I want to mark certain areas in my plot, which I have solved by using rectangle. Now the problem is that I would also like to have the grid on but this is covered by the rectangles. Is there a way to color different areas of the background or have the grid moved forward so that it is plotted above the rectangles?
My code looks like this:
rectangle('Position', [0 0.46 3*l+1 1.6], 'FaceColor', bg2);
rectangle('Position', [0 0.6 3*l+1 1.6], 'FaceColor', bg);
hold on
grid on
plot(Isr(:, 2), abs(Isr(:, 1)), 'r');
plot(Isx(:, 2), abs(Isx(:, 1)), 'b');
title('Current');
axis([0 3*l 0 1.6]);
legend(Legend);
xlabel('$k$');
ylabel('$I_F/I_{3\Phi}$');
Here is a minimal example:
l = 10;
x = linspace(0,l);
rectangle('Position', [0 0.46 l+1 1.6], 'FaceColor', 'm');
rectangle('Position', [0 0.6 l+1 1.6], 'FaceColor', 'g');
hold on
grid on
plot(x, sin(x), 'r');
plot(x, cos(x), 'b');
axis([0 l 0 1.6]);
legend('Sine', 'Cosine');
You should be able to accomplish this by modifying the Layer property of your axes, which modifies the placement of grid lines and tick marks in relation to graphic objects:
l = 10;
x = linspace(0,l);
myax = axes;
rectangle('Position', [0 0.46 l+1 1.6], 'FaceColor', 'm');
rectangle('Position', [0 0.6 l+1 1.6], 'FaceColor', 'g');
hold on
grid on
plot(x, sin(x), 'r');
plot(x, cos(x), 'b');
axis([0 l 0 1.6]);
legend('Sine', 'Cosine');
myax.Layer = 'top';
Note that I also added an axes call to give me a handle to the axes. The same behavior can be accomplished without this by using gca with set: set(gca, 'Layer', 'top');
Related
Is it possible to get principal point (cx, cy) from a 4x4 projection matrix? This is the same matrix asked in this question: Getting focal length and focal point from a projection matrix
(SCNMatrix4)
s = (m11 = 1.83226573,
m12 = 0,
m13 = 0,
m14 = 0,
m21 = 0,
m22 = 2.44078445,
m23 = 0,
m24 = 0,
m31 = -0.00576340035,
m32 = -0.0016724075,
m33 = -1.00019991,
m34 = -1,
m41 = 0,
m42 = 0,
m43 = -0.20002,
m44 = 0)
The values I'm trying to calculate in this 3x3 camera matrix is x0 and y0.
I recently confronted this problem, and quite astonished I couldn't find a relevant solution on Internet, because it seems to be a simple mathematics problem.
After a few days of struggling with matrices, I found a solution.
Let's define two Cartesian coordinate system, the camera coordinate system with x', y', z' axes, and the world coordinate system with x, y, z axes. The camera(or the eye) is positioned at the origin of the camera coordinate system and the image plane(a plane containing the screen) is z' = -n, where n is the focal length and the focal point is the position of the camera. I am using the convention of OpenGL and n is the nearVal argument of the glFrustum().
You can define a 4x4 transformation matrix M in a homogeneous coordinate system to deal with a projection. The M transforms a coordinate (x, y, z) in the world coordinate system into a coordinate (x', y', z') in the camera coordinate system like the following, where # means a matrix multiplication.
[
[x_prime_h],
[y_prime_h],
[z_prime_h],
[w_prime_h],
] = M # [
[x_h],
[y_h],
[z_h],
[w_h],
]
[x, y, z] = [x_h, y_h, z_h] / w_h
[x_prime, y_prime, z_prime] = [x_prime_h, y_prime_h, z_prime_h] / w_prime_h
Now assume you are given M = P V, where P is a perspective projection matrix and V is a view transformation matrix. The theoretical projection matrix is like the following.
P_theoretical = [
[n, 0, 0, 0],
[0, n, 0, 0],
[0, 0, n, 0],
[0, 0, -1, 0],
]
In OpenGL, an augmented matrix like the following is used to cover the normalization and nonlinear scaling on z coordinates, where l, r, b, t, n, f are the left, right, bottom, top, nearVal, farVal arguments of the glFrustum().(The resulting z' coordinate is not actually the coordinate of a projected point, but a value used for Z-buffering.)
P = [
[2*n/(r-l), 0, (r+l)/(r-l), 0],
[0, 2*n/(t-b), (t+b)/(t-b), 0],
[0, 0, -(f+n)/(f-n), -2*n*f/(f-n)],
[0, 0, -1, 0],
]
The transformation V is like the following, where r_ij is the element at i-th row and j-th column of the 3x3 rotational matrix R and (c_0, c_1, c_2) is the coordinate of the camera.
