I try to interpolate curves in the gaps of some already existing curves using matlab (and the interp1-function).
Start of edit 1
I already have the data of 5 Torque over rpm curves that I obtained with a number of simulations for each curve. Since simulation time is precious, I would like to save time with the interpolation of curves that "fill the gap" between the already existing ones.
I'm looking to form the following:
Further thoughts of me down below.
End of edit 1
I tried to follow the steps from the question in the thread Interpolation between two curves (matlab) but it does not seem to work with my code. I am not sure if the code is actually applicable since the curves might overlap...
I tried to edit the code from the link above like the following:
% Save the data in one array each. The original data is stored in the
% arrays "x/yOriginal" row-wise.
curve1_data = [xOriginal(:,1) yOriginal(:,1)];
curve2_data = [xOriginal(:,2) yOriginal(:,2)];
% This was to try if the code from the link works
yy = [0:1:60];
xx1 = interp1(curve1_data(:,2),curve1_data(:,1),yy,'spline');
xx2 = interp1(curve2_data(:,2),curve2_data(:,1),yy,'spline');
m = 3; % Curve_Offset
mm(:,m) = xx1 + (xx2-xx1)*(m/(8750-5000));
% With the following I tried to interpolate over x
xx = (xOriginal(1,1):1:xOriginal(1,2));
yy1 = interp1(curve1_data(:,1),curve1_data(:,2),xx,'spline');
yy2 = interp1(curve2_data(:,1),curve2_data(:,2),xx,'spline');
m = 3; % Curve_Offset
% From the original code:
mm(:,m) = xx1 + (xx2-xx1)*(m/(8750-5000));
% Interpolation over x
yINT = yy1 + (yy2-yy1)*(m/8750-5000);
None of the interpolation-techniques worked, the y-values are either 90% negative (with the code from the link) or way too high (10e8 with the interpolation over x).
What I expected was that it creates a curve a little less steep than the curve "to its left" and a bit steeper than the curve "to its right".
My further thoughts:
The existing curves are the product of big 3-dimensional arrays. I.e. it could be that it is more the way to go to interpolate the arrays and then "read out" the Torque-over-rpm-Curves. On the other hand, I don't see a way to interpolate between two 1001-by-7001-by-5 arrays...
Moreover, for the next steps with the programm, the curve-interpolation needs to be quite fine (it is necessary to have way more than 1 interpolated curve between two existing curves) which makes the problem even more difficult.
If I understand right, the plot is generated with something like this:
plot(xOriginal(:,1),yOriginal(:,1))
plot(xOriginal(:,2),yOriginal(:,2))
% ...
If so, you can plot an intermediate curve with
plot((xOriginal(:,1) + xOriginal(:,2))/2, (yOriginal(:,1) + yOriginal(:,2))/2)
That is, the average between each pair of coordinates forms a curve that is exactly half-way between the two original curves.
Use weighted averages to generate more of these curves. This is linear interpolation.
d = 0.2;
plot(d*xOriginal(:,1) + (1-d)*xOriginal(:,2), d*yOriginal(:,1) + (1-d)*yOriginal(:,2))
Setting d = 0.5 we go back to the half-way case above.
Example:
xOriginal(:,1) = linspace(0.2,0.5,100);
yOriginal(:,1) = 3 * cos(xOriginal(:,1)*10-2.5) + 3;
xOriginal(:,2) = linspace(0.6,0.8,100);
yOriginal(:,2) = cos(xOriginal(:,2)*20-13.5) + 1;
clf; hold on
plot(xOriginal(:,1),yOriginal(:,1))
plot(xOriginal(:,2),yOriginal(:,2))
plot((xOriginal(:,1)+xOriginal(:,2))/2,(yOriginal(:,1)+yOriginal(:,2))/2)
(the interpolated line is in orange)
Related
Say I've got a number of curves with different length (number of points in each curve and points distance are all vary). Could I find a curve in 3D space that fit best for this group of lines?
Code example in Matlab would be appreciated.
example data set:
the 1st curve has 10 points.
18.5860 18.4683 18.3576 18.2491 18.0844 17.9016 17.7709 17.6401 17.4617 17.2726
91.6178 91.5711 91.5580 91.5580 91.5701 91.6130 91.5746 91.5050 91.3993 91.2977
90.6253 91.1090 91.5964 92.0845 92.5565 93.0199 93.5010 93.9785 94.4335 94.8851
the 2nd curve has 8 points.
