I have a point point=[x y] and a vector vec=[X Y], where X and Y are vectors containing x,y values for many point. I have calculated the euclidean distances of all the points in vec from point with the following code:
diff=vec-point;
squared=diff.*diff;
distances=sqrt(sum(squared,2));
I have seen the pdist() function, but could not find a good way to use it in my code. Is there a more elegant way to do so?
You can use something like m = [point; vec] and then distances=pdist(m, 'euclidean'), however it will compute O((n+1)ˆ2) distances rather than the O(n) you need. If the code is not performance critical I wouldn't worry about it and just use the code which is more elegant and easier to understand.
Related
I have a 2D 2401*266 matrix K which corresponds to x values (t: stored in a 1*266 array) and y values(z: stored in a 1*2401 array).
I want to extrapolate the matrix K to predict some future values (corresponding to t(1,267:279). So far I have extended t so that it is now a 1*279 matrix using a for loop:
for tq = 267:279
t(1,tq) = t(1,tq-1)+0.0333333333;
end
However I am stumped on how to extrapolate K without fitting a polynomial to each individual row?
I feel like there must be a more efficient way than this??
There are countless of extrapolation methods in the literature, "fitting a polynomial to each row" would be just one of them, not necessarily invalid, not sure why you mention that you do no wan't to do it. For 2D data perhaps fitting a surface would lead to better results though.
However, if you want an easy, simple way (that might or might not work with your problem), you can always use the function interp2, for interpolation. If you chose spline or makima as interpolation functions, it will also extrapolate for any query point outside the domain of K.
I have an image of a cytoskeleton. There are a lot of small objects inside and I want to calculate the length between all of them in every axis and to get a matrix with all this data. I am trying to do this in matlab.
My final aim is to figure out if there is any axis with a constant distance between the object.
I've tried bwdist and to use connected components without any luck.
Do you have any other ideas?
So, the end goal is that you want to globally stretch this image in a certain direction (linearly) so that the distances between nearest pairs end up the closest together, hopefully the same? Or may you do more complex stretching ? (note that with arbitrarily complex one you can always make it work :) )
If linear global one, distance in x' and y' is going to be a simple multiplication of the old distance in x and y, applied to every pair of points. So, the final euclidean distance will end up being sqrt((SX*x)^2 + (SY*y)^2), with SX being stretch in x and SY stretch in y; X and Y are distances in X and Y between pairs of points.
If you are interested in just "the same" part, solution is not so difficult:
Find all objects of interest and put their X and Y coordinates in a N*2 matrix.
Calculate distances between all pairs of objects in X and Y. You will end up with 2 matrices sized N*N (with 0 on the diagonal, symmetric and real, not sure what is the name for that type of matrix).
Find minimum distance (say this is between A an B).
You probably already have this. Now:
Take C. Make N-1 transformations, which all end up in C->nearestToC = A->B. It is a simple system of equations, you have X1^2*SX^2+Y1^2*SY^2 = X2^2*SX^2+Y2*SY^2.
So, first say A->B = C->A, then A->B = C->B, then A->B = C->D etc etc. Make sure transformation is normalized => SX^2 + SY^2 = 1. If it cannot be found, the only valid transformation is SX = SY = 0 which means you don't have solution here. Obviously, SX and SY need to be real.
Note that this solution is unique except in case where X1 = X2 and Y1 = Y2. In this case, grab some other point than C to find this transformation.
For each transformation check the remaining points and find all nearest neighbours of them. If distance is always the same as these 2 (to a given tolerance), great, you found your transformation. If not, this transformation does not work and you should continue with the next one.
If you want a transformation that minimizes variations between distances (but doesn't require them to be nearly equal), I would do some optimization method and search for a minimum - I don't know how to find an exact solution otherwise. I would pick this also in case you don't have linear or global stretch.
If i understand your question correctly, the first step is to obtain all of the objects center of mass points in the image as (x,y) coordinates. Then, you can easily compute all of the distances between all points. I suggest taking a look on a histogram of those distances which may provide some information as to the nature of distance distribution (for example if it is uniformly random, or are there any patterns that appear).
Obtaining the center of mass points is not an easy task, consider transforming the image into a binary one, or some sort of background subtraction with blob detection or/and edge detector.
For building a histogram you can use histogram.
I am using pdist2(x(i), y(j), 'euclidean') formula to find euclidean distance between x and y instead of the manual method
sqrt((x(i)-y(i))^2).
And to find co relation coefficient I am using corrcoeff( x(i), y(j)) . Is this the correct way to find correlation coefficient and euclidean distance between x and y matrix?
I get different answers when I use formula and manual method.
I think it is not correct.
I suppose x and y are matrixes, so you are using pdist2 to compute the distance between the observation i (from x(i)) and j (from y(j)). In the case of the manual method you are using the i index in both of them. Maybe the differences are due to the incorrect use of the indexes i and j. Show us your code so we can confirm it.
By the way, as #Luis pointed out, it is better to use pdist2 to compute all the distances at the same time (it is much faster). So, if you have two matrixes x and y, use: pdist2(x,y).