V = [
[r_00, r_01, r_02, -(r_00*c_0 + r_01*c_1 + r_02*c_2)],
[r_10, r_11, r_12, -(r_10*c_0 + r_11*c_1 + r_12*c_2)],
[r_20, r_21, r_22, -(r_20*c_0 + r_21*c_1 + r_22*c_2)],
[0, 0, 0, 1],
]
The P and V can be represented with block matrices like the following.
C = [
[c_0],
[c_1],
[c_2],
]
A = [
[2*n/(r-l), 0, (r+l)/(r-l)],
[0, 2*n/(t-b), (t+b)/(t-b)],
[0, 0, -(f+n)/(f-n)],
]
B = [
[0],
[0],
[-2*n*f/(f-n)],
]
P = [
[A,B],
[[0, 0, -1], [0]],
]
V = [
[R, -R # C],
[[0, 0, 0], [1]],
]
M = P # V = [
[A # R, -A # R # C + B],
[[0, 0, -1] # R, [0, 0, 1] # R # C],
]
Let m_ij be the element of M at i-th row and j-th column. Taking the first element of the second row of the above block notation of M, you can solve for the elementary z' vector of the camera coordinate system, the opposite direction from the camera point to the intersection point between the image plane and its normal line passing through the focal point.(The intersection point is the principal point.)
e_z_prime = [0, 0, 1] # R = -[m_30, m_31, m_32]
Taking the second column of the above block notation of M, you can solve for C like the following, where inv(X) is an inverse of a matrix X.
C = - inv([
[m_00, m_01, m_02],
[m_10, m_11, m_12],
[m_30, m_31, m_32],
]) # [
[m_03],
[m_13],
[m_33],
]
Let p_ij be the element of P at i-th row and j-th column.
Now you can solve for p_23 = -2nf/(f-n) like the following.
B = [
[m_03],
[m_13],
[m_23],
] + [
[m_00, m_01, m_02],
[m_10, m_11, m_12],
[m_20, m_21, m_22],
] # C
p_23 = B[2] = m_23 + (m_20*c_0 + m_21*c_1 + m_22*c_2)
Now using the fact p_20 = p_21 = 0, you can get p_22 = -(f+n)/(f-n) like the following.
p_22 * e_z_prime = [m_20, m_21, m_22]
p_22 = -(m_20*m_30 + m_21*m_31 + m_22*m_32)
Now you can get n and f from p_22 and p_23 like the following.
n = p_23/(p_22-1)
= -(m_23 + m_20*c_0+m_21*c_1+m_22*c_2) / (m_20*m_30+m_21*m_31+m_22*m_32 + 1)
f = p_23/(p_22+1)
= -(m_23 + m_20*c_0+m_21*c_1+m_22*c_2) / (m_20*m_30+m_21*m_31+m_22*m_32 - 1)
From the camera position C, the focal length n and the elementary z' vector e_z_prime, you can get the principal point, C - n * e_z_prime.
As a side note, you can prove the input matrix of inv() in the formula for getting C is nonsingular. And you can also find elementary x' and y' vectors of the camera coordinate system, and find the l, r, b, t using these vectors.(There will be two valid solutions for the (e_x_prime, e_y_prime, l, r, b, t) tuple, due to the symmetry.) And finally this solution can be expanded when the transformation matrix is mixed with the world transformation which does an anisotropic scaling, that is when M = P V W and W can have unequal eigenvalues.
I have a table of data points to be subtracted by another table of data points. I have succeed in subtracting the y values of each data point successfully, but it is the x values of each data point in that I have trouble.
m = 10;
DataList = Table[{}, {i, 1, m}];
BGData = Import["BatchData-Background.txt", "Table"];
BGPlot = ListPlot[BGData, FrameLabel -> {"Time (s)", "Voltage [V]"}, PlotStyle -> Black]
Do[{DataList[[i]] = Import["BatchData-B" <> ToString[i] <> "V.txt", "Table"];
DataPlot = ListPlot[DataList[[i]], FrameLabel -> {"Time (s)", "Voltage (V)"}, PlotStyle -> Gray]; Print["B = ", i, "Volts"]; Print[DataPlot];}, {i, 1, m}];
m = 10;
SubDataList = Table[{DataList[[i, All, 2]] - BGData[[All, 2]]}, {i, 1, m}];
Do[{SubDataPlot = ListPlot[SubDataList[[i]]];}, {i, 1, m}] `
m = 10;
SubDataList = Table[{DataList[[i,All,2]]-BGData[[All,2]]},{i,1,m}];
Do[{SubDataPlot=ListPlot[SubDataList[[i]]];},{i,1,m}]
This is the code that I am having trouble with. However, my plot will get y values correct, but my x-axis is automatically set to 0 to 1400 in steps of 200. However, my data points shows that my x-axis should be from 0 to 0.07 in steps of 0.01.
