I have a piece of numpy code.
distance = np.ones((N,)) * 1e10
mask = dist < distance
distance[mask] = dist[mask]
l think the distance and dist is a array.through comparing the value of distance and dist,I want to find a smaller value and store that in distance.
In c++,I can only do this with a for loop.Does eigen have a function to implement this operation.thanks!!
I can only do this with a for loop. I hope to find a function of eigen to implement this operation.
Related
Edit: So I found a page related to rasterizing a trapezoid https://cse.taylor.edu/~btoll/s99/424/res/ucdavis/GraphicsNotes/Rasterizing-Polygons/Rasterizing-Polygons.html but am still trying to figure out if I can just do the edges
I am trying to generate points for the corners of an arbitrary N-gon. Where N is a non-zero positive integer. But I am trying to do so efficiently without the need of division and trig. I am thinking that there is probably some of sort of Bresenham's type algorithm for this but I cannot seem to find anything.
The only thing I can find on stackoverflow is this but the internal angle increments are found by using 2*π/N:
How to draw a n sided regular polygon in cartesian coordinates?
Here was the algorithm from that page in C
float angle_increment = 2*PI / n_sides;
for(int i = 0; i < n_sides; ++i)
{
float x = x_centre + radius * cos(i*angle_increment +start_angle);
float y = y_centre + radius * sin(i*angle_increment +start_angle);
}
So is it possible to do so without any division?
There's no solution which avoids calling sin and cos, aside from precomputing them for all useful values of N. However, you only need to do the computation once, regardless of N. (Provided you know where the centre of the polygon is. You might need a second computation to figure out the coordinates of the centre.) Every vertex can be computed from the previous vertex using the same rotation matrix.
The lookup table of precomputed values is not an unreasonable solution, since at some sufficiently large value of N the polygon becomes indistinguishable from a circle. So you probably only need a lookup table of a few hundred values.
I am trying to convert a Matlab code into Python. To do a Multisignal 1-D wavelet decomposition I use mdwtdec() in Matlab. The closest function I found in python is in pywt lib, wavedec().
While the matlab function take 4 arguments mdwtdec(DIRDEC,X,LEV,WNAME): here X is the signal, LEV is the level, WNAME is the wavelet name and DIRDEC is Direction indicator: r (row) or c (column).
mdwtdec(DIRDEC,X,LEV,WNAME) returns the wavelet decomposition at level LEV of each row (if DIRDEC = r) or each column (if DIRDEC = c) of matrix X, using the wavelet WNAME.
In pywavelets, wavedec(X,LEV,WNAME) takes X, LEV and WNAME similar to Matlab. There is no option for the direction, so by default it is set to 'r'. I need 'c' as an option too, How do I achieve it in Python?
pywt's wavedec assumes that the signal is an array. If the signal isn't an array, then the results are less convincing (for square matrices the result appears to correspond to a rowwise operation, but the same cannot be said for non-square matrices).
Fortunately you can run wavedec for each row or column using numpy's apply_along_axis. With a little bit of extra wrapping, you can define a Matlab approximation (Matlab's version probably has better error handling) to mdwtdec such as:
import pywt
import numpy as np
def mdwtdec(dirdec,x,lev,wavelet):
"""
Multisignal 1D Discrete Wavelet decomposition.
Parameters
----------
dirdec : char
Direction indicator: 'r' (row) or 'c' (column)
x : matrix
Input matrix
wavelet : Wavelet object or name string
Wavelet to use
lev : int
Decomposition level (must be >= 0). If level is None then it
will be calculated using the ``dwt_max_level`` function.
"""
return np.apply_along_axis(
lambda y: np.concatenate(pywt.wavedec(y, wavelet, level=lev)),
axis={'c':0, 'r':1}[dirdec], arr=x)
in which I've used:
np.concatenate to keep the approximation and details coefficients as a tight row/column vector
a lambda expression to specify the executed function along with the mapping of input arguments
I have a scientific code that uses both sine and cosine of the same argument (I basically need the complex exponential of that argument). I was wondering if it were possible to do this faster than calling sine and cosine functions separately.
Also I only need about 0.1% precision. So is there any way I can find the default trig functions and truncate the power series for speed?