15.2091 15.0894 14.9765 14.8567 14.7360 14.6144 14.4695 14.3017
90.1138 89.9824 89.8683 89.7716 89.6889 89.6040 89.4928 89.3624
99.4393 99.9066 100.3802 100.8559 101.3340 101.8115 102.2770 102.7296
a desired curve is one that could represent these two exist curves.
I have thinking of make these curves as points scatters and fit a line out of them. But only a straight line can I get from many code snippet online.
So did I missing something or could someone provide some hint. Thanks.
Hard to come up with a bulletproof solution without more details, but here's an approach that works for the sample data provided. I found the line of best fit for all the points, and then parameterized all the points along that line of best fit. Then I did least-squares polynomial fitting for each dimension separately. This produced a three-dimensional parametric curve that seems to fit the data just fine.
Note that curve fitting approaches other than polynomial least-squares might be better suited to some cases---just substitute the preferred fitting function for polyfit and polyval.
Hope this is helpful!
clear;
close all;
pts1=[18.5860 18.4683 18.3576 18.2491 18.0844 17.9016 17.7709 17.6401 17.4617 17.2726;
91.6178 91.5711 91.5580 91.5580 91.5701 91.6130 91.5746 91.5050 91.3993 91.2977;
90.6253 91.1090 91.5964 92.0845 92.5565 93.0199 93.5010 93.9785 94.4335 94.8851]';
pts2=[ 15.2091 15.0894 14.9765 14.8567 14.7360 14.6144 14.4695 14.3017;
90.1138 89.9824 89.8683 89.7716 89.6889 89.6040 89.4928 89.3624;
99.4393 99.9066 100.3802 100.8559 101.3340 101.8115 102.2770 102.7296]';
%Combine all of our curves into a single point cloud
X = [pts1;pts2];
%=======================================================
%We want to first find the line of best fit
%This line will provide a parameterization of the points
%See accepted answer to http://stackoverflow.com/questions/10878167/plot-3d-line-matlab
% calculate centroid
x0 = mean(X)';
% form matrix A of translated points
A = [(X(:, 1) - x0(1)) (X(:, 2) - x0(2)) (X(:, 3) - x0(3))];
% calculate the SVD of A
[~, S, V] = svd(A, 0);
% find the largest singular value in S and extract from V the
% corresponding right singular vector
[s, i] = max(diag(S));
a = V(:, i);
%=======================================================
a=a / norm(a);
%OK now 'a' is a unit vector pointing along the line of best fit.
%Now we need to compute a new variable, 't', for each point in the cloud
%This 't' value will parameterize the curve of best fit.
%Essentially what we're doing here is taking the dot product of each
%shifted point (contained in A) with the normal vector 'a'
t = A * a;
tMin = min(t);
tMax = max(t);
%This variable represents the order of our polynomial fit
%Use the smallest number that produces a satisfactory result
polyOrder = 8;
%Polynomial fit all three dimensions separately against t
pX = polyfit(t,X(:,1),polyOrder);
pY = polyfit(t,X(:,2),polyOrder);
pZ = polyfit(t,X(:,3),polyOrder);
%And that's our curve fit: (pX(t),pY(t),pZ(t))
%Now let's plot it.
tFine = tMin:.01:tMax;
fitXFine = polyval(pX,tFine);
fitYFine = polyval(pY,tFine);
fitZFine = polyval(pZ,tFine);
figure;
scatter3(X(:,1),X(:,2),X(:,3));
hold on;
plot3(fitXFine,fitYFine,fitZFine);
hold off;
I have got a cell, which is like this : Data={[2,3],[5,6],[1,4],[6,7]...}
The number in every square brackets represent x and y of a point respectively. There will be a new coordinate into the cell in every loop of my algorithm.
I want to plot these points into a time-changing curve, which will tell me the trajectory of the point.
As a beginner of MATLAB, I have no idea of this stage. Thanks for your help.
Here is some sample code to get you started. It uses some basic Matlab functionalities that you will hopefully find useful as you continue using it. I added come data points to you cell array for illustrative purposes.
The syntax to access elements into the cell array might seem weird but is important. Look here for details about cell array indexing.