Both DataList and BGData are table of its own and the specifications that I have done for each just extracts the y-values in each table and subtracts them.
Read a bit more of the documentation of ListPlot. The expression
ListPlot[SubDataList]
plots, as you observe, the values in SubDataList from 1 to m. If you supply a second list of values to the function, perhaps
ListPlot[{SubDataList, BGData}]
treats the first list as a list of x-coordinates and the second as a list of y-coordinates.
Making some example data points for subDataList from 0 to 1400 and plotting them on an x-scale from 0 to 0.07:-
subDataList = Table[Sin[x], {x, 0, 4 Pi, 4 Pi/1400}];
ListPlot[Transpose[{Prepend[Range[1400]*0.00005, 0], subDataList}]]
I have 2 vectors that are x and y coordinates of the 8 vertexes of a polygon
x=[5 5 7 7 9 9 5 7]
y=[8 6 6 8 6 8 10 10]
I wanna sort them (clockwise) to obtain the right vectors (to draw the polygon correctly)
x=[5 7 9 9 7 7 5 5]
y=[6 6 6 8 8 10 10 8]
Step 1: Find the unweighted mean of the vertices:
cx = mean(x);
cy = mean(y);
Step 2: Find the angles:
a = atan2(y - cy, x - cx);
Step 3: Find the correct sorted order:
[~, order] = sort(a);
Step 4: Reorder the coordinates:
x = x(order);
y = y(order);
Python version (numpy) for Ben Voigt's algorithm:
def clockwise(points):
x = points[0,:]
y = points[1,:]
cx = np.mean(x)
cy = np.mean(y)
a = np.arctan2(y - cy, x - cx)
order = a.ravel().argsort()
x = x[order]
y = y[order]
return np.vstack([x,y])
Example:
In [281]: pts
Out[281]:
array([[7, 2, 2, 7],
[5, 1, 5, 1]])
In [282]: clockwise(pts)
Out[282]:
array([[2, 7, 7, 2],
[1, 1, 5, 5]])
I tried the solutions by #ben-voight and #mclafee, but I think they are sorting the wrong way.
When using atan2 the angles are stated in the following way:
Matlab Atan2
The angle is positive for counter-clockwise angles (upper half-plane,
y > 0), and negative for clockwise angles (lower half-plane, y < 0).
Wikipedia Atan2
This means that using ascending sort() of Numpy or Matlab will progress counterclockwise.
This can be verified using the Shoelace equation
Wikipedia Shoelace
Python Shoelace
So, adjusting the answers mentioned above to use descending sorting the correct solution in Matlab is
cx = mean(x);
cy = mean(y);
a = atan2(y - cy, x - cx);
[~, order] = sort(a, 'descend');
x = x(order);
y = y(order);
The solution in numpy is
import numpy as np
def clockwise(points):
x = points[0,:]
y = points[1,:]
cx = np.mean(x)
cy = np.mean(y)
a = np.arctan2(y - cy, x - cx)
order = a.ravel().argsort()[::-1]
x = x[order]
y = y[order]
return np.vstack([x,y])
pts = np.array([[7, 2, 2, 7],
[5, 1, 5, 1]])
clockwise(pts)
pts = np.array([[1.0, 1.0],
[-1.0, -1.0],
[1.0, -1.0],
[-1.0, 1.0]]).transpose()
clockwise(pts)
Output:
[[7 2 2 7]
[5 1 5 1]]
[[2 7 7 2]
[5 5 1 1]]
[[ 1. -1. 1. -1.]
[ 1. -1. -1. 1.]]
[[-1. 1. 1. -1.]
[ 1. 1. -1. -1.]]
Please notice the [::-1] used to invert arrays / lists.
This algorithm does not apply to non-convex polygons.
Instead, consider using MATLAB's poly2cw()
Friends, I have function COs[x^{2}+2 n] + n, how can I plot this function on single plot with x,0,10 and n,1,9. What I need, is x-range on x-axis and function value as y-axis. It will like a filled density plot. Thanks in anticipations.
Not a filled density plot...
