I have 3 graphs of an IV curve (monotonic increasing function. consider a positive quadratic function in the 1st quadrant. Photo attached.) at 3 different temperatures that are not obtained linearly. That is, one is obtained at 25C, one at 125C and one at 150C.
What I want to make is an interpolated 2D array to fill in the other temperatures. My current method to build a meshgrid-type array is as follows:
H = 5;
W = 6;
[Wmat,Hmat] = meshgrid(1:W,1:H);
X = [1:W; 1:W];
Y = [ones(1,W); H*ones(1,W)];
Z = [vecsatIE25; vecsatIE125];
img = griddata(X,Y,Z,Wmat,Hmat,'linear')
This works to build a 6x6 array, which I can then index one row from, then interpolate from that 1D array.
This is really not what I want to do.
For example, the rows are # temps = 25C, 50C, 75C, 100C, 125C and 150C. So I must select a temperature of, say, 50C when my temperature is actually 57.5C. Then I can interpolate my I to get my V output. So again for example, my I is 113.2A, and I can actually interpolate a value and get a V for 113.2A.
When I take the attached photo and digitize the plot information, I get an array of points. So my goal is to input any Temperature and any current to get a voltage by interpolation. The type of interpolation is not as important, so long as it produces reasonable values - I do not want nearest neighbor interpolation, linear or something similar is preferred. If it is an option, I will try different kinds of interpolation later (cubic, linear).
I am not sure how I can accomplish this, ideally. The meshgrid array does not need to exist. I simply need the 1 value.
Thank you.
If I understand the question properly, I think what you're looking for is interp2:
Vq = interp2(X,Y,V,Xq,Yq) where Vq is the V you want, Xq and Yq are the temperature and current, and X, Y, and V are the input arrays for temperature, current, and voltage.
As an option, you can change method between 'linear', 'nearest', 'cubic', 'makima', and 'spline'
Related
I'm quite new to MatLab and this problem really drives me insane:
I have a huge array of 2 column and about 31,000 rows. One of the two columns depicts a spatial coordinate on a grid the other one a dependent parameter. What I want to do is the following:
I. I need to split the array into smaller parts defined by the spatial column; let's say the spatial coordinate are ranging from 0 to 500 - I now want arrays that give me the two column values for spatial coordinate 0-10, then 10-20 and so on. This would result in 50 arrays of unequal size that cover a spatial range from 0 to 500.
II. Secondly, I would need to calculate the average values of the resulting columns of every single array so that I obtain per array one 2-dimensional point.
III. Thirdly, I could plot these points and I would be super happy.
Sadly, I'm super confused since I miserably fail at step I. - Maybe there is even an easier way than to split the giant array in so many small arrays - who knows..
I would be really really happy for any suggestion.
Thank you,
Arne
First of all, since you wish a data structure of array of different size you will need to place them in a cell array so you could try something like this:
res = arrayfun(#(x)arr(arr(:,1)==x,:), unique(arr(:,1)), 'UniformOutput', 0);
The previous code return a cell array with the array splitted according its first column with #(x)arr(arr(:,1)==x,:) you are doing a function on x and arrayfun(function, ..., 'UniformOutput', 0) applies function to each element in the following arguments (taken a single value of each argument to evaluate the function) but you must notice that arr must be numeric so if not you should map your values to numeric values or use another way to select this values.
In the same way you could do
uo = 'UniformOutput';
res = arrayfun(#(x){arr(arr(:,1)==x,:), mean(arr(arr(:,1)==x,2))), unique(arr(:,1)), uo, 0);
You will probably want to flat the returning value, check the function cat, you could do:
res = cat(1,res{:})
Plot your data depends on their format, so I can't help if i don't know how the data are, but you could try to plot inside a loop over your 'res' variable or something similar.
Step I indeed comes with some difficulties. Once these are solved, I guess steps II and III can easily be solved. Let me make some suggestions for step I:
You first define the maximum value (maxValue = 500;) and the step size (stepSize = 10;). Now it is possible to iterate through all steps and create your new vectors.
for k=1:maxValue/stepSize
...
end
As every resulting array will have different dimensions, I suggest you save the vectors in a cell array:
Y = cell(maxValue/stepSize,1);
Use the find function to find the rows of the entries for each matrix. At each step k, the range of values of interest will be (k-1)*stepSize to k*stepSize.
row = find( (k-1)*stepSize <= X(:,1) & X(:,1) < k*stepSize );
You can now create the matrix for a stepk by
Y{k,1} = X(row,:);
Putting everything together you should be able to create the cell array Y containing your matrices and continue with the other tasks. You could also save the average of each value range in a second column of the cell array Y:
Y{k,2} = mean( Y{k,1}(:,2) );
I hope this helps you with your task. Note that these are only suggestions and there may be different (maybe more appropriate) ways to handle this.
I have a large data set with two arrays, say x and y. The arrays have over 1 million data points in size. Is there a simple way to do a scatter plot of only 2000 of these points but have it be representative of the entire set?
I'm thinking along the lines of creating another array r ; r = max(x)*rand(2000,1) to get a random sample of the x array. Is there a way to then find where a value in r is equal to, or close to a value in x ? They wouldn't have to be in the same indexed location but just throughout the whole matrix. We could then plot the y values associated with those found x values against r
I'm just not sure how to code this. Is there a better way than doing this?
I'm not sure how representative this procedure will be of your data, because it depends on what your data looks like, but you can certainly code up something like that. The easiest way to find the closest value is to take the min of the abs of the difference between your test vector and your desired value.
r = max(x)*rand(2000,1);
for i = 1:length(r)
[~,z(i)] = min(abs(x-r(i)));
end
plot(x(z),y(z),'.')
Note that the [~,z(i)] in the min line means we want to store the index of the minimum value in vector z.
You might also try something like a moving average, see this video: http://blogs.mathworks.com/videos/2012/04/17/using-convolution-to-smooth-data-with-a-moving-average-in-matlab/
Or you can plot every n points, something like (I haven't tested this, so no guarantees):
n = 1000;
plot(x(1:n:end),y(1:n:end))
Or, if you know the number of points you want (again, untested):
npoints = 2000;
interval = round(length(x)/npoints);
plot(x(1:interval:end),y(1:interval:end))
Perhaps the easiest way is to use round function and convert things to integers, then they can be compared. For example, if you want to find points that are within 0.1 of the values of r, multiply the values by 10 first, then round:
r = max(x) * round(2000,1);
rr = round(r / 0.1);
xx = round(x / 0.1);
inRR = ismember(xx, rr)
plot(x(inRR), y(inRR));
By dividing by 0.1, any values that have the same integer value are within 0.1 of each other.
ismember returns a 1 for each value of xx if that value is in rr, otherwise a 0. These can be used to select entries to plot.
I am new to Matlab.
Lets say I have an array a = [1:1:1000]
I have to divide this into 50 parts 1-20; 21-40 .... 981-1000.
I am trying to do it this way.
E=1000X
a=[1:E]
n=50
d=E/n
b=[]
for i=0:n
b(i)=a[i:d]
end
But I am unable to get the result.
And the second part I am working on is, depending on another result, say if my answer is 3, the first split array should have a counter and that should be +1, if the answer is 45 the 3rd split array's counter should be +1 and so on and in the end I have to make a histogram of all the counters.
You can do all of this with one function: histc. In your situation:
X = (1:1:1000)';
Edges = (1:20:1000)';
Count = histc(X, Edges);
Essentially, Count contains the number of elements in X that fall into the categories defined in Edges, where Edges is a monotonically increasing vector whose elements define the boundaries of sequential categories. A more common example might be to construct X using a probability density, say, the uniform distribution, eg:
X = 1000 * rand(1000, 1);
Play around with specifications for X and Edges and you should get the idea. If you want the actual histogram plot, look into the hist function.
As for the second part of your question, I'm not really sure what you're asking.
I have an array of 32766 values, which I would like to upsample to fit other arrays of 65534 values.
I could also cycle in a way to take multiple times the same value, but I have to use it several times.
There is a way to increase the number of samples? I've seen the resample function, but it seems for a specific type of object data...
Edit
I was looking for the wrong term: I've found the function interp that upsamples for an integer number, and now I've used it and adapted the array replicating the last two values to fit the other; there is a way to automatically achieve the same size?
You can use interp1:
x = 1:10;
y = x.*x;
%The x values that you want to be interpolated;
xi = 1:0.25:10;
yi = interp1(x,y,xi);
I'm applying the function histcnd to a matrix of size 744x2. This function calculates frequencies of values within certain edges. I want to set the edges to, for example, groups of 5 values, but I can't seem to be able to do it.
The syntax of the function is histcnd(X,edges), where edges must have the same length to the number of columns of X. How do I define 'edges' as a 2-column vector, so that it will group values of each column every 5 values?
What about using something like this:
X = randn(744,2);
[a,b] = size(X);
edges = num2cell([linspace(min(X(1,:)),max(X(1,:)),a/5); linspace(min(X(2,:)), max(X(2,:)),a/5)],2);
# not sure if it's the same histcnd, the one I found wants edges to be a cell array
H = histcnd(X, edges);
You can probably pick the min/max values for each axis in a more intelligent way if you know something about your data.