I am using K-Means algorithm to classify SIFT vector features.
My K-Means skeleton is
choose first K points of the data as initial centers
do{
// assign each point to the corresponding cluster
data assignment;
// get the recalculated center of each cluster
// suppose point is multi-dimensional data, e.g.(x1, x2, x3...)
// the center is composed by average value of each dimension.
// e.g. ((x1 + y1 + ...)/n, (x2 + y2 + ...)/n ...)
new centroid;
sum total memberShipChanged;
}while(some point still changes their membership, i.e. total memberShipChanged != 0)
We all know that K-Means aims to get the minimal of within cluster sum of square, illustrated as below snapshot.
And we can use a do while iteration to reach the target. Now I prove why after every iteration the within cluster sum of square is smaller.
Proof:
For simplicity, I only consider 2 iterations.
After data assignment process, every descriptor vector has its new nearest cluster center, so the within cluster sum of square decrease after this process. After all, the within cluster of square is sum of every vector to one center, if every vector choose it own new nearest neighbor, there is no doubt that the sum decreases.
In new centroid process, I use arithmetic mean to calculate the new center vector, the local sum of one cluster must decrease.
So the within cluster sum of square decrease twice in one iteration. And after several iterations, every descriptor vector doesn't change its membership, and within cluster sum of square reaches the local minimum.
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Now My question comes:
my SURF data is derived from 3000 images, every descriptor vector is 128-dimension, and there are 1,296,672 vectors in total. And in my code, I print
1)vector number of each clsuter
2)total memeberShipChanged in one iteration
3)within cluster sum of square before one iteration.
Here is output:
sum of square : 8246977014860
90504 228516 429755 266828 1653711 398631 193081 240072
memberShipChanged : 3501098
sum of square : 4462579627000
244521 284626 448700 228211 1361902 303864 317464 311810
memberShipChanged : 975442
sum of square : 4561378972772
323746 457785 388988 228431 993328 304606 473668 330546
memberShipChanged : 828709
sum of square : 4678353976030
359537 480818 346767 222646 789858 332876 612672 355924
memberShipChanged : 563256
......
I only list 4 iteration output of it. From the output, we can see that after first iteration. within cluster sum of square really decrease from 8246977014860 to 4462579627000. But other iterations are nearly of on use in minimize it, but we can still observe the memberShipChanged is converging. I don't know why this happen. I think the first k-means iteration is overwhelming important.
Besides, what should I set the new center coordinate of a empty cluster when the memberShipChanged still doesn't converge to 0 yet? Now I use (0, 0, 0, 0, 0 ...). But is this accurate, perhaps the within cluster of sum increases due to it.
Related
I made some recordings from a head tracker which provides the 4 values of the quaternions, which are saved in a csv (each row is a set of quaternion plus a timestamp).
I need to calculate for the whole recording how much the head moved. This is needed for an experiment where I would like to see whether under a condition the head moved more or less compared to another condition.
What is the best way to get a single quantity for each recording?
I have some proposals but I do not know how much appropriate they are:
PROPOSAL 1) I calculate the cumulative sum of the absolute value of the derivatives for each quaternion value, then I sum the 4 sums together to get a single value
PROPOSAL 2) I calculate the cumulative sum of the absolute value of the derivatives of the norm
Sounds like you just want a rough estimate of total angular movement as a single value. One way is to assume minimum rotation angle between quaternion samples and then just add up those angles. E.g., suppose two consecutive quaternion samples are q1 and q2. Then calculate the quaternion multiply q = q1 * inv(q2) and your delta-angle for that step is 2*acos(abs(qw)). Do this for each step and add up all the delta angles.
I have a network that has to be divided into a grid of square cells (200 X 200 m). each cell includes sub-segments of the edges.