One other thing I have in mind is, is there any way to perform the remainder operation such that the result is always positive? In my own algorithm I used x=fmod(x,2*pi); but then I would need to add 2pi if x is negative (smaller domain means I can use a shorter power series)
EDIT: LUT turned out to be the best approach for this, however I am glad I learned about other approximation techniques. I will also advise using an explicit midpoint approximation. This is what I ended up doing:
const int N = 10000;//about 3e-4 error for 1000//3e-5 for 10 000//3e-6 for 100 000
double *cs = new double[N];
double *sn = new double[N];
for(int i =0;i<N;i++){
double A= (i+0.5)*2*pi/N;
cs[i]=cos(A);
sn[i]=sin(A);
}
The following part approximates (midpoint) sincos(2*pi*(wc2+t[j]*(cotp*t[j]-wc)))
double A=(wc2+t[j]*(cotp*t[j]-wc));
int B =(int)N*(A-floor(A));
re += cs[B]*f[j];
im += sn[B]*f[j];
Another approach could have been using the chebyshev decomposition. You can use the orthogonality property to find the coefficients. Optimized for exponential, it looks like this:
double fastsin(double x){
x=x-floor(x/2/pi)*2*pi-pi;//this line can be improved, both inside this
//function and before you input it into the function
double x2 = x*x;
return (((0.00015025063885163012*x2-
0.008034350857376128)*x2+ 0.1659789684145034)*x2-0.9995812174943602)*x;} //7th order chebyshev approx
If you seek fast evaluation with good (but not high) accuracy with powerseries you should use an expansion in Chebyshev polynomials: tabulate the coefficients (you'll need VERY few for 0.1% accuracy) and evaluate the expansion with the recursion relations for these polynomials (it's really very easy).
References:
Tabulated coefficients: http://www.ams.org/mcom/1980-34-149/S0025-5718-1980-0551302-5/S0025-5718-1980-0551302-5.pdf
Evaluation of chebyshev expansion: https://en.wikipedia.org/wiki/Chebyshev_polynomials
You'll need to (a) get the "reduced" argument in the range -pi/2..+pi/2 and consequently then (b) handle the sign in your results when the argument actually should have been in the "other" half of the full elementary interval -pi..+pi. These aspects should not pose a major problem:
determine (and "remember" as an integer 1 or -1) the sign in the original angle and proceed with the absolute value.
use a modulo function to reduce to the interval 0..2PI
Determine (and "remember" as an integer 1 or -1) whether it is in the "second" half and, if so, subtract pi*3/2, otherwise subtract pi/2. Note: this effectively interchanges sine and cosine (apart from signs); take this into account in the final evaluation.
This completes the step to get an angle in -pi/2..+pi/2
After evaluating sine and cosine with the Cheb-expansions, apply the "flags" of steps 1 and 3 above to get the right signs in the values.
Just create a lookup table. The following will let you lookup the sin and cos of any radian value between -2PI and 2PI.
// LOOK UP TABLE
var LUT_SIN_COS = [];
var N = 14400;
var HALF_N = N >> 1;
var STEP = 4 * Math.PI / N;
var INV_STEP = 1 / STEP;
// BUILD LUT
for(var i=0, r = -2*Math.PI; i < N; i++, r += STEP) {
LUT_SIN_COS[2*i] = Math.sin(r);
LUT_SIN_COS[2*i + 1] = Math.cos(r);
}
You index into the lookup table by:
var index = ((r * INV_STEP) + HALF_N) << 1;
var sin = LUT_SIN_COS[index];
var cos = LUT_SIN_COS[index + 1];
Here's a fiddle that displays the % error you can expect from different sized LUTS http://jsfiddle.net/77h6tvhj/
EDIT Here's an ideone (c++) with a ~benchmark~ vs the float sin and cos. http://ideone.com/SGrFVG For whatever a benchmark on ideone.com is worth the LUT is 5 times faster.
One way to go would be to learn how to implement the CORDIC algorithm. It is not difficult and pretty interesting intelectually. This gives you both the cosine and the sine. Wikipedia gives a MATLAB example that should be easy to adapt in C++.
Note that you can augment speed and reduce precision simply by lowering the parameter n.
About your second question, it has already been asked here (in C). It seems that there is no simple way.
You can also calculate sine using a square root, given the angle and the cosine.
The example below assumes the angle ranges from 0 to 2π:
double c = cos(angle);
double s = sqrt(1.0-c*c);
if(angle>pi)s=-s;
For single-precision floats, Microsoft uses 11-degree polynomial approximation for sine, 10-degree for cosine: XMScalarSinCos.
They also have faster version, XMScalarSinCosEst, that uses lower-degree polynomials.
If you aren’t on Windows, you’ll find same code + coefficients on geometrictools.com under Boost license.
At the moment i'm working with a camera to detect a marker. I use opencv and the Aruco Libary.
Only I'm stuck with a problem right now. I need to detect if the distance between 2 marker is less than a specific value. I have a function to calculate the distance, I can compare everything. But I'm looking for the most efficient way to keep track of all the markers (around 5/6) and how close they are together.
There is a list with markers but I cant find a efficient way to compare all of them.
I have a
Vector <Marker>
I also have a function called getDistance.
double getDistance(cv::Point2f punt1, cv::Point2f punt2)
{
float xd = punt2.x-punt1.x;
float yd = punt2.y-punt1.y;
double Distance = sqrtf(xd*xd + yd*yd);
return Distance;
}
The Markers contain a Point2f, so i can compare them easily.
