1.8,0.7,0,0,0,0,0],
[0,0,0,0,0.4,1.4,4,5.4,2.2,0.4,0,0,0,0],
[0,0,0.7,1.1,0.4,1.1,3.2,3.6,1.1,0,0,0,0,0],
[0,0.4,2.9,3.6,1.1,0.4,0.7,0.7,0.4,0.4,0,0,0,0],
[0,0.4,2.5,3.2,1.8,0.7,0.4,0.4,0.4,1.4,0.7,0,0,0],
[0,0,0.7,3.6,5.8,2.9,1.4,2.2,1.4,1.8,1.1,0,0,0],
[0,0,1.1,5,6.8,3.2,4,6.1,1.8,0.4,0.4,0,0,0],
[0,0,0.4,1.1,1.8,1.8,4.3,3.2,0.7,0,0,0,0,0],
[0,0,0,0,0,0.4,0.7,0.4,0,0,0,0,0,0]])
arr = []
for i in xrange(grid.shape[0] - 1):
for j in xrange(grid.shape[1] - 1):
tot = grid[i][j] + grid[i+1][j] + grid[i][j+1] + grid[i+1][j+1]
arr.append([(i,j),tot])
best = []
arr.sort(key = lambda x: x[1])
for i in xrange(5):
best.append(arr.pop())
badpos = set([(best[-1][0][0]+x,best[-1][0][1]+y)
for x in [-1,0,1] for y in [-1,0,1] if x != 0 or y != 0])
for j in xrange(len(arr)-1,-1,-1):
if arr[j][0] in badpos:
arr.pop(j)
for item in best:
print grid[item[0][0]:item[0][0]+2,item[0][1]:item[0][1]+2]
I basically just make an array with the position of the upper-left and the sum of each 2x2 square and sort it by the sum. I then take the 2x2 square with the highest sum out of contention, put it in the best array, and remove all other 2x2 squares that used any part of this just removed 2x2 square.
It seems to work fine except with the last paw (the one with the smallest sum on the far right in your first picture), it turns out that there are two other eligible 2x2 squares with a larger sum (and they have an equal sum to each other). One of them is still selects one square from your 2x2 square, but the other is off to the left. Fortunately, by luck we see to be choosing more of the one that you would want, but this may require some other ideas to be used to get what you actually want all of the time.

I am not sure this answers the question, but it seems like you can just look for the n highest peaks that don't have neighbors.
Here is the gist. Note that it's in Ruby, but the idea should be clear.
require 'pp'
NUM_PEAKS = 5
NEIGHBOR_DISTANCE = 1
data = [[1,2,3,4,5],
[2,6,4,4,6],
[3,6,7,4,3],
]
def tuples(matrix)
tuples = []
matrix.each_with_index { |row, ri|
row.each_with_index { |value, ci|
tuples << [value, ri, ci]
}
}
tuples
end
def neighbor?(t1, t2, distance = 1)
[1,2].each { |axis|
return false if (t1[axis] - t2[axis]).abs > distance
}
true
end
# convert the matrix into a sorted list of tuples (value, row, col), highest peaks first
sorted = tuples(data).sort_by { |tuple| tuple.first }.reverse
# the list of peaks that don't have neighbors
non_neighboring_peaks = []
sorted.each { |candidate|
# always take the highest peak
if non_neighboring_peaks.empty?
non_neighboring_peaks << candidate
puts "took the first peak: #{candidate}"
else
# check that this candidate doesn't have any accepted neighbors
is_ok = true
non_neighboring_peaks.each { |accepted|
if neighbor?(candidate, accepted, NEIGHBOR_DISTANCE)
is_ok = false
break
end
}
if is_ok
non_neighboring_peaks << candidate
puts "took #{candidate}"
else
puts "denied #{candidate}"
end
end
}
pp non_neighboring_peaks

Maybe a naive approach is sufficient here: Build a list of all 2x2 squares on your plane, order them by their sum (in descending order).
First, select the highest-valued square into your "paw list". Then, iteratively pick 4 of the next-best squares that don't intersect with any of the previously found squares.

What if you proceed step by step: you first locate the global maximum, process if needed the surrounding points given their value, then set the found region to zero, and repeat for the next one.