I have generated the simulation data output and used sumolib to extract edge-based output. I have to calculate the average traffic volume in each cell (not edge) measured in (vehicles/second).
this is part of the script I have written:
extract edge-based density and speed values:
for interval in sumolib.output.parse('cairo.edgeDataOutput.xml','interval'):
for edge in interval.edge:
edgeDataOutput[edge.id]= (edge.density,edge.speed)
after saving density and speed into edgeDataOutput, I have to aggregate into cells and calculate avg.traffic volume in each cell:
for cellID in ids:
density=0
speed=0
n=0 #avg.traffic vol
for edgeID in cell_edgeMap[cellID].keys():
if edgeID in edgeDataOutput.keys():
density+= float(edgeDataOutput[edgeID][1])
speed+= float(edgeDataOutput[edgeID][2])
n += (float(edgeDataOutput[edgeID][1])/1000) * float(edgeDataOutput[edgeID][2]) #traffic vol = (density/1000)*speed
densities.append(int((density / len(cell_edgeMap[cellID].keys()))+0.5))
speeds.append(int((speed / len(cell_edgeMap[cellID].keys()))+0.5))
numOfVehicles.append(int(n/len(cell_edgeMap[cellID].keys())))
as you can see from the code, I sum up the density, speed values of each edge that is in the cell then divide by the number of edges inside the cell to get the mean value.
density at cell(veh/Km) = sum(density at each edge inside cell)/num of edges inside cell.
speed at cell(m/s) = sum(speed at each edge inside cell)/num of edges in cell.
and I am using the following formula to calculate the traffic volume at each cell:
avg.traffic volume at cell(veh/s) = sum(avg.traffic volume at each edge inside cell)/num of edges inside cell.
avg.traffic volume at edge(veh/s) = density at edge(veh/Km) * speed at edge(m/s) / 1000.
I just want to make sure that I am using the write formula.
It is not easy to answer whether you do the right thing because averaging over both time and space is always hard. Furthermore it is not clear what you are trying to measure. The traffic volume usually denotes the total (or average) number of vehicles and is measured without unit (so only number of vehicles). The traffic flow is measured in vehicles per time unit but is usually applied only to a cross section not to an area. If you want the average number of cars in the cell it should suffice to divide the sum of the number of sampleSeconds by the length of the interval. The second value needs a more in depth discussion but I would probably at least multiply it with the edge length when summing up.
so I am making a simple simulation of different planets with individual velocity flying around space and orbiting each other.
I plan to simulate their pull on each other by considering each planet as projecting their own "gravity vector field." Each time step I'm going to add the vectors outputted from each planets individual vector field equation (V = -xj + (-yj) or some notation like it) except the one being effected in the calculation, and use the effected planets position as input to the equations.
However this would inaccurate, and does not consider the gravitational pull as continuous and constant. Bow do I calculate the movement of my planets if each is continuously effecting the others?
Thanks!
In addition to what Blender writes about using Newton's equations, you need to consider how you will be integrating over your "acceleration field" (as you call it in the comment to his answer).
The easiest way is to use Euler's Method. The problem with that is it rapidly diverges, but it has the advantage of being easy to code and to be reasonably fast.
If you are looking for better accuracy, and are willing to sacrifice some performance, one of the Runge-Kutta methods (probably RK4) would ordinarily be a good choice. I'll caution you that if your "acceleration field" is dynamic (i.e. it changes over time ... perhaps as a result of planets moving in their orbits) RK4 will be a challenge.
Update (Based on Comment / Question Below):
If you want to calculate the force vector Fi(tn) at some time step tn applied to a specific object i, then you need to compute the force contributed by all of the other objects within your simulation using the equation Blender references. That is for each object, i, you figure out how all of the other objects pull (apply force) and those vectors when summed will be the aggregate force vector applied to i. Algorithmically this looks something like:
for each object i
Fi(tn) = 0
for each object j ≠ i
Fi(tn) = Fi(tn) + G * mi * mj / |pi(tn)-pj(tn)|2
Where pi(tn) and pj(tn) are the positions of objects i and j at time tn respectively and the | | is the standard Euclidean (l2) normal ... i.e. the Euclidean distance between the two objects. Also, G is the gravitational constant.
