I would like to use Smooth.ppp in spatstat to calculate a sort of "moving average" according to a specific function. The specific distance-dependent weights I would like to use are given by a function wt; for simplicity
wt=function(x,y) exp(-1e5*(x-y)^2)
In the extreme case where wt=kernel, I'd expect no smoothing (ie input marks = smoothed estimates). I'm wondering what I am mis-understanding here about the kernel and how it is applied?
remotes::install_github("spatstat/spatstat.core")
n=4; PPP=ppp(rep(1:n,each=n),rep(1:n,n), c(1,n),c(1,n), marks=1:n^2);
smo=Smooth.ppp(PPP,cutoff=2,kernel=wt,at="points")
rbind(marks(PPP),smo)
(I'm using the latest spatstat build to allow estimates at points using a custom kernel)
This example may have been misinterpreted.
The kernel should be a function(x, y) in the R language which gives the value, at a spatial location (x,y), of the kernel centred at the origin (0,0). Generally the kernel takes its largest values when (x,y) is close to (0,0), and drops to zero when (x,y) is far from (0,0).
The function wt defined in your example has values close to 1 along the diagonal line x = y, and drops to zero rapidly away from the diagonal.
That is unusual. It means that a data point at location (a,b) will be 'smoothed' along the infinite line through the data point with unit slope, with equation y = x + b-a, rather than being smoothed over a region close to (a,b) as it normally would.
The example point pattern PPP consists of points along the diagonal y=x.
The smoothed value at a data point is the weighted average of the mark values at all data points, with weights proportional to the kernel value. In your example, the kernel value for each pair of data points, wt(x1-x2, y1-y2), is equal to 1 because all the data and query points lie on the same line with slope 1.
The kernel weights are all equal in this example, so the smoothed values should all be equal to the average mark value, if leaveoneout=FALSE, and if leaveoneout=TRUE then the smoothed value at data point i is the average of the mark values at the data points excluding point i.
Related
I am using scipy.signal.find_peaks function (link) to determine the peaks of a signal provided.
The signal is loaded in a dataframe like this
x = df["sig_coord"] # the x coordinate of a signal , time msec
Y = df["sig_value"] # value of a signal f(x) - Measured Voltage at the moment x, V
The dataframe has abot 10k points for one signal.
peaks = signal.find_peaks(Y, prominence=2, distance = 40)
the parameter distance actually the measure of how far peaks can be distanced from each other, as I undersood. But the dimension of this parameter must be set in discrete number values - or point number of a signal.
My x-scale is non-monotonic, the distance between pints is changed in a special manner (non-equidistant measurements) so it's unconvenient to use discrete numbers as a dimension of a distance in this case... It will be much better to use a distance provided in a time dimesion (msec in my case). Is it possible to do that?
Maybe there is a way to use find_peaks functionality providing the distance in coordinates that have physical meaning, not strictly discrets number..?
Or maybe it's possible to recalc those values in a simple manner?
Because I have various ratio msec/signal points at various signal parts..
I am working on some shaders, and I need to transform normals.
I read in few tutorials the way you transform normals is you multiply them with the transpose of the inverse of the modelview matrix. But I can't find explanation of why is that so, and what is the logic behind that?
It flows from the definition of a normal.
Suppose you have the normal, N, and a vector, V, a tangent vector at the same position on the object as the normal. Then by definition N·V = 0.
Tangent vectors run in the same direction as the surface of an object. So if your surface is planar then the tangent is the difference between two identifiable points on the object. So if V = Q - R where Q and R are points on the surface then if you transform the object by B:
V' = BQ - BR
= B(Q - R)
= BV
The same logic applies for non-planar surfaces by considering limits.
In this case suppose you intend to transform the model by the matrix B. So B will be applied to the geometry. Then to figure out what to do to the normals you need to solve for the matrix, A so that:
(AN)·(BV) = 0
Turning that into a row versus column thing to eliminate the explicit dot product:
[tranpose(AN)](BV) = 0
Pull the transpose outside, eliminate the brackets:
transpose(N)*transpose(A)*B*V = 0
So that's "the transpose of the normal" [product with] "the transpose of the known transformation matrix" [product with] "the transformation we're solving for" [product with] "the vector on the surface of the model" = 0
But we started by stating that transpose(N)*V = 0, since that's the same as saying that N·V = 0. So to satisfy our constraints we need the middle part of the expression — transpose(A)*B — to go away.