In order to give nice colors to the points, I generated an array based on the jet colormap built-in in Matlab. Basically issuing the command
Colors = jet(N)
create a N x 3 matrix in which every row is a 3-element color ranging from blue to red. That way you can see which points were detected before other (i.e. blue before red). Of course you can change that to anything you want (look here if you're interested).
So here is the code. If something is unclear please ask for clarifications.
clear
clc
%// Get data
Data = {[2,3],[5,6],[1,4],[6,7],[8,1],[5,2],[7,7]};
%// Set up a matrix to color the points. Here I used a jet colormap
%// available from MATLAB but that could be anything.
Colors = jet(numel(Data));
figure;
%// Use "hold all" to prevent the content of the figure to be overwritten
%// at every iterations.
hold all
for k = 1:numel(Data)
%// Note the syntax used to access the content of the cell array.
scatter(Data{k}(1),Data{k}(2),60,Colors(k,:),'filled');
%// Trace a line to link consecutive points
if k > 1
line([Data{k-1}(1) Data{k}(1)],[Data{k-1}(2) Data{k}(2)],'LineStyle','--','Color','k');
end
end
%// Set up axis limits
axis([0 10 0 11])
%// Add labels to axis and add a title.
xlabel('X coordinates','FontSize',16)
ylabel('Y coordinates','FontSize',16)
title('This is a very boring title','FontSize',18)
Which outputs the following:
This would be easier to achieve if all of your data was stored in a n by 2 (or 2 by n) matrix. In this case, each row would be a new entry. For example:
Data=[2,3;
5,6;
1,4;
6,7];
plot(Data(:, 1), Data(:, 2))
Would plot your points. Fortunately, Matlab is able to handle matrices which grow on every iteration, though it is not recommended.
If you really wanted to work with cells, there are a couple of ways you could do it. Firstly, you could assign the elements to a matrix and repeat the above method:
NumPoints = numel(Data);
DataMat = zeros(NumPoints, 2);
for I = 1:NumPoints % Data is a cell here
DataMat(I, :) = cell2mat(Data(I));
end
You could alternatively plot the elements straight from the cell, though this would limit your plot options.
NumPoints = numel(Data);
hold on
for I = 1:NumPoints
point = cell2mat(Data(I));
plot(point(1), point(2))
end
hold off
With regards to your time changing curve, if you find that Matlab starts to slow down after it stores lots of points, it is possible to limit your viewing window in time with clever indexing. For example:
index = 1;
SamplingRate = 10; % How many times per second are we taking a sample (Hertz)?
WindowTime = 10; % How far into the past do we want to store points (seconds)?
NumPoints = SamplingRate * WindowTime
Data = zeros(NumPoints, 2);
while running
% Your code goes here
Data(index, :) = NewData;
index = index + 1;
index = mod(index-1, NumPoints)+1;
plot(Data(:, 1), Data(:, 2))
drawnow
end
Will store your data in a Matrix of fixed size, meaning Matlab won't slow down.
I have a set of data with over 4000 points. I want to exclude grooves from them, ideally from the point from which they start. The data look for example like this:
The problem with this is the noise I get at the top of the plateaus. I have an idea, in which I would take an average value of the most common within some boundaries (again, ideally sth like the red line here:
and then I would construct a temporary matrix, which would fill up one by one with Y if they are less than this average. If the Y(i) would rise above average, the matrix would find its minima and compare it with the global minima. If the temporary matrix's minima wouldn't be sth like 80% of the global minima, it would be discarded as noise.
I've tried using mean(Y), interpolating and fitting it in a polynomial (the green line) - none of those method would cut it to the point I would be satisfied.
I need this to be extremely robust and it doesn't need to be quick. The top and bottom values can vary a lot, as well as the shape of the plateaus. The groove width is more or less the same.
Do you have any ideas? Again, the point is to extract the values that would make the groove.
How about a median filter?
