If I have the following vectors:
55 degrees, 200 magnitude, start at origin
35 degrees, 130 magnitude, start at head of vector 1
How do I visualize them?
I would expect there to be a more elegant way to express this, but while
this does not put the vectors "tip-to-tail" (sort of), it does compute and display the resulting vector:
vector (200*cos(55),200*sin(55))+(130*cos(35),130*sin(35))
Related
I have x,y data points.
Using these points, I am trying to create a function to fit 50 (y points)points to generate the corresponding x coordinates.
But in my plot, when I try to zoom, the plot, I can see the 50 points provided is fitting the curve, but data points are slightly deviating from the plot. There is a small change from data point (in the range on delta=.001) with respect to the line generated from the 50 points if I zoom.
How do I generate a perfect curve which fits the data points along with the 50 points provided.
please refer the screenshot of the code
To cover 50 points perfectly you need to increase the order of the polynom. So instead of polyfit(x, y, 10) try polyfit(x, y, 49) ?
See https://arachnoid.com/polysolve/
A "perfect" fit (one in which all the data points are matched) can often be gotten by setting the degree of the regression to the number of data pairs minus one.
I just digged a bit through the Tensorflow Object Detection API code especially the eval_util part, as I wanted to implement the COCO metrics.
But I noticed that the metrics are solely calculated using the bounding boxes which have normalized coordinates between [0, 1].
There are no aspect ratios or absolute coordinates used.
So, doesn't this mean that the intersection over unions calculated on these results are incorrect?
Let's take an 200x100 image pixel as an example.
If the box would be off by 20px to the left, that's 0.1 in normalized coordinates.
But if it would be off by 20px to the top, that would be 0.2 in normalized coordinates.
Doesn't that mean, being off to the top is harder penalizing the score than being off to the side?
I believe the predicted coordinates are resized to the absolute image coordinates in the eval binary.
But the other thing I would say is that IOU is scale invariant in the sense that if you scale two boxes by some factor, they will still have the same IOU overlap. As an example if we scale by 2 in the x-direction and scale by 3 in the y direction:
If A is (x1, y1, x2, y2) and B is (u1, v1, u2, v2), then IOU((A, B))
= IOU((2*x1, 3*y1, 2*x2, 3*y2), (2*u1, 3*v1, 2*u2, 3*v2))
What this means is that evaluating in normalized coordinates should give the same result as evaluating in absolute coordinates.
Is there a way to chose the x/y output axes range from np.fft2 ?
I have a piece of code computing the diffraction pattern of an aperture. The aperture is defined in a 2k x 2k pixel array. The diffraction pattern is basically the inner part of the 2D FT of the aperture. The np.fft2 gives me an output array same size of the input but with some preset range of the x/y axes. Of course I can zoom in by using the image viewer, but I have already lost detail. What is the solution?
Thanks,
Gert
import numpy as np
import matplotlib.pyplot as plt
r= 500
s= 1000
y,x = np.ogrid[-s:s+1, -s:s+1]
mask = x*x + y*y <= r*r
aperture = np.ones((2*s+1, 2*s+1))
aperture[mask] = 0
plt.imshow(aperture)
plt.show()
ffta= np.fft.fft2(aperture)
plt.imshow(np.log(np.abs(np.fft.fftshift(ffta))**2))
plt.show()
Unfortunately, much of the speed and accuracy of the FFT come from the outputs being the same size as the input.
The conventional way to increase the apparent resolution in the output Fourier domain is by zero-padding the input: np.fft.fft2(aperture, [4 * (2*s+1), 4 * (2*s+1)]) tells the FFT to pad your input to be 4 * (2*s+1) pixels tall and wide, i.e., make the input four times larger (sixteen times the number of pixels).
Begin aside I say "apparent" resolution because the actual amount of data you have hasn't increased, but the Fourier transform will appear smoother because zero-padding in the input domain causes the Fourier transform to interpolate the output. In the example above, any feature that could be seen with one pixel will be shown with four pixels. Just to make this fully concrete, this example shows that every fourth pixel of the zero-padded FFT is numerically the same as every pixel of the original unpadded FFT:
# Generate your `ffta` as above, then
N = 2 * s + 1
Up = 4
fftup = np.fft.fft2(aperture, [Up * N, Up * N])
relerr = lambda dirt, gold: np.abs((dirt - gold) / gold)
print(np.max(relerr(fftup[::Up, ::Up] , ffta))) # ~6e-12.