Euler's Method breaks the simulation into discrete time slices. It looks at the current state and in the case of your example, considers all of the forces applied in aggregate to all of the objects within your simulation and then applies those forces as a constant over the period of the time slice. When using
ai(tn) = Fi(tn)/mi
(ai(tn) = acceleration vector at time tn applied to object i, Fi(tn) is the force vector applied to object i at time tn, and mi is the mass of object i), the force vector (and therefore the acceleration vector) is held constant for the duration of the time slice. In your case, if you really have another method of computing the acceleration, you won't need to compute the force, and can instead directly compute the acceleration. In either event, with the acceleration being held as constant, the position at time tn+1, p(tn+1) and velocity at time tn+1, v(tn+1), of the object will be given by:
pi(tn+1) = 0.5*ai(tn)*(tn+1-tn)2 + vi(tn)*(tn+1-tn)+pi(tn)
vi(tn+1) = ai(tn+1)*(tn+1-tn) + vi(tn)
The RK4 method fits the driver of your system to a 2nd degree polynomial which better approximates its behavior. The details are at the wikipedia site I referenced above, and there are a number of other resources you should be able to locate on the web. The basic idea is that instead of picking a single force value for a particular timeslice, you compute four force vectors at specific times and then fit the force vector to the 2nd degree polynomial. That's fine if your field of force vectors doesn't change between time slices. If you're using gravity to derive the vector field, and the objects which are the gravitational sources move, then you need to compute their positions at each of the four sub-intervals in order compute the force vectors. It can be done, but your performance is going to be quite a bit poorer than using Euler's method. On the plus side, you get more accurate motion of the objects relative to each other. So, it's a challenge in the sense that it's computationally expensive, and it's a bit of a pain to figure out where all the objects are supposed to be for your four samples during the time slice of your iteration.
There is no such thing as "continuous" when dealing with computers, so you'll have to approximate continuity with very small intervals of time.
That being said, why are you using a vector field? What's wrong with Newton?
And the sum of the forces on an object is that above equation. Equate the two and solve for a
So you'll just have to loop over all the objects one by one and find the acceleration on it.
Suppose you have a list of 2D points with an orientation assigned to them. Let the set S be defined as:
S={ (x,y,a) | (x,y) is a 2D point, a is an orientation (an angle) }.
Given an element s of S, we will indicate with s_p the point part and with s_a the angle part. I would like to know if there exist an efficient data structure such that, given a query point q, is able to return all the elements s in S such that
(dist(q_p, s_p) < threshold_1) AND (angle_diff(q_a, s_a) < threshold_2) (1)
where dist(p1,p2), with p1,p2 2D points, is the euclidean distance, and angle_diff(a1,a2), with a1,a2 angles, is the difference between angles (taken to be the smallest one). The data structure should be efficient w.r.t. insertion/deletion of elements and the search as defined above. The number of vectors can grow up to 10.000 and more, but take this with a grain of salt.
Now suppose to change the above requirement: instead of using the condition (1), let's request all the elements of S such that, given a distance function d, we want all elements of S such that d(q,s) < threshold. If i remember well, this last setup is called range-search. I don't know if the first case can be transformed in the second.
For the distance search I believe the accepted best method is a Binary Space Partition tree. This can be stored as a series of bits. Each two bits (for a 2D tree) or three bits (for a 3D tree) subdivides the space one more level, increasing resolution.
Using a BSP, locating a set of objects to compare distances with is pretty easy. Just find the smallest set of squares or cubes which contain the edges of your distance box.
For the angle, I don't know of anything. I suppose that you could store each object in a second list or tree sorted by its angle. Then you would find every object at the proper distance using the BSP, every object at the proper angles using the angle tree, then do a set intersection.
You have effectively described a "three dimensional cyclindrical space", ie. a space that is locally three dimensional but where one dimension is topologically cyclic. In other words, it is locally flat and may be modeled as the boundary of a four-dimensional object C4 in (x, y, z, w) defined by
z^2 + w^2 = 1
where
a = arctan(w/z)
With this model, the space defined by your constraints is a 2-dimensional cylinder wrapped "lengthwise" around a cross section wedge, where the wedge wraps around the 4-d cylindrical space with an angle of 2 * threshold_2. This can be modeled using a "modified k-d tree" approach (modified 3-d tree), where the data structure is not a tree but actually a graph (it has cycles). You can still partition this space into cells with hyperplane separation, but traveling along the curve defined by (z, w) in the positive direction may encounter a point encountered in the negative direction. The tree should be modified to actually lead to these nodes from both directions, so that the edges are bidirectional (in the z-w curve direction - the others are obviously still unidirectional).
