I'm using the curve fitting app in MATLAB. If I understand correctly the "b1" component in the left box is the mean of function i.e. the x point where y=50% and my x data is [-0.8 -0.7 -0.5 0 0.3 0.5 0.7], so why is this number in this example so big (631)?
General model Gauss1:
f(x) = a1*exp(-((x-b1)/c1)^2)
Coefficients (with 95% confidence bounds):
a1 = 3.862e+258 (-Inf, Inf)
b1 = 631.2 (-1.117e+06, 1.119e+06)
c1 = 25.83 (-2.287e+04, 2.292e+04)
Your data looks like cdf and not pdf. You can use this code for your solution
xi=[-0.8,-0.7,-0.5, 0.0, 0.3, 0.5, 0.7];
yi= [0.2, 0.0, 0.2, 0.2, 0.5, 1.0, 1.0];
fun=#(v) normcdf(xi,v(1),v(2))-yi;
[v]=lsqnonlin(fun,[1,1]); %[1,2]
mu=v(1); sigma=v(2);
x=linspace(-1.5,1.5,100);
y=normcdf(x,mu,sigma);
figure(1);clf;plot(xi,yi,'x',x,y);
annotation('textbox',[0.2,0.7,0.1,0.1], 'String',sprintf('mu=%f\nsigma=%f',mu,sigma),'FitBoxToText','on','FontSize',16);
you will get: mu=0.24537, sigma=0.213
And if you still want to fit to pdf, just change the function 'normcdf' in 'fun' (and 'y') to 'normpdf'.
Related
I have some data fitted to an exponential function (y1) using the curve fitting app to calculate the coefficients. I then imported the model and the prediction intervals (p11; a 1441x2 matrix). (t and Asiii are imported datasets, just generated some data to explain the code)
S=[0.95,0.87,0.83,0.73,0.62,0.51,0.41,0.31,0.22,0.04,0.002,0,0,0,0];
t=[1,3,5,10,20,30,40,50,60,90,120,180,240,360,1440];
hold on
s2=scatter(t,Asiii,250,[0.1216, 0.8, 0.0157],'^','filled');
x = 0:1440;
y1 = 1.016*exp(-0.02349*x)
p1=plot(x,y1,'Color',[0.1216, 0.8, 0.0157],...
'LineWidth',1.5);
p11 = predint(model,x,0.95,'observation','on');
plot(x,p11,'--','Color',[0.4235, 0.9608, .6],...
'LineWidth',1.5);
%Patch
patch([x;x],[p11(:,1)';p11(:,2)'],'g');
%Fill
fill(x, p11,p11,'facecolor', 'red', 'edgecolor', 'none', 'facealpha', 0.4);
I want to have the area between the prediciton bounds/confidence intervals filled with a different color, but it seems I can only get a solid black shading using 'patch'.
I also tried using the 'fill' function but the polygons go all the way to the top of y-axis. Is there a way to adjust the Edge thickness/color?
Right now I am doing a parameter sweep and I am trying to convert my data to a 3D graph to show the results in a very nice fashion. The problem is that I don't quite know how to plot it as I am having an issue with the result variable.
mute_rate = [0.5, 0.25, 0.125, 0.0625, 0.03125, 0.015625]
mute_step = linspace(-2.5, 2.5, 6)
results = [949.58, 293.53, 57.69, 53.65, 293.41, 1257.49;
279.19, 97.94, 32.60, 29.52, 90.52, 286.94;
32.96, 28.06, 19.56, 6.44, 13.47, 55.80;
2.01, 1.52, 5.38, 1.00, 0.89, 1.41;
0.61, 0.01, 18.59, 0.03, 0.56, 1.22;
1.85, 1.51, 18.64, 18.57, 18.54, 6.90]
So the first row in the result variable presents the results of the mute rate and mute step performed on the population from my genetic algorithm. For example:
0.5, -2.5 = 949.58,
0.5, -1.5 = 293.53,
0.5, -0.5 = 57.69
etc
It sounds like you want something akin to:
mesh(mute_step, mute_rate, results);
shading interp;
Other styles of plot would be surf or pcolor (for a 2d view).