One way to increase performance is to keep all the distances squared and avoid using the square root function. If you square the specific value you are checking against then this should work fine.
There isn't really a lot to recommend. If I understand the question and I'm counting the pairs correctly, you'll need to calculate 10 distances when you have 5 points, and 15 distances when you have 6 points. If you need to determine all of the distances, then you have no choice but to calculate all of the distances. I don't see any way around that. The only advice I can give is to make sure you calculate the distance between each pair only once (e.g., once you know the distance between points A and B, you don't need to calculate the distance between B and A).
It might be possible to sort the vector in such a way that you can short circuit your loop. For instance, if you sort it correctly and the distance between point A and point B is larger than your threshold, then the distances between A and C and A and D will also be larger than the threshold. But keep in mind that sorting isn't free, and it's likely that for small sets of points it would be faster to just calculate all distances ("Fancy algorithms are slow when n is small, and n is usually small. Fancy algorithms have big constants. Until you know that n is frequently going to be big, don't get fancy. ... For example, binary trees are always faster than splay trees for workaday problems.").
Newer versions of the C and C++ standard library have a hypot function for calculating distance between points:
#include <cmath>
double getDistance(cv::Point2f punt1, cv::Point2f punt2)
{
return std::hypot(punt2.x - punt1.x, punt2.y - punt1.y);
}
It's not necessarily faster, but it should be implemented in a way that avoids overflow when the points are far apart.
One minor optimization is to simply check if the change in X or change in Y exceeds the threshold. If it does, you can ignore the distance between those two points because the overall distance will also exceed the threshold:
const double threshold = ...;
std::vector<cv::Point2f> points;
// populate points
...
for (auto i = points.begin(); i != points.end(); ++i) {
for (auto j = i + 1; j != points.end(); ++j) {
double dx = std::abs(i->x - j->x), dy = std::abs(i->y - j->y);
if (dx > threshold || dy > threshold) {
continue;
}
double distance = std::hypot(dx, dy);
if (distance > threshold) {
continue;
}
...
}
}
If you're dealing with large amounts of data inside your vector you may want to consider some multithreading using future.
Vector <Marker> could be chunked into X chunks which are asynchronously computed together and stored inside std::future<>, putting to use #Sesame's suggestion will also increase your speed as well.
I'm studying about the CSS algorithm and I don't get the hang of the concept of 'Arc Length Parameter'.
According to the literature, planar curve Gamma(u)=(x(u),y(u)) and they say this u is the arc length parameter and apparently, Gaussian Kernel g is also parameterized by this u here.
Stop me if I got something wrong but, aren't x and y location of the pixel? How is it represented by another parameter?
I had no idea when I first saw it on the literature so, I looked up the code. and apparently, I got puzzled even more.
here is the portion of the code
void getGaussianDerivs(double sigma, int M, vector<double>& gaussian,
vector<double>& dg, vector<double>& d2g) {
int L = (M - 1) / 2;
double sigma_sq = sigma * sigma;
double sigma_quad = sigma_sq*sigma_sq;
dg.resize(M); d2g.resize(M); gaussian.resize(M);
Mat_<double> g = getGaussianKernel(M, sigma, CV_64F);
for (double i = -L; i < L+1.0; i += 1.0) {
int idx = (int)(i+L);
gaussian[idx] = g(idx);
// from http://www.cedar.buffalo.edu/~srihari/CSE555/Normal2.pdf
dg[idx] = (-i/sigma_sq) * g(idx);
d2g[idx] = (-sigma_sq + i*i)/sigma_quad * g(idx);
}
}
so, it seems the code uses simple 1D Gaussian Kernel Aperture size of M and it is trying to compute its 1st and 2nd derivatives. As far as I know, 1D Gaussian kernel has parameter of x which is a horizontal coordinate and sigma which is scale. it seems like that 'arc length parameter u' is equivalent to the variable of x. That doesn't make any sense because later in the code, it directly convolutes the set of x and y on the contour.
what is this u?
PS. since I replied to the fellow who tried to answer my question, I think I should modify my question, so, here we go.
What I'm confusing is, how is this parameter 'u' implemented in codes? I think I understood the full code above -of course, I inserted only a portion of the code- but the problem is, I have no idea about what it would be for the 'improved' version of the algorithm. It says it's using 'affine length parameter' instead of this 'arc length parameter' and I'm not so sure how I implement the concept into the code.
According to the literature, the main difference between arc length parameter and affine length parameter is it's sampling interval and arc length parameter uses 1 for the vertical and horizontal direction and root of 2 for the diagonal direction. It makes sense since the portion of the code above is using for loop to compute 1st and 2nd derivatives of the 1d Gaussian and it directly inserts the value of interval 1 but, how is it gonna be with different interval with different variable? Is it possible that I'm not able to use 'for loop' for it?