Let's define some noisy data similar to yours, and plot it in blue:
x = .2*sin((0:9999)/1000); %// signal
x(1000:1099) = x(1000:1099) + sin((0:99)/50*pi); %// noise: spike
x(5000:5199) = x(5000:5199) - sin((0:199)/100*pi); %// noise: wider spike
x = x + .05*sin((0:9999)/10); %// noise: high-freq ripple
plot(x)
Now apply the median filter (using medfilt2 from the Image Processing Toolbox) and plot in red. The parameter k controls the filter memory. It should chosen to be large compared to noise variations, and small compared to signal variations:
k = 500; %// filter memory. Choose as needed
y = medfilt2(x,[1 k]);
hold on
plot(y, 'r', 'linewidth', 2)
In case you don't have the image processing toolbox and can't use medfilt2 a method that's more manual. Skip the extreme values, and do a curve fit with sin1 as curve type. Note that this will only work if the signal is in fact a sine wave!
x = linspace(0,3*pi,1000);
y1 = sin(x) + rand()*sin(100*x).*(mod(round(10*x),5)<3);
y2 = 20*(mod(round(5*x),5) == 0).*sin(20*x);
y = y1 + y2; %// A messy sine-wave
yy = y; %// Store the messy sine-wave
[~, idx] = sort(y);
y(idx(1:round(0.15*end))) = y(idx(round(0.15*end))); %// Flatten out the smallest values
y(idx(round(0.85*end):end)) = y(idx(round(0.85*end)));%// Flatten out the largest values
[foo goodness output] = fit(x.',y.', 'sin1'); %// Do a curve fit
plot(foo,x,y) %// Plot it
hold on
plot(x,yy,'black')
Might not be perfect, but it's a step in the right direction.
I am having difficulty with calculating 2D area of contours produced from a Kernel Density Estimation (KDE) in Matlab. I have three variables:
X and Y = meshgrid which variable 'density' is computed over (256x256)
density = density computed from the KDE (256x256)
I run the code
contour(X,Y,density,10)
This produces the plot that is attached. For each of the 10 contour levels I would like to calculate the area. I have done this in some other platforms such as R but am having trouble figuring out the correct method / syntax in Matlab.
C = contourc(density)
I believe the above line would store all of the values of the contours allowing me to calculate the areas but I do not fully understand how these values are stored nor how to get them properly.
This little script will help you. Its general for contour. Probably working for contour3 and contourf as well, with adjustments of course.
[X,Y,Z] = peaks; %example data
% specify certain levels
clevels = [1 2 3];
C = contour(X,Y,Z,clevels);
xdata = C(1,:); %not really useful, in most cases delimters are not clear
ydata = C(2,:); %therefore further steps to determine the actual curves:
%find curves
n(1) = 1; %n: indices where the certain curves start
d(1) = ydata(1); %d: distance to the next index
ii = 1;
while true
n(ii+1) = n(ii)+d(ii)+1; %calculate index of next startpoint
if n(ii+1) > numel(xdata) %breaking condition
n(end) = []; %delete breaking point
break
end
d(ii+1) = ydata(n(ii+1)); %get next distance
ii = ii+1;
end
%which contourlevel to calculate?
value = 2; %must be member of clevels
sel = find(ismember(xdata(n),value));
idx = n(sel); %indices belonging to choice
L = ydata( n(sel) ); %length of curve array
% calculate area and plot all contours of the same level
for ii = 1:numel(idx)
x{ii} = xdata(idx(ii)+1:idx(ii)+L(ii));
y{ii} = ydata(idx(ii)+1:idx(ii)+L(ii));
figure(ii)
patch(x{ii},y{ii},'red'); %just for displaying purposes
%partial areas of all contours of the same plot
areas(ii) = polyarea(x{ii},y{ii});
end
% calculate total area of all contours of same level
totalarea = sum(areas)
Example: peaks (by Matlab)
Level value=2 are the green contours, the first loop gets all contour lines and the second loop calculates the area of all green polygons. Finally sum it up.
If you want to get all total areas of all levels I'd rather write some little functions, than using another loop. You could also consider, to plot just the level you want for each calculation. This way the contourmatrix would be much easier and you could simplify the process. If you don't have multiple shapes, I'd just specify the level with a scalar and use contour to get C for only this level, delete the first value of xdata and ydata and directly calculate the area with polyarea
Here is a similar question I posted regarding the usage of Matlab contour(...) function.
The main ideas is to properly manipulate the return variable. In your example
c = contour(X,Y,density,10)
the variable c can be returned and used for any calculation over the isolines, including area.
I apologize for the ambiguous title, but I am not entirely sure how to phrase this one. So bear with me.
I have a matrix of data. Each column and row represents a certain vector (column 1 = row 1, column 2 = row 2, etc.), and every cell value is the cosine similarity between the corresponding vectors. So every value in the matrix is a cosine.
There are a couple of things I want to do with this. First, I want to create a figure that shows all of the vectors on it. I know the cosine of the angle between every vector, and I know the magnitude of each vector, but that is the only information I have - is there some algorithm I can implement that will run through all of the various pair-wise angles and display it graphically? That is, I don't know where all the vectors are in relation to each other, and there are too many data points to do this by hand (e.g. if I only had three vectors, and the angles between them all were 45, 12, and 72 degrees it would be trivial). So how do I go about doing this? I don't even have the slightest idea what sort of mathematical function I would need to do this. (I have 83 vectors, so that's thousands of cosine values). So basically this figure (it could be either 2D or multidimensional, and to be honest I would like to do both) would show all of the vectors and how they relate to each other in space (so I could compare both angles and relative magnitudes).
The other thing I would like to do is simpler but I am having a hard time figuring it out. I can convert the cosine values into Cartesian coordinates and display them in a scatter plot. Is there a way to connect each of the points of a scatter plot to (0,0) on the plot?
Finally, in trying to figure out how to do some of the above on my own I have run into some inconsistencies. I calculated the mean angles and Cartesian coordinates for each of the 83 vectors. The math for this is easy, and I have checked and double-checked it. However, when I try to plot it, different plotting methods give me radically different things. So, if I plot the Cartesian coordinates as a scatter plot I get this:
If I plot the mean angles in a compass plot I get this:
And if I use a quiver plot I get something like this (I transformed this a little by shifting the origin up and to the right just so you can see it better):
Am I doing something wrong, or am I misunderstanding the plotting functions I am using? Because these results all seem pretty inconsistent. The mean angles on the compass plot are all <30 degrees or so, but on the quiver plot some seem to exceed 90 degrees, and on the scatter plot they extend above 30 as well. What's going on here?
(Here is my code:)
cosine = load('LSA.txt');
[rows,columns]=size(cosine);
p = cosine.^2;
pp = bsxfun(#minus, 1, p);
sine = sqrt(pp);
tangent = sine./cosine;
Xx = zeros(rows,1);
Yy = zeros(rows,1);
for i = 1:columns
x = cosine(:,i);
y = sine(:,i);
Xx(i,1) = sum(x) * (1/columns);
Yy(i,1) = sum(y) * (1/columns);
end
scatter(Xx,Yy);
Rr = zeros(rows,1);
Uu = zeros(rows,1);
for j = 1:rows
Rr(j,1) = sqrt(Xx(j,1).^2 + Yy(j,1).^2);
Uu(j,1) = atan2(Xx(j,1),Yy(j,2));
end
%COMPASS PLOT
[theta,rho] = pol2cart(Uu,1);
compass(theta,rho);
%QUIVER PLOT
r = 7;
sx = ones(size(cosine))*2; sy = ones(size(cosine))*2;
pu = r * cosine;
pv = r * sine;
h = quiver(sx,sy,pu,pv);
set(gca, 'XLim', [1 10], 'YLim', [1 10]);
You can exactly solve this problem. The dot product calculates the cosine. This means your matrix is actually M=V'*V
This should be solvable through eigenvalues. And you said you also have the length.
Your only problem - as your original matrix the vectors will be 83 dimensional. Not easy to plot in 2 or 3 dimensions. I think you are over simplifying by just using the average angle. There are some techniques called dimensionality reduction - here's a toolbox. I would suggest a sammon projection on 1-cosine (as this would be the distance of points on the unit ball) to calculate the vectors for such a plot.
In the quiver plot, you are plotting all of the data in the cosine and sine matrices. In the other plots, you are only plotting the means. The first two plots appear to match up, so no problem there.
A few other things. I notice that in
Uu(j,1) = atan2(Xx(j,1),Yy(j,2));
Yy(j,2) is not actually defined, so it seems like this code should fail.
Furthermore, you could define Yy and Xx as:
Xx = mean(cosine,2);
Yy = mean(sine,2);
And also get rid of the other for loop:
Rr = sqrt(Xx.^2 + Yy.^2)
Uu = atan2(Xx,Yy)
I still have to think about your first question, but I hope this was helpful.