I am working on loop-closure detection problem in two different seasons, e.g., summer, and fall. I need to make precision-recall curves. Suppose, I have taken 500 image from summer and 500 image from fall season. I have distance matrix. enter image description here
But I am totally confused, how to make precison recall curves. Like, for each image from one season, I will get 500 nearest images in ascending (distance) order. I know the definition of precision and recall, but i can't get close to the solution of this problem. Looking forward for any kind of help or comments or advice. thanks in advance.
In precision-recall plots each point is a pair of precision and recall values. In your case, I guess, you'd need to compute those values for each image and then average them.
Imagine you have 1000 images in total and only 100 images that belong to summer. If you take 500 closest images to some "summer" image, precision in the best case (when the first images always belong to the class) would be:
precision(summer) = 100 / (100 + 400) = (retrieved summer images) / (retrieved summer images + other retrieved images) = 0.2
And recall:
recall(summer) = 100 / (100 + 0) = (retrieved summer images) / (retrieved summer images + not retrieved summer images) = 1
As you can see, it has high recall because all the summer images were retrieved, but low precision, because there are only 100 images, and 400 other images don't belong to the class.
Now, if you take the first 100 images instead of 500, both recall and precision would equal 1.
If you take 50 first images, then precision would be still 1, but recall would drop to 0.5.
So, by varying the number of images you can get points for the precision-recall curve. For the above-described example these points would be (0.2, 1), (1, 1), (1, 0.5).
You could compute these values for each of the 1000 images using different thresholds.
Related
I'm trying to plot some sine waves (code example in plain js here).
When freq is "low" (freq = 10 hz in that case), the plot is quite nice:
The problem is when I increase the freq (try to set var freq = 50 for example):
lots of ripples, it becomes distorted and not so good as plot. If I increment it more, even worse (var freq = 8030 for example is terrible).
When I see those kind of graph on pro systems, they are displayed just fine.
How would you improve it? FFT, splines, whatever? Which is the right approch?
I don't really need accurancy (i.e. for waveform analysis or whatever), just plot it nicely (as in Desmos https://www.desmos.com/calculator/eodkjlywjh, for example).
Like Matt said, the problem here is the same as in plotting a discrete-time signal shows amplitude modulation: "At higher frequencies, when the peak falls between two samples, the sampled points can be a lot lower than the peak." This causes the peaks of the waveform to vary, making a kind of ripple visual effect at the top and bottom of the plot.
Increasing the sampling resolution helps. If you can't increase sampling resolution by much, adjust the sampling resolution to the closest integer multiple of the sine's frequency and set the sampling phase so that it samples the waveform peaks. For instance for a 100 Hz wave cos(2 π (100*x + 0.3)), you could sample at 400 Hz at x[n] = n/400 - 0.003. That way there are four samples per period, with x[0], x[4], x[8], ... sampling exactly on the peaks.
Another thought: plotting with a thicker line can help to smooth over some defects. It looks like the desmos example that you linked is using something like a 3-pixel line width.
I'm working on some gridded temperature data, which I have categorised into a matrix where each cell can be one of two classes - let's say 0 or 1 for simplicity. For each class I want to calculate patch statistics, taking inspiration from FRAGSTATS, which is used in landscape ecology to characterise the shape and size of habitat patches.
For my purposes, a patch is a cluster of adjacent cells of the same class. Here's an example matrix, mat:
mat <-
matrix(c(0,1,0,
1,1,1,
1,0,1), nrow = 3, ncol = 3,
byrow = TRUE)
0 1 0
1 1 1
1 0 1
All the 1s in mat form a single patch (we'll ignore the 0s), and in order to calculate various different shape metrics I need to be able to calculate the perimeter (i.e. number of outside edges).
EDIT
Sorry I apparently can't post an image because I don't have enough reputation, but you can see in the black lines of G5W's answer below that the outside borders of 1's represent the outside edges I'm referring to.
Manually I can count that the patch of 1s has 14 outside edges and I know the area (i.e. number of cells) is 6. Based on a paper by He et al. and this other question I've figured out how to calculate the number of inside edges (5 in this example), but I'm really struggling to do the same for the outside edges! I think it's something to do with how the patch shape compares to the largest integer square that has a smaller area (in this case, a 2 x 2 square), but so far my research and pondering have been to no avail.
N.B. I am aware of the package SDMTools, which can calculate various FRAGSTATS metrics. Unfortunately the metrics returned are too processed e.g. instead of just Aggregation Index, I need to know the actual numbers used to calculate it (number of observed shared edges / maximum number of shared edges).