(That relerr is just a simple relative error, which you want to be close to machine precision, around 2e-16. The largest error between every 4th sample of the zero-padded FFT and the unpadded FFT is 6e-12 which is quite close to machine precision, meaning these two arrays are nearly numerically equivalent.) End aside
Zero-padding is the most straightforward way around your problem. But it does cost you a lot of memory. And it is frustrating because you might only care about a tiny, tiny part of the transform. There's an algorithm called the chirp z-transform (CZT, or colloquially the "zoom FFT") which can do this. If your input is N (for you 2*s+1) and you want just M samples of the FFT's output evaluated anywhere, it will compute three Fourier transforms of size N + M - 1 to obtain the desired M samples of the output. This would solve your problem too, since you can ask for M samples in the region of interest, and it wouldn't require prohibitively-much memory, though it would need at least 3x more CPU time. The downside is that a solid implementation of CZT isn't in Numpy/Scipy yet: see the scipy issue and the code it references. Matlab's CZT seems reliable, if that's an option; Octave-forge has one too and the Octave people usually try hard to match/exceed Matlab.
But if you have the memory, zero-padding the input is the way to go.
I have a set of first 25 Zernike polynomials. Below are shown few in Cartesin co-ordinate system.
z2 = 2*x
z3 = 2*y
z4 = sqrt(3)*(2*x^2+2*y^2-1)
:
:
z24 = sqrt(14)*(15*(x^2+y^2)^2-20*(x^2+y^2)+6)*(x^2-y^2)
I am not using 1st since it is piston; so I have these 24 two-dim ANALYTICAL functions expressed in X-Y Cartesian co-ordinate system. All are defined over unit circle, as they are orthogonal over unit circle. The problem which I am describing here is relevant to other 2D surfaces also apart from Zernike Polynomials.
Suppose that origin (0,0) of the XY co-ordinate system and the centre of the unit circle are same.
Next, I take linear combination of these 24 polynomials to build a 2D wavefront shape. I use 24 random input coefficients in this combination.
w(x,y) = sum_over_i a_i*z_i (i=2,3,4,....24)
a_i = random coefficients
z_i = zernike polynomials
Upto this point, everything is analytical part which can be done on paper.
Now comes the discretization!
I know that when you want to re-construct a signal (1Dim/2Dim), your sampling frequency should be at least twice the maximum frequency present in the signal (Nyquist-Shanon principle).
Here signal is w(x,y) as mentioned above which is nothing but a simple 2Dim
function of x & y. I want to represent it on computer now. Obviously I can not take all infinite points from -1 to +1 along x axis and same for y axis.
I have to take finite no. of data points (which are called sample points or just samples) on this analytical 2Dim surface w(x,y)
I am measuring x & y in metres, and -1 <= x <= +1; -1 <= y <= +1.
e.g. If I divide my x-axis from -1 to 1, in 50 sample points then dx = 2/50= 0.04 metre. Same for y axis. Now my sampling frequency is 1/dx i.e. 25 samples per metre. Same for y axis.
But I took 50 samples arbitrarily; I could have taken 10 samples or 1000 samples. That is the crux of the matter here: how many samples points?How will I determine this number?
There is one theorem (Nyquist-Shanon theorem) mentioned above which says that if I want to re-construct w(x,y) faithfully, I must sample it on both axes so that my sampling frequency (i.e. no. of samples per metre) is at least twice the maximum frequency present in the w(x,y). This is nothing but finding power spectrum of w(x,y). Idea is that any function in space domain can be represented in spatial-frequency domain also, which is nothing but taking Fourier transform of the function! This tells us how many (spatial) frequencies are present in your function w(x,y) and what is the maximum frequency out of these many frequencies.
Now my question is first how to find out this maximum sampling frequency in my case. I can not use MATLAB fft2() or any other tool since it means already I have samples taken across the wavefront!! Obviously remaining option is find it analytically ! But that is time consuming and difficult since I have 24 polynomials & I will have to use then continuous Fourier transform i.e. I will have to go for pen and paper.
Any help will be appreciated.
Thanks
Key Assumptions
You want to use the "Nyquist-Shanon" theorem to determine sampling frequency
Obviously remaining option is find it analytically ! But that is time
consuming and difficult since I have 21 polynomials & I have to use
continuous Fourier transform i.e. done by analytically.
Given the assumption I have made (and noting that consideration of other mathematical techniques is out of scope for StackOverflow), you have no option but to calculate the continuous Fourier Transform.
However, I believe you haven't considered all the options for calculating the transform other than a laborious paper exercise e.g.
Numerical approximation of the continuous F.T. using code
Symbolic Integration e.g. Wolfram Alpha
Surely a numerical approximation of the Fourier Transform will be adequate for your solution?
I am assuming this is for coursework or research rather, so all you really care about as a physicist is a solution that is the quickest solution that is accurate within the scope of your problem.
So to conclude, IMHO, don't waste time searching for a more mathematically elegant solution or trick and just solve the problem with one of the above methods
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.