I had a raster with values from 0 to 0,3 which I transformed into an image. Then i gave all values <=0.3 the value 1. I thought it makes it easier if i calculate this for a single value. I applied Gaussian smoothing to the image and then converted it back to a raster. For the smoothing I used the Smooth.Im function of the spatstat package. However, I do not know which unit my scale has. Does it have something to do with pixel density or how can I understand the unit? I have attached an image as an example
Thank you and best regards
Smooth.im is a function in the spatstat package family. It performs kernel smoothing of the input image.
The value of the output at a pixel i, say, is equal to a weighted average of the values of the input at pixels j with weights w(i,j). The weights sum to 1 (i.e. for any i, the sum of w(i,j) over all j is equal to 1.) The weights w(i,j) get smaller as the distance between i and j increases. So, the output pixel value is basically an average of the input pixel values in a neighbourhood.
If the input image pixel values were measurements expressed in some unit (say weights expressed kilograms), then the output image pixel values are expressed in the same unit, and are averages of the input values.
If I understand your question, your input image has only the pixel values 0 and 1. The output image pixel values are weighted averages of these 0/1 values, which may lie anywhere between 0 and 1.
For further explanation see Chapter 6 of the spatstat book
ıf we talk about image filtering etc., for example blurring or deblurring, the applied filters obey the rule of energy conservation which is satisfied when the sum of the members of the filter is equal to '1'. So, if your filter obey this rule, ı dont think your unit is changed after smoothing process.
Related
I'm having trouble interpreting the results I got from relrisk. My data is a multiple point process containing two marks (two rodents species AA and RE), I want to know if they are spatially segregated or not.
> summary(REkm)
Marked planar point pattern: 46 points
Average intensity 0.08101444 points per square unit
*Pattern contains duplicated points*
Coordinates are given to 3 decimal places
i.e. rounded to the nearest multiple of 0.001 units
Multitype:
frequency proportion intensity
AA 15 0.326087 0.02641775
RE 31 0.673913 0.05459669
Window: rectangle = [4, 38] x [0.3, 17] units
x 16.7 units)
Window area = 567.8 square units
relkm <- relrisk(REkm)
plot(relkm, main="Relrisk default")
The bandwidth of this relrisk estimation is automatically selection by default(bw.relrisk), but when I tried other numeric number using sigma= 0.5 or 1, the results are somehow kind of weird.
How did this happened? Was it because the large proportion of blank space of my ppp?
According to chapter.14 of Spatial Point Patterns books and the previous discussion, I assume the default of relrisk will show the ratio of intensities (case divided by control, in my case: RE divided by AA), but if I set casecontrol=FALSE, I can get the spatially-varying probability of each type.
Then why the image of type RE in the Casecontrol=False looks exactly same as the relrisk estimation by default? Or they both estimate p(RE)=λRE/ λRE+λAA for each sites?
Any help will be appreciated! Thanks a lot!
That's two questions.
Why does the image for RE when casecontrol=FALSE look the same as the default output from relrisk?
The definitive source of information about spatstat functions is the online documentation in the help files. The help file for relrisk.ppp gives full details of the behaviour of this function. It says that the calculation of probabilities and risks is controlled by the argument relative. If relative=FALSE (the default), the code calculates the spatially varying probability of each type. If relative=TRUE it calculates the relative risk of each type i, defined as the ratio of the probability of type i to the probability of type c where c is the type designated as the control. If you wanted the relative risk then you should set relative=TRUE.
Very different results obtained when setting sigma=0.5 compared to the automatically selected bandwidth.
Your example output says that the window is 34 by 17 units. A smoothing bandwidth of sigma=0.5 is very small for this region. Imagine each data point being replaced by a blurry circle of radius 0.5; there would be a lot of empty space. The smoothing procedure is encountering numerical problems which are causing the funky artefacts.
You could try a range of different values of sigma, say from 1 to 15, and decide which value produces the most satisfactory result.
The plot of relrisk(REkm, casecontrol=FALSE) suggests that the automatic bandwidth selector bw.relriskppp chose a much larger value of sigma, perhaps about 10. You can investigate this by
b <- bw.relriskppp(REkm)
print(b)
plot(b)
The print command will print the chosen value of sigma that was used in the default calculation. The plot command will show the cross-validation criterion which was maximised to select the bandwidth. This gives you an idea of the range of values of sigma that are acceptable according to the automatic selector.
Read the help file for bw.relriskppp about the different options available for bandwidth selection method. Maybe a different choice of method would give you a more acceptable result from your viewpoint.
I am trying to learn some facial landmark detection model, and notice that many of them use NME(Normalized Mean Error) as performance metric:
The formula is straightforward, it calculate the l2 distance between ground-truth points and model prediction result, then divided it by a normalized factor, which vary from different dataset.
However, when adopting this formula on some landmark detector that some one developed, i have to deal with this non-trivial situation, that is some detector may not able to generate enough number landmarks for some input image(might because of NMS/model inherited problem/image quality etc). Thus some of ground-truth points might not have their corresponding one in the prediction result.
So how to solve this problem, should i just add such missing point result to "failure result set" and use FR to measure the model, and ignore them when doing the NME calculation?
If you have as output of neural network an vector 10x1 as example
that is your points like [x1,y1,x2,y2...x5,y5]. This vector will be fixed length cause of number of neurons in your model.
If you have missing points - this is because (as example you have 4 from 5 points) some points are go beyond the image width and height. Or are with minus (negative) like [-0.1, -0.2, 0.5,0.7 ...] there first 2 points you can not see on image like they are mission but they will be in vector and you can callculate NME.
In some custom neural nets that can be possible, because missing values will be changed to biggest error points.
I am trying to convert an heightmap into a matrix of normals using central differencing which will later correspond to the steepness of a giving point.
I found several links with correct results but without explaining the math behind.
T
L O R
B
From this link I realised I can just do:
Vec3 normal = Vec3(2*(R-L), 2*(B-T), -4).Normalize();
The thing is that I don't know where the 2* and -4 comes from.
In this explanation of central differencing I see that we should divide that value by 2, but I still don't know how to connect all of this.
What I really want to know is the linear algebra definition behind this.
I have an heightmap, I want to measure the central differences and I want to obtain the normal vector to use later to measure the steepness.
PS: the Z-axis is the height.
From vector calculus, the normal of a surface is given by the gradient operator:
A height map h(x, y) is a special form of the function f:
For a discretized height map, assuming that the grid size is 1, the first-order approximations to the two derivative terms above are given by:
Since the x step from L to R is 2, and same for y. The above is exactly the formula you had, divided through by 4. When this vector is normalized, the factor of 4 is canceled.
(No linear algebra was harmed in the writing of this answer)
Given some x data points in an N dimensional space, I am trying to find a fixed length representation that could describe any subset s of those x points? For example the mean of the s subset could describe that subset, but it is not unique for that subset only, that is to say, other points in the space could yield the same mean therefore mean is not a unique identifier. Could anyone tell me of a unique measure that could describe the points without being number of points dependent?
In short - it is impossible (as you would achieve infinite noiseless compression). You have to either have varied length representation (or fixed length with length being proportional to maximum number of points) or dealing with "collisions" (as your mapping will not be injective). In the first scenario you simply can store coordinates of each point. In the second one you approximate your point clouds with more and more complex descriptors to balance collisions and memory usage, some posibilities are:
storing mean and covariance (so basically perofming maximum likelihood estimation over Gaussian families)
performing some fixed-complexity density estimation like Gaussian Mixture Model or training a generative Neural Network
use set of simple geometrical/algebraical properties such as:
number of points
mean, max, min, median distance between each pair of points
etc.
Any subset can be identified by a bit mask of length ceiling(lg(x)), where bit i is 1 if the corresponding element belongs to the subset. There is no fixed-length representation that is not a function of x.
EDIT
I was wrong. PCA is a good way to perform dimensionality reduction for this problem, but it won't work for some sets.
However, you can almost do it. Where "almost" is formally defined by the Johnson-Lindenstrauss Lemma, which states that for a given large dimension N, there exists a much lower dimension n, and a linear transformation that maps each point from N to n, while keeping the Euclidean distance between every pair of points of the set within some error ε from the original. Such linear transformation is called the JL Transform.
In other words, your problem is only solvable for sets of points where each pair of points are separated by at least ε. For this case, the JL Transform gives you one possible solution. Moreover, there exists a relationship between N, n and ε (see the lemma), such that, for example, if N=100, the JL Transform can map each point to a point in 5D (n=5), an uniquely identify each subset, if and only if, the minimum distance between any pair of points in the original set is at least ~2.8 (i.e. the points are sufficiently different).
Note that n depends only on N and the minimum distance between any pair of points in the original set. It does not depend on the number of points x, so it is a solution to your problem, albeit some constraints.
For the question "Ellipse around the data in MATLAB", in the answer given by Amro, he says the following:
"If you want the ellipse to represent
a specific level of standard
deviation, the correct way of doing is
by scaling the covariance matrix"
and the code to scale it was given as
STD = 2; %# 2 standard deviations
conf = 2*normcdf(STD)-1; %# covers around 95% of population
scale = chi2inv(conf,2); %# inverse chi-squared with dof=#dimensions
Cov = cov(X0) * scale;
[V D] = eig(Cov);
I don't understand the first 3 lines of the above code snippet. How is the scale calculated by chi2inv(conf,2), and what is the rationale behind multiplying it with the covariace matrix?
Additional Question:
I also found that if I scale it with 1.5 STD, i.e. 86% tiles, the ellipse can cover all of the points, my points set are clumping together, at almost all the cases. On the other hand, if I scale it with 3 STD, i.e. 99%tiles, the ellipse is far too big. Then how can I choose a STD to just tightly cover the clumping points?
Here is an example:
The inner ellipse corresponds to 1.5 STD and outer to 2.5 STD. why 1.5 STD is tightly cover the clumping white points? Is there any approach or reason to define it?
The objective of displaying an ellipse around the data points is to show the confidence interval, or in other words, "how much of the data is within a certain standard deviation way from the mean"
In the above code, he has chosen to display an ellipse that covers 95% of the data points. For a normal distribution, ~67% of the data is 1 s.d. away from the mean, ~95% within 2 s.d. and ~99% within 3 s.d. (the numbers are off the top of my head, but you can easily verify this by calculating the area under the curve). Hence, the value STD=2; You'll find that conf is approx 0.95.
The distance of the data points from the centroid of the data goes something like (xi^2+yi^2)^0.5, ignoring coefficients. Sums of squares of random variables follow a chi-square distribution and hence to get the corresponding 95 percentile, he uses the inverse chi-square function, with d.o.f. 2, as there are two variables.
Lastly, the rationale behind multiplying the scaling constant follows from the fact that for a square matrix A with eigenvalues a1,...,an, the eigenvalues of a matrix kA, where k is a scalar is simply ka1,...,kan. The eigenvalues give the corresponding lengths of the major/minor axis of the ellipse, and so scaling the ellipse or the eigenvalues to the 95%tile is equivalent to multiplying the covariance matrix with the scaling factor.
EDIT
Cheng, although you might already know this, I suggest that you also read this answer to a question on randomness. Consider a Gaussian random variable with zero mean, unit variance. The PDF of a collection of such random variables looks like this
Now, if I were to take two such collections of random variables, square them separately and add them to form a single collection of a new random variable, its distribution looks like this
This is the chi-square distribution with 2 degrees of freedom (since we added two collections).
The equation of the ellipse in the above code can be written as x^2/a^2 +y^2/b^2=k, where x,y are the two random variables, a and b are the major/minor axes, and k is some scaling constant that we need to figure out. As you can see, the above can be interpreted as squaring and adding two collections of Gaussian random variables, and we just saw above what its distribution looks like. So, we can say that k is a random variable that is chi-square distributed with 2 degrees of freedom.
Now all that needs to be done is to find a value for k such that 95%ile of the data is within it. Just like the 1s.d, 2s.d, 3s.d. percentiles that we're familiar with Gaussians, the 95%tile for chi-square with 2 degrees of freedom is around 6.18. This is what Amro obtains from the chi2inv function. He could have just as well written scale=chi2inv(0.95,2) and it would have been the same. It's just that talking in terms of n s.d. away from the mean is intuitive.
Just to illustrate, here's a PDF of the chi-square distribution above, with 95% of the area < some x shaded in red. This x is ~6.18.
Hope this helped.