This is my first post on here so I hope it's detailed enough! Thanks in advance :)
If you know the area and the number of inside edges, it is simple to calculate the number of outside edges. Every patch has four edges so in some way, the total number of edges is 4 * area. But that is not quite right because every inside edge is shared between two patches. So the right number of total edges is
4*area - inside
The number of outside edges is the total edges minus the inside edges, so
outside = total - inside = (4*area- inside) - inside = 4*area - 2*inside.
You can see that the area is made up of 6 squares each of which has 4 sides. The inside edges (the red ones) are shared by two adjacent squares.
I want to apply heat transfer ( heat conduction and convection) for a hemisphere. It is a transient homogeneous heat transfer in spherical coordinates. There is no heat generation. Boundary conditions of hemisphere is in the beginning at Tinitial= 20 degree room temperature. External-enviromental temperature is -22 degree. You can imagine that hemisphere is a solid material. Also, it is a non-linear model, because thermal conductivity is changing after material is frozen, and this is going to change the temperature profile.
I want to find the temperature profile of this solid during a certain time until center temperature reach to -22 degree.
In this case, Temperature depends on 3 parameters : T(r,theta,t). radius, angle, and time.
1/α(∂T(r,θ,t))/∂t =1/r^2*∂/∂r(r^2(∂T(r,θ,t))/∂r)+ 1/(r^2*sinθ )∂/∂θ(sinθ(∂T(r,θ,t))/∂θ)
I applied finite difference method using matlab, However, boundary conditions have issues. There are convection on surface of the hemisphere, and conduction in the inner nodes, bottom of the hemisphere has constant temperature which is air temperature (-22). You can see the scripts which i am using for BCs in the matlab file.
% Temperature at surface of hemisphere solid boundary node
for i=nodes
for j=1:1:(nodes-1)
Qcd_ot(i,j)= ((k(i,j)+ k(i-1,j))/2)*A(i-1,j)*(( Told(i,j)-Told(i-1,j))/dr); % heat conduction out of node
Qcv(i,j) = h*(Tair-Told(i,j))*A(i,j); % heat transfer through convectioin on surface
Tnew(i,j) = ((Qcv(i,j)-Qcd_ot(i,j))/(mass(i,j)*cp(i,j))/2)*dt + Told(i,j);
end % end of for loop
end
% Temperature at inner nodes
for i=2:1:(nodes-1)
for j=2:1:(nodes-1)
Qcd_in(i,j)= ((k(i,j)+ k(i+1,j))/2)*A(i,j) *((2/R)*(( Told(i+1,j)-Told(i,j))/(2*dr)) + ((Told(i+1,j)-2*Told(i,j)+Told(i-1,j))/(dr^2)) + ((cot(y)/(R^2))*((Told(i,j+1)-Told(i,j-1))/(2*dy))) + (1/(R^2))*(Told(i,j+1)-2*Told(i,j)+ Told(i,j-1))/(dy^2));
Qcd_out(i,j)= ((k(i,j)+ k(i-1,j))/2)*A(i-1,j)*((2/R)*(( Told(i,j)-Told(i-1,j))/(2*dr)) +((Told(i+1,j)-2*Told(i,j)+Told(i-1,j))/(dr^2)) + ((cot(y)/(R^2))*((Told(i,j+1)-Told(i,j-1))/(2*dy))) + (1/(R^2))*(Told(i,j+1)-2*Told(i,j)+ Told(i,j-1))/(dy^2));
Tnew(i,j) = ((Qcd_in(i,j)-Qcd_out(i,j))/(mass(i,j)*cp(i,j)))*dt + Told(i,j);
end %end for loop
end % end for loop
%Temperature for at center line nodes
for i=2:1:(nodes-1)
for j=1
Qcd_line(i,j)=((k(i,j)+ k(i+1,j))/2)*A(i,j)*(Told(i+1,j)-Told(i,j))/dr;
Qcd_lineout(i,j)=((k(i,j)+ k(i-1,j))/2)*A(i-1,j)*(Told(i,j)-Told(i-1,j))/dr;
Tnew(i,j)= ((Qcd_line(i,j)-Qcd_lineout(i,j))/(mass(i,j)*cp(i,j)))*dt + Told(i,j);
end
end
% Temperature at bottom point (center) of the hemisphere solid
for i=1
for j=1:1:(nodes-1)
Qcd_center(i,j)=(((k(i,j)+k(i+1,j))/2)*A(i,j)*(Told(i+1,j)-Tair)/dr);
Tnew(i,j)= ((Qcd_center(i,j))/(mass(i,j)*cp(i,j)))*dt + Told(i,j);
end
end
% Temperature at all bottom points of the hemisphere
Tnew(:,nodes)=-22;
Told=Tnew;
t=t+dt;
Tnew temperatures values are getting bigger exponentially after program is run, and then becoming NaN. It supposed to show me cooling and freezing temperature profile of solid until it reaches to Tair temperature. I could not figure out the reasons why it is changing like that.
I would like to hear your suggestions for BCs implementation to this program, or how should i change them according to this conditions. Thanks in advance !!
Your code is too long to read and understand completely, but it looks like you are using a simple forward Euler scheme, is that correct? If so, try to reduce the time-step dt, maybe by a lot, since this method can become numerically unstable if dt is too big. This might slow down the speed of the computation (again by a lot), but that is the price you pay for such a simple algorithm. There are alternatives methods that do not suffer from instability, but they are much harder to implement, since you need to solve a system of equations.
I did some thermal simulations using this simple scheme a long time ago. I found that the stability criteria was dt < (dx)^2 * c_p * rho / (6 * k), which should be valid for a simulation on a 3D cartesian grid, where dx is the spatial step, c_p is the specific heat, rho the density and k the thermal conductivity of the material. I don't know how to convert this to your case with spherical coordinates. The thing I learned then was to choose small time-steps, but more importantly as large dx as possible: when you reduce dx by a factor 2, you also need to reduce dt by a factor 4 to keep things stable. At the same time, for a 3D problem, the number of elements will increase by a factor 8. So the total simulation time scales with 1 / (dx)^5!!!
I want to know how to measure the distance between two pixels in dicom . already done some google found pixel spacing (0028,0030) need to find the distance . could some one clearly explain ....
thanks
Assuming that you're trying to measure distances in the subject/animal/phantom/whatever, it all depends on whether you want to measure distances between different slices or just in the same slice.
Volumetric DICOM series typically have a slice spacing (0012,0088) in addition to the pixel spacing which you need to take into account. Note that there is also such a thing as slice thickness, which is distinct and should not be used for calculating distances, as there can be a gap or overlap between consecutive slices.
It is helpful to define a voxelspacing vector as follows (pseudocode):
voxelspacing.x = first element of PixelSpacing (0028,0030), i.e. before "\"
voxelspacing.y = second element of PixelSpacing (0028,0030), i.e. after "\"
voxelspacing.z = SliceSpacing (0018,0088) or 0 if 2D and/or not specified
Some brain-dead manufacturers and de-identification tools break the slice spacing tag in which case you'll have to calculate it from another source, such as difference in consecutive slice location, patient image position, etc, but that's another matter.
Moving on, you now have the distance in millimeters between voxels for each dimension. You can then calculate the real-world euclidean distance given voxel coordinates in pointA and pointB:
delta = (pointA - pointB) * voxelspacing
distance = sqrt(delta.x^2 + delta.y^2 + delta.z^2);
Where all the operators are element-wise. It is critical to individually multiply the voxel coordinates with their respective spacings before computing distance, because voxels are typically not isotropic.
You need to know the dot pitch of the monitor. For example a jumbotron has huge pixels (guessing), so the distance is larger than it would be for a typical desktop monitor. Ask the manufacturer of the monitor for this information. After that use pythogorean theorum. sqrt(a^2 + b^2) = c c being the total distance and a/b are x and y distances. to find a and be you would find the coordinates of one pixel and subtract from the other. a = (x1-x2) b = (
I am new to this forum and not a native english speaker, so please be nice! :)
Here is the challenge I face at the moment:
I want to calculate the (approximate) relative coordinates of yet unknown points in a 3D euclidean space based on a set of given distances between 2 points.
In my first approach I want to ignore possible multiple solutions, just taking the first one by random.
e.g.:
given set of distances: (I think its creating a pyramid with a right-angled triangle as a base)
P1-P2-Distance
1-2-30
2-3-40
1-3-50
1-4-60
2-4-60
3-4-60
Step1:
Now, how do I calculate the relative coordinates for those points?
I figured that the first point goes to 0,0,0 so the second one is 30,0,0.
After that the third points can be calculated by finding the crossing of the 2 circles from points 1 and 2 with their distances to point 3 (50 and 40 respectively). How do I do that mathematically? (though I took these simple numbers for an easy representation of the situation in my mind). Besides I do not know how to get to the answer in a correct mathematical way the third point is at 30,40,0 (or 30,0,40 but i will ignore that).