I need to draw a smooth curve through some points, which I then want to show as an SVG path. So I create a B-Spline with scipy.interpolate, and can access some arrays that I suppose fully define it. Does someone know a reasonably simple way to create Bezier curves from these arrays?
import numpy as np
from scipy import interpolate
x = np.array([-1, 0, 2])
y = np.array([ 0, 2, 0])
x = np.r_[x, x[0]]
y = np.r_[y, y[0]]
tck, u = interpolate.splprep([x, y], s=0, per=True)
cx = tck[1][0]
cy = tck[1][1]
print( 'knots: ', list(tck[0]) )
print( 'coefficients x: ', list(cx) )
print( 'coefficients y: ', list(cy) )
print( 'degree: ', tck[2] )
print( 'parameter: ', list(u) )
The red points are the 3 initial points in x and y. The green points are the 6 coefficients in cx and cy. (Their values repeat after the 3rd, so each green point has two green index numbers.)
Return values tck and u are described scipy.interpolate.splprep documentation
knots: [-1.0, -0.722, -0.372, 0.0, 0.277, 0.627, 1.0, 1.277, 1.627, 2.0]
# 0 1 2 3 4 5
coefficients x: [ 3.719, -2.137, -0.053, 3.719, -2.137, -0.053]
coefficients y: [-0.752, -0.930, 3.336, -0.752, -0.930, 3.336]
degree: 3
parameter: [0.0, 0.277, 0.627, 1.0]
Not sure starting with a B-Spline makes sense: form a catmull-rom curve through the points (with the virtual "before first" and "after last" overlaid on real points) and then convert that to a bezier curve using a relatively trivial transform? E.g. given your points p0, p1, and p2, the first segment would be a catmull-rom curve {p2,p0,p1,p2} for the segment p1--p2, {p0,p1,p2,p0} will yield p2--p0, and {p1, p2, p0, p1} will yield p0--p1. Then you trivially convert those and now you have your SVG path.
As demonstrator, hit up https://editor.p5js.org/ and paste in the following code:
var points = [{x:150, y:100 },{x:50, y:300 },{x:300, y:300 }];
// add virtual points:
points = points.concat(points);
function setup() {
createCanvas(400, 400);
tension = createSlider(1, 200, 100);
}
function draw() {
background(220);
points.forEach(p => ellipse(p.x, p.y, 4));
for (let n=0; n<3; n++) {
let [c1, c2, c3, c4] = points.slice(n,n+4);
let t = 0.06 * tension.value();
bezier(
// on-curve start point
c2.x, c2.y,
// control point 1
c2.x + (c3.x - c1.x)/t,
c2.y + (c3.y - c1.y)/t,
// control point 2
c3.x - (c4.x - c2.x)/t,
c3.y - (c4.y - c2.y)/t,
// on-curve end point
c3.x, c3.y
);
}
}
Which will look like this:
Converting that to Python code should be an almost effortless exercise: there is barely any code for us to write =)
And, of course, now you're left with creating the SVG path, but that's hardly an issue: you know all the Bezier points now, so just start building your <path d=...> string while you iterate.
A B-spline curve is just a collection of Bezier curves joined together. Therefore, it is certainly possible to convert it back to multiple Bezier curves without any loss of shape fidelity. The algorithm involved is called "knot insertion" and there are different ways to do this with the two most famous algorithm being Boehm's algorithm and Oslo algorithm. You can refer this link for more details.
Here is an almost direct answer to your question (but for the non-periodic case):
import aggdraw
import numpy as np
import scipy.interpolate as si
from PIL import Image
# from https://stackoverflow.com/a/35007804/2849934
def scipy_bspline(cv, degree=3):
""" cv: Array of control vertices
degree: Curve degree
"""
count = cv.shape[0]
degree = np.clip(degree, 1, count-1)
kv = np.clip(np.arange(count+degree+1)-degree, 0, count-degree)
max_param = count - (degree * (1-periodic))
spline = si.BSpline(kv, cv, degree)
return spline, max_param
# based on https://math.stackexchange.com/a/421572/396192
def bspline_to_bezier(cv):
cv_len = cv.shape[0]
assert cv_len >= 4, "Provide at least 4 control vertices"
spline, max_param = scipy_bspline(cv, degree=3)
for i in range(1, max_param):
spline = si.insert(i, spline, 2)
return spline.c[:3 * max_param + 1]
def draw_bezier(d, bezier):
path = aggdraw.Path()
path.moveto(*bezier[0])
for i in range(1, len(bezier) - 1, 3):
v1, v2, v = bezier[i:i+3]
path.curveto(*v1, *v2, *v)
d.path(path, aggdraw.Pen("black", 2))
cv = np.array([[ 40., 148.], [ 40., 48.],
[244., 24.], [160., 120.],
[240., 144.], [210., 260.],
[110., 250.]])