Hence we can conclude that:
transpose(A)*B = identity
=> transpose(A) = identity*inverse(B)
=> transpose(A) = inverse(B)
=> A = transpose(inverse(B))
My favorite proof is below where N is the normal and V is a tangent vector. Since they are perpendicular their dot product is zero. M is any 3x3 invertible transformation (M-1 * M = I). N' and V' are the vectors transformed by M.
To get some intuition, consider the shear transformation below.
Note that this does not apply to tangent vectors.
Take a look at this tutorial:
https://paroj.github.io/gltut/Illumination/Tut09%20Normal%20Transformation.html
You can imagine that when the surface of a sphere stretches (so the sphere is scaled along one axis or something similar) the normals of that surface will all 'bend' towards each other. It turns out you need to invert the scale applied to the normals to achieve this. This is the same as transforming with the Inverse Transpose Matrix. The link above shows how to derive the inverse transpose matrix from this.
Also note that when the scale is uniform, you can simply pass the original matrix as normal matrix. Imagine the same sphere being scaled uniformly along all axes, the surface will not stretch or bend, nor will the normals.
If the model matrix is made of translation, rotation and scale, you don't need to do inverse transpose to calculate normal matrix. Simply divide the normal by squared scale and multiply by model matrix and we are done. You can extend that to any matrix with perpendicular axes, just calculate squared scale for each axes of the matrix you are using instead.
I wrote the details in my blog: https://lxjk.github.io/2017/10/01/Stop-Using-Normal-Matrix.html
Don't understand why you just don't zero out the 4th element of the direction vector before multiplying with the model matrix. No inverse or transpose needed. Think of the direction vector as the difference between two points. Move the two points with the rest of the model - they are still in the same relative position to the model. Take the difference between the two points to get the new direction, and the 4th element, cancels out to zero. Lot cheaper.
I need to offset a curve, which by the simplest way is just shifting the points perpendicularly. I can access each point to calculate angle of each line along given path, for now I use atan2. Then I take those two angle and make average of it. It returns the shortest angle, not what I need in this case.
How can I calculate angle of each connection? Concerning that I am not interested in the shortest angle but the one that would create parallel offset curve.
Assuming 2D case...
So do a cross product of direction vectors of 2 neighboring lines the sign of z coordinate of the result will tell you if the lines are CW/CCW
So if you got 3 consequent control points on the polyline: p0,p1,p2 then:
d1 = p1-p0
d2 = p2-p1
if you use some 3D vector math then convert them to 3D by setting:
d1.z=0;
d2.z=0;
now compute 3D cross:
n = cross(d1,d2)
which returns vector perpendicular to both vectors of size equals to the area of quad (parallelogram) constructed with d1,d2 as base vectors. The direction (from the 2 possible) is determined by the winding rule of the p0,p1,p2 so inspecting z of the result is enough.
The n.x,n.y are not needed so you can compute directly without doing full cross product:
n.z=(d1.x*d2.y)-(d1.y*d2.x)
if (n.z>0) case1
if (n.z<0) case2
if the case1 is CW or CCW depends on your coordinate system properties (left/right handness). This approach is very commonly used in CG fur back face culling of polygons ...
if n.z is zero it means that your vectors/lines are either parallel or at lest one of them is zero.
I think these might interest you:
draw outline for some connected lines
How can I create an internal spiral for a polygon?
Also in 2D you do not need atan2 to get perpendicular vector... You can do instead this:
u = (x,y)
v = (-y,x)
w = (x,-y)
so u is any 2D vector and v,w are the 2 possible perpendicular vectors to u in 2D. they are the result of:
cross((x,y,0),(0,0,1))
cross((0,0,1),(x,y,0))
When I blur an object in Inkscape by let's say 10%, it get's a filter with a feGaussionBlur with a stdDeviation of 10% * size / 2.
However the filter has a size of 124% (it is actually that big, Inkscape doesn't add a bit just to be on the safe-side).
Where does this number come from? My guess would be 100% + 2.4 * (2*stdDeviation/size), but then where does this 2.4 come from?
From the SVG 1.1 spec:
This filter primitive performs a Gaussian blur on the input image.
The Gaussian blur kernel is an approximation of the normalized convolution:
G(x,y) = H(x)I(y)
where
H(x) = exp(-x2/ (2s2)) / sqrt(2* pis2)
and
I(y) = exp(-y2/ (2t2)) / sqrt(2 pi*t2)
with 's' being the standard deviation in the x direction and 't' being the standard deviation in the y direction, as specified by ‘stdDeviation’.
The value of ‘stdDeviation’ can be either one or two numbers. If two numbers are provided, the first number represents a standard deviation value along the x-axis of the current coordinate system and the second value represents a standard deviation in Y. If one number is provided, then that value is used for both X and Y.
Even if only one value is provided for ‘stdDeviation’, this can be implemented as a separable convolution.
For larger values of 's' (s >= 2.0), an approximation can be used: Three successive box-blurs build a piece-wise quadratic convolution kernel, which approximates the Gaussian kernel to within roughly 3%.
let d = floor(s * 3*sqrt(2*pi)/4 + 0.5)
... if d is odd, use three box-blurs of size 'd', centered on the output pixel.
... if d is even, two box-blurs of size 'd' (the first one centered on the pixel boundary between the output pixel and the one to the left, the second one centered on the pixel boundary between the output pixel and the one to the right) and one box blur of size 'd+1' centered on the output pixel.
Note: the approximation formula also applies correspondingly to 't'.*
I am trying to generate a certain amount of random uniform points inside a rectangle (I know the pair of coordinates for each corner).
Let our rectangle be
ABCD
My idea is:
Divide the rectangle into two triangles by the AC diagonal. Find the slope and the intercept of the diagonal.
Then, generate two random numbers from [0,1] interval, let them be a,b.
Evaluate x = aAB and y = bAD (AB, AD, distances). If A is not (0,0), then we can add to x and y A's coordinates.
Now we have a point (x,y). If it is not in the lower triangle (ABC), skip to the next step.
Else, add the point to our plot and also add the symmetric of (x,y) vs. the AC diagonal so that we can fill the upper triangle (ADC) too.
I have implemented this, but I highly doubt that the points are uniformly generated (judging from the plot). How should I modify my algorithm? I guess that the issue is related to how I pick the triangle and the symmetric thing.
Why not just generate x=random([A.x, B.x]) and y=random([B.y, C.y]) and put them together as (x,y)? A n-dimensional uniform distribution is simply the product of the n uniform distributions of the components.
This is referred to as point picking and other similar terms. You seem to be on the right track in that the points should come from the uniform distribution. Your plot looks reasonably random to me.
What are you doing with upper and lower triangles? They seem unnecessary and would certainly make things less random. Is this some form variance reduction along the lines of antithetic variates? If #Paddy3118 is right an you really just need random-ish points to fill the space, then you should look into low-discrepancy sequences. The Halton sequence generalizes the van der Corput sequence to multiple dimensions. If you have Matlab's Statistics Toolbox check out the sobolset and haltonset functions or qrandstream and qrand.
This approach (from #Xipan Xiao & #bonanova.) should be reproducible in many languages. MATLAB code below.
a = 0; b = 1;
n = 2000;
X = a + (b-a)*rand(n,1);
Y = a + (b-a)*rand(n,1);
Newer versions of MATLAB can make use of the makedist and random commands.
pdX = makedist('Uniform',a,b);
pdY = makedist('Uniform',a,b);
X = random(pdX,n,1);
Y = random(pdY,n,1);
The points (X,Y) will be uniformly in the rectangle with corner points (a,a), (a,b), (b,a), (b,b).
For verification, we can observe the marginal distributions for X and Y and see that those are uniform as well.
scatterhist(X,Y,'Marker','.','Direction','out')
Update: Using haltonset (suggested by #horchler)
p = haltonset(2);
XY = net(p,2000);
scatterhist(XY(:,1),XY(:,2),'Marker','.','Direction','out')
If you are after a more uniform density then you might consider a Van der Corput sequence. The sequence finds use in Monte-Carlo simulations and Wolfram Mathworld calls them a quasi-random sequence.
Generate two random numbers in the interval [0,1] translate and scale them to your rectangle as x and y.
There is just my thought, i haven't test with code yet.
1.Divide the rectangle to grid with N x M cells, depends on variable density.
2.loop through the cell and pick a random point in the cell until it reached your target point quantity.