The same for the correlation.
The correlation between the two matrices can be calculated as:
r = corr2(x,y)
Now, if you are after the element-wise distance, how about:
dist=gsqrt((x-y).^2);
I need to pre-compute the histogram intersection kernel matrices for using LIBSVM in MATLAB.
Assume x, y are two vectors. The kernel function is K(x, y) = sum(min(x, y)). In order to be efficient, the best practice in most cases is to vectorize the operations.
What I want to do is like calculate the kernel matrices like calculating the euclidean distance between two matrices, like pdist2(A, B, 'euclidean'). After defining function 'intKernel', I could calculate the intersection kernel by calling pdist2(A, B, intKernel).
I know the function 'pdist2' may be an option. But I have no idea how to write the self-defined distance function. While, I do not know how to code the intersection kernel between vector(1-by-M) and matrix(M-by-N) in one condense expression.
'repmat' may not be feasible, because the matrix is really large, let us say, 20000-by-360000.
Any help would be appreciated.
Regards,
Peiyun
I think pdist2 is a good option, so I help you to define your distance function.
According to the doc, the self-defined distance function must have 2 inputs: first one is a 1-by-N vector; second one is a M-by-N matrix (be careful of the order!).
To avoid the use of repmat which is indeed memory-consumant, you can use bsxfun to apply some basic operations on data with expansion over singleton dimensions. In your case, you can do the following thing:
distance_kernel = #(x,Y) sum(bsxfun(#min,x,Y),2);
Summation is done over the columns to get a column vector as output.
Then just call pdist2 and you are done.
I would like to use a MATLAB function to find the minimum length between a point and a curve? The curve is described by a complicated function that is not quite smooth. So I hope to use an existing tool of matlab to compute this. Do you happen to know one?
When someone says "its complicated" the answer is always complicated too, since I never know exactly what you have. So I'll describe some basic ideas.
If the curve is a known nonlinear function, then use the symbolic toolbox to start with. For example, consider the function y=x^3-3*x+5, and the point (x0,y0) =(4,3) in the x,y plane.
Write down the square of the distance. Euclidean distance is easy to write.
(x - x0)^2 + (y - y0)^2 = (x - 4)^2 + (x^3 - 3*x + 5 - 3)^2
So, in MATLAB, I'll do this partly with the symbolic toolbox. The minimal distance must lie at a root of the first derivative.
sym x
distpoly = (x - 4)^2 + (x^3 - 3*x + 5 - 3)^2;
r = roots(diff(distpoly))
r =
-1.9126
-1.2035
1.4629
0.82664 + 0.55369i
0.82664 - 0.55369i
I'm not interested in the complex roots.
r(imag(r) ~= 0) = []
r =
-1.9126
-1.2035
1.4629
Which one is a minimzer of the distance squared?
subs(P,r(1))
ans =
35.5086
subs(P,r(2))
ans =
42.0327
subs(P,r(3))
ans =
6.9875
That is the square of the distance, here minimized by the last root in the list. Given that minimal location for x, of course we can find y by substitution into the expression for y(x)=x^3-3*x+5.
subs('x^3-3*x+5',r(3))
ans =
3.7419
So it is fairly easy if the curve can be written in a simple functional form as above. For a curve that is known only from a set of points in the plane, you can use my distance2curve utility. It can find the point on a space curve spline interpolant in n-dimensions that is closest to a given point.
For other curves, say an ellipse, the solution is perhaps most easily solved by converting to polar coordinates, where the ellipse is easily written in parametric form as a function of polar angle. Once that is done, write the distance as I did before, and then solve for a root of the derivative.
A difficult case to solve is where the function is described as not quite smooth. Is this noise or is it a non-differentiable curve? For example, a cubic spline is "not quite smooth" at some level. A piecewise linear function is even less smooth at the breaks. If you actually just have a set of data points that have a bit of noise in them, you must decide whether to smooth out the noise or not. Do you wish to essentially find the closest point on a smoothed approximation, or are you looking for the closest point on an interpolated curve?
For a list of data points, if your goal is to not do any smoothing, then a good choice is again my distance2curve utility, using linear interpolation. If you wanted to do the computation yourself, if you have enough data points then you could find a good approximation by simply choosing the closest data point itself, but that may be a poor approximation if your data is not very closely spaced.
If your problem does not lie in one of these classes, you can still often solve it using a variety of methods, but I'd need to know more specifics about the problem to be of more help.
There's two ways you could go about this.
The easy way that will work if your curve is reasonably smooth and you don't need too high precision is to evaluate your curve at a dense number of points and simply find the minimum distance:
t = (0:0.1:100)';
minDistance = sqrt( min( sum( bxsfun(#minus, [x(t),y(t)], yourPoint).^2,2)));
The harder way is to minimize a function of t (or x) that describes the distance
distance = #(t)sum( (yourPoint - [x(t),y(t)]).^2 );
%# you can use the minimum distance from above as a decent starting guess
tAtMin = fminsearch(distance,minDistance);
minDistanceFitte = distance(tAtMin);