im = Image.fromarray(np.ones((400, 400, 3), dtype=np.uint8) * 255)
bezier = bspline_to_bezier(cv)
d = aggdraw.Draw(im)
draw_bezier(d, bezier)
d.flush()
# show/save im
I didn't look much into the periodic case, but hopefully it's not too difficult.
I'm plotting a surface in matlab, where the x and y values are 0.15-0.85, with 0.05 gaps (0.15,0.2,0.25...).
However, when I plot the surface, the axis are on a 0.0 - 1.0 range, with 0.1 jumps (0.0, 0.1, 0.2...) so the entire surface is distorted. How do I set the axis ranges and deltas?
So as written in my comments you could solve your problem like this:
figure(1)
surf(data)
axis([0.15 0.85 0.15 0.85])
set(gca, 'XTick',0.20:0.05:0.80)
set(gca, 'YTick',0.20:0.05:0.80)
I am trying to migrate some code from Matlab to Opencv and need an exact replica of the gradient function. I have tried the cv::Sobel function but for some reason the values in the resulting cv::Mat are not the same as the values in the Matlab version. I need the X and Y gradient in separate matrices for further calculations.
Any workaround that could achieve this would be great
Sobel can only compute the second derivative of the image pixel which is not what we want.
(f(i+1,j) + f(i-1,j) - 2f(i,j)) / 2
What we want is
(f(i+i,j)-f(i-1,j)) / 2
So we need to apply
Mat kernelx = (Mat_<float>(1,3)<<-0.5, 0, 0.5);
Mat kernely = (Mat_<float>(3,1)<<-0.5, 0, 0.5);
filter2D(src, fx, -1, kernelx)
filter2D(src, fy, -1, kernely);
Matlab treats border pixels differently from inner pixels. So the code above is wrong at the border values. One can use BORDER_CONSTANT to extent the border value out with a constant number, unfortunately the constant number is -1 by OpenCV and can not be changed to 0 (which is what we want).
So as to border values, I do not have a very neat answer to it. Just try to compute the first derivative by hand...
You have to call Sobel 2 times, with arguments:
xorder = 1, yorder = 0
and
xorder = 0, yorder = 1
You have to select the appropriate kernel size.
See documentation
It might still be that the MatLab implementation was different, ideally you should retrieve which kernel was used there...
Edit:
If you need to specify your own kernel, you can use the more generic filter2D. Your destination depth will be CV_16S (16bit signed).
Matlab computes the gradient differently for interior rows and border rows (the same is true for the columns of course). At the borders, it is a simple forward difference gradY(1) = row(2) - row(1). The gradient for interior rows is computed by the central difference gradY(2) = (row(3) - row(1)) / 2.
I think you cannot achieve the same result with just running a single convolution filter over the whole matrix in OpenCV. Use cv::Sobel() with ksize = 1, then treat the borders (either manually or by applying a [ 1 -1 ] filter).
Pei's answer is partly correct. Matlab uses these calculations for the borders:
G(:,1) = A(:,2) - A(:,1);
G(:,N) = A(:,N) - A(:,N-1);
so used the following opencv code to complete the gradient:
static cv::Mat kernelx = (cv::Mat_<double>(1, 3) << -0.5, 0, 0.5);
static cv::Mat kernely = (cv::Mat_<double>(3, 1) << -0.5, 0, 0.5);
cv::Mat fx, fy;
cv::filter2D(Image, fx, -1, kernelx, cv::Point(-1, -1), 0, cv::BORDER_REPLICATE);
cv::filter2D(Image, fy, -1, kernely, cv::Point(-1, -1), 0, cv::BORDER_REPLICATE);
fx.col(fx.cols - 1) *= 2;
fx.col(0) *= 2;
fy.row(fy.rows - 1) *= 2;
fy.row(0) *= 2;
Jorrit's answer is partly correct.
In some cases, the value of the directional derivative may be negative, and MATLAB will retain these negative numbers, but OpenCV Mat will set the negative number to 0.