But getting the fourth point is not as easy as that. I thought I have to use 3 spheres in calculate the crossing to get the point, but how do I do that?
Step2:
After I figured out how to calculate this "simple" example I want to use more unknown points... For each point there is minimum 1 given distance to another point to "link" it to the others. If the coords can not be calculated because of its degrees of freedom I want to ignore all possibilities except one I choose randomly, but with respect to the known distances.
Step3:
Now the final stage should be this: Each measured distance is a bit incorrect due to real life situation. So if there are more then 1 distances for a given pair of points the distances are averaged. But due to the imprecise distances there can be a difficulty when determining the exact (relative) location of a point. So I want to average the different possible locations to the "optimal" one.
Can you help me going through my challenge step by step?
You need to use trigonometry - specifically, the 'cosine rule'. This will give you the angles of the triangle, which lets you solve the 3rd and 4th points.
The rules states that
c^2 = a^2 + b^2 - 2abCosC
where a, b and c are the lengths of the sides, and C is the angle opposite side c.
In your case, we want the angle between 1-2 and 1-3 - the angle between the two lines crossing at (0,0,0). It's going to be 90 degrees because you have the 3-4-5 triangle, but let's prove:
50^2 = 30^2 + 40^2 - 2*30*40*CosC
CosC = 0
C = 90 degrees
This is the angle between the lines (0,0,0)-(30,0,0) and (0,0,0)- point 3; extend along that line the length of side 1-3 (which is 50) and you'll get your second point (0,50,0).
Finding your 4th point is slightly trickier. The most straightforward algorithm that I can think of is to firstly find the (x,y) component of the point, and from there the z component is straightforward using Pythagoras'.
Consider that there is a point on the (x,y,0) plane which sits directly 'below' your point 4 - call this point 5. You can now create 3 right-angled triangles 1-5-4, 2-5-4, and 3-5-4.
You know the lengths of 1-4, 2-4 and 3-4. Because these are right triangles, the ratio 1-4 : 2-4 : 3-4 is equal to 1-5 : 2-5 : 3-5. Find the point 5 using trigonometric methods - the 'sine rule' will give you the angles between 1-2 & 1-4, 2-1 and 2-4 etc.
The 'sine rule' states that (in a right triangle)
a / SinA = b / SinB = c / SinC
So for triangle 1-2-4, although you don't know lengths 1-4 and 2-4, you do know the ratio 1-4 : 2-4. Similarly you know the ratios 2-4 : 3-4 and 1-4 : 3-4 in the other triangles.
I'll leave you to solve point 4. Once you have this point, you can easily solve the z component of 4 using pythagoras' - you'll have the sides 1-4, 1-5 and the length 4-5 will be the z component.
I'll initially assume you know the distances between all pairs of points.
As you say, you can choose one point (A) as the origin, orient a second point (B) along the x-axis, and place a third point (C) along the xy-plane. You can solve for the coordinates of C as follows:
given: distances ab, ac, bc
assume
A = (0,0)
B = (ab,0)
C = (x,y) <- solve for x and y, where:
ac^2 = (A-C)^2 = (0-x)^2 + (0-y)^2 = x^2 + y^2
bc^2 = (B-C)^2 = (ab-x)^2 + (0-y)^2 = ab^2 - 2*ab*x + x^2 + y^2
-> bc^2 - ac^2 = ab^2 - 2*ab*x
-> x = (ab^2 + ac^2 - bc^2)/2*ab
-> y = +/- sqrt(ac^2 - x^2)
For this to work accurately, you will want to avoid cases where the points {A,B,C} are in a straight line, or close to it.
Solving for additional points in 3-space is similar -- you can expand the Pythagorean formula for the distance, cancel the quadratic elements, and solve the resulting linear system. However, this does not directly help you with your steps 2 and 3...
Unfortunately, I don't know a well-behaved exact solution for steps 2 and 3, either. Your overall problem will generally be both over-constrained (due to conflicting noisy distances) and under-constrained (due to missing distances).
You could try an iterative solver: start with a random placement of all your points, compare the current distances with the given ones, and use that to adjust your points in such a way as to improve the match. This is an optimization technique, so I would look up books on numerical optimization.
If you know the distance between the nodes (fixed part of system) and the distance to the tag (mobile) you can use trilateration to find the x,y postion.
I have done this using the Nanotron radio modules which have a ranging capability.