These cycles do not change the effectiveness of the data structure in locating nearby points or allowing your constraint search. In fact, for the most part, those algorithms are only slightly modified (the simplest approach being to hold a visited node data structure to prevent cycles in the search - you test the next neighbors about to be searched).
This will work especially well for your criteria, since the region you define is effectively bounded by these axis-defined hyperplane-bounded cells of a k-d tree, and so the search termination will leave a region on average populated around pi / 4 percent of the area.
I have a 3D datacube, with two spatial dimensions and the third being a multi-band spectrum at each point of the 2D image.
H[x, y, bands]
Given a wavelength (or band number), I would like to extract the 2D image corresponding to that wavelength. This would be simply an array slice like H[:,:,bnd]. Similarly, given a spatial location (i,j) the spectrum at that location is H[i,j].
I would also like to 'smooth' the image spectrally, to counter low-light noise in the spectra. That is for band bnd, I choose a window of size wind and fit a n-degree polynomial to the spectrum in that window. With polyfit and polyval I can find the fitted spectral value at that point for band bnd.
Now, if I want the whole image of bnd from the fitted value, then I have to perform this windowed-fitting at each (i,j) of the image. I also want the 2nd-derivative image of bnd, that is, the value of the 2nd-derivative of the fitted spectrum at each point.
Running over the points, I could polyfit-polyval-polyder each of the x*y spectra. While this works, this is a point-wise operation. Is there some pytho-numponic way to do this faster?
If you do least-squares polynomial fitting to points (x+dx[i],y[i]) for a fixed set of dx and then evaluate the resulting polynomial at x, the result is a (fixed) linear combination of the y[i]. The same is true for the derivatives of the polynomial. So you just need a linear combination of the slices. Look up "Savitzky-Golay filters".
EDITED to add a brief example of how S-G filters work. I haven't checked any of the details and you should therefore not rely on it to be correct.
So, suppose you take a filter of width 5 and degree 2. That is, for each band (ignoring, for the moment, ones at the start and end) we'll take that one and the two on either side, fit a quadratic curve, and look at its value in the middle.
So, if f(x) ~= ax^2+bx+c and f(-2),f(-1),f(0),f(1),f(2) = p,q,r,s,t then we want 4a-2b+c ~= p, a-b+c ~= q, etc. Least-squares fitting means minimizing (4a-2b+c-p)^2 + (a-b+c-q)^2 + (c-r)^2 + (a+b+c-s)^2 + (4a+2b+c-t)^2, which means (taking partial derivatives w.r.t. a,b,c):
4(4a-2b+c-p)+(a-b+c-q)+(a+b+c-s)+4(4a+2b+c-t)=0
-2(4a-2b+c-p)-(a-b+c-q)+(a+b+c-s)+2(4a+2b+c-t)=0
(4a-2b+c-p)+(a-b+c-q)+(c-r)+(a+b+c-s)+(4a+2b+c-t)=0
or, simplifying,
22a+10c = 4p+q+s+4t
10b = -2p-q+s+2t
10a+5c = p+q+r+s+t
so a,b,c = p-q/2-r-s/2+t, (2(t-p)+(s-q))/10, (p+q+r+s+t)/5-(2p-q-2r-s+2t).
And of course c is the value of the fitted polynomial at 0, and therefore is the smoothed value we want. So for each spatial position, we have a vector of input spectral data, from which we compute the smoothed spectral data by multiplying by a matrix whose rows (apart from the first and last couple) look like [0 ... 0 -9/5 4/5 11/5 4/5 -9/5 0 ... 0], with the central 11/5 on the main diagonal of the matrix.
So you could do a matrix multiplication for each spatial position; but since it's the same matrix everywhere you can do it with a single call to tensordot. So if S contains the matrix I just described (er, wait, no, the transpose of the matrix I just described) and A is your 3-dimensional data cube, your spectrally-smoothed data cube would be numpy.tensordot(A,S).
This would be a good point at which to repeat my warning: I haven't checked any of the details in the few paragraphs above, which are just meant to give an indication of how it all works and why you can do the whole thing in a single linear-algebra operation.