Show[Table[
Plot[Cos[x^2 + 2 n] + n, {x, 0, 10}],
{n, 1, 9}], PlotRange -> All]
I have a image(png format) in hand. The lines that bound the ellipses (represent the nucleus) are over straight which are impractical. How could i extract the lines from the image and make them bent, and with the precondition that they still enclose the nucleus.
The following is the image:
After bending
EDIT: How can i translate the Dilation And Filter part in answer2 into Matlab language? I can't figure it out.
Ok, here is a way involving several randomization steps needed to get a "natural" non symmetrical appearance.
I am posting the actual code in Mathematica, just in case someone cares translating it to Matlab.
(* A preparatory step: get your image and clean it*)
i = Import#"http://i.stack.imgur.com/YENhB.png";
i1 = Image#Replace[ImageData[i], {0., 0., 0.} -> {1, 1, 1}, {2}];
i2 = ImageSubtract[i1, i];
i3 = Inpaint[i, i2]
(*Now reduce to a skeleton to get a somewhat random starting point.
The actual algorithm for this dilation does not matter, as far as we
get a random area slightly larger than the original elipses *)
id = Dilation[SkeletonTransform[
Dilation[SkeletonTransform#ColorNegate#Binarize#i3, 3]], 1]
(*Now the real random dilation loop*)
(*Init vars*)
p = Array[1 &, 70]; j = 1;
(*Store in w an image with a different color for each cluster, so we
can find edges between them*)
w = (w1 =
WatershedComponents[
GradientFilter[Binarize[id, .1], 1]]) /. {4 -> 0} // Colorize;
(*and loop ...*)
For[i = 1, i < 70, i++,
(*Select edges in w and dilate them with a random 3x3 kernel*)
ed = Dilation[EdgeDetect[w, 1], RandomInteger[{0, 1}, {3, 3}]];
(*The following is the core*)
p[[j++]] = w =
ImageFilter[ (* We apply a filter to the edges*)
(Switch[
Length[#1], (*Count the colors in a 3x3 neighborhood of each pixel*)
0, {{{0, 0, 0}, 0}}, (*If no colors, return bkg*)
1, #1, (*If one color, return it*)
_, {{{0, 0, 0}, 0}}])[[1, 1]] (*If more than one color, return bkg*)&#
Cases[Tally[Flatten[#1, 1]],
Except[{{0.`, 0.`, 0.`}, _}]] & (*But Don't count bkg pixels*),
w, 1,
Masking -> ed, (*apply only to edges*)
Interleaving -> True (*apply to all color chanels at once*)]
]
The result is:
Edit
For the Mathematica oriented reader, a functional code for the last loop could be easier (and shorter):
NestList[
ImageFilter[
If[Length[#1] == 1, #1[[1, 1]], {0, 0, 0}] &#
Cases[Tally[Flatten[#1, 1]], Except[{0.` {1, 1, 1}, _}]] & , #, 1,
Masking -> Dilation[EdgeDetect[#, 1], RandomInteger[{0, 1}, {3, 3}]],
Interleaving -> True ] &,
WatershedComponents#GradientFilter[Binarize[id,.1],1]/.{4-> 0}//Colorize,
5]
What you have as input is the Voronoi diagram. You can recalculate it using another distance function instead of the Euclidean one.
Here is an example in Mathematica using the Manhattan Distance (i3 is your input image without the lines):
ColorCombine[{Image[
WatershedComponents[
DistanceTransform[Binarize#i3,
DistanceFunction -> ManhattanDistance] ]], i3, i3}]
Edit
I am working with another algorithm (preliminary result). What do you think?
Here is what I came up with, it is not a direct translation of #belisarius code, but should be close enough..
%# read image (indexed image)
[I,map] = imread('http://i.stack.imgur.com/YENhB.png');
%# extract the blobs (binary image)
BW = (I==1);
%# skeletonization + dilation
BW = bwmorph(BW, 'skel', Inf);
BW = imdilate(BW, strel('square',2*1+1));
%# connected components
L = bwlabel(BW);
imshow(label2rgb(L))
%# filter 15x15 neighborhood
for i=1:13
L = nlfilter(L, [15 15], #myFilterFunc);
imshow( label2rgb(L) )
end
%# result
L(I==1) = 0; %# put blobs back
L(edge(L,'canny')) = 0; %# edges
imshow( label2rgb(L,#jet,[0 0 0]) )
myFilterFunc.m
function p = myFilterFunc(x)
if range(x(:)) == 0
p = x(1); %# if one color, return it
else
p = mode(x(x~=0)); %# else, return the most frequent color
end
end
The result:
and here is an animation of the process: