I have a 4032 X 102 matrix (first 2 columns are the coordinates). I would like to interpolate every column by a 48 X 84 meshgrid. It's working column-by-column, but it would be great if it can be done by one command (with a for loop maybe).
x = 1:84; y = 1:48;
[X,Y] = meshgrid(x,y);
Z = griddata(data(:,1),data(:,2),(:,3:102),X,Y'v4');
The input data is also grid data but not in matrix form. My goal is to create map(grid again) from this. So there are X, Y coordinates and values. Each column represent data of a map, and the values along the coordinates.
First 2 columns contain the coordinates of the data point. These are the first 2 row of a map with 4 cell spacing. Z is the gridded data matrix along the 48 x 84 grid.
Many thanks!
You can reshape a 4032x1 vector into a 48x84 2-D matrix:
reshape(vector, 48,84)
Since you have 102 of them, and they are already stored in a single matrix variable, you can now store each single matrix in the first two dimensions (1st dim has 48 elements and 2nd has 84) and all 102 matrices are indexed in the third dimension.
reshape(data, 48,84,102)
Related
I have some time/frequency data and I try to interpolate it using the interp2 function of Matlab. The data [F,T,data] is obtained from a different Matlab routine of Matlab, spectrogram in case you are interested.
[~,F,T,data] = spectrogram(...)
data = 10*log10(data);
I can plot the data using surf. The data is fine, I believe. However interpolating the data seems to be a problem. Even using interp2(F,T,data,F,T) (so actually no interpolating) gives the error below.
What is going wrong here?
I have the data which I use here: https://www.dropbox.com/s/zr5zpfhp6qyarzw/test.mat
interp2(F,T,data,f,t)
Error using griddedInterpolant
The grid vectors do not define a grid of points that match the given values.
Error in interp2>makegriddedinterp (line 228)
F = griddedInterpolant(varargin{:});
Error in interp2 (line 128)
F = makegriddedinterp({X, Y}, V, method,extrap);
>> size(F),size(T),size(data),size(f),size(t)
ans =
129 1
ans =
1 52
ans =
129 52
ans =
200 1
ans =
1 121
The problem is that you should swap F and T:
interp2(T,F,data,t,f);
The first argument corresponds with the columns of the matrix and the second argument with the rows as documented here:
If X and Y are grid vectors, then V must be a matrix containing
length(Y) rows and length(X) columns.
As an alternative, you may take the transpose of data:
interp2(F,T,data',f,t);
Reasoning behind (strange) argument order
interp2(X,Y,V,Xq,Yq) is interpreted as the interpolation of a function, represented by the matrix V, i.e. the sample values. The problem is that the arguments/indexes of a function/matrix are rather supplied in an opposite order:
matrix(row, column)
versus
function(x,y)
x (first argument) often represents the horizontal axes and therefore corresponds with the column (second argument) argument and idem for y and row.
Let us have a 4D matrix (tensor), output:
[X,Y,Z] = ndgrid(-50:55,-55:60,-50:60);
a = 1:4;
output = zeros([size(X),length(a)]);
Next, we determine the area inside the ellipsoid:
position = [0,0,0];
radius = [10,20,10];
test_func = #(X,Y,Z) ((X-position(1))/radius(1)).^2 ...
+ ((Y-position(2))/radius(2)).^2 ...
+ ((Z-position(3))/radius(3)).^2 <= 1;
condition = test_func(X,Y,Z);
I need to fill the matrix output inside the ellipsoid for the first 3 dimensions. But for the fourth dimension I need to fill a. I need to do something like this:
output(condition,:) = a;
But it does not work. How to do it? Any ideas please!
If I understand your question correctly, you have a 4D matrix where each temporal blob of pixels in 3D for each 4D slice is filled with a number... from 1 up to 4 where each number tells you which slice you're in.
You can cleverly use bsxfun and permute to help you accomplish this task:
output = bsxfun(#times, double(condition), permute(a, [1 4 3 2]));
This takes a bit of imagination to imagine how this works but it's quite simple. condition is a 3D array of logical values where each location in this 3D space is either 0 if it doesn't or 1 if it does belong to a point inside an ellipsoid. a is a row vector from 1 through 4 or however many elements you want this to end with. Let's call this N to be more general.
permute(a, [1 4 3 2]) shuffles the dimensions such that we create a 4D vector where we have 1 row, 1 column and 1 slice but we have 4 elements going into the fourth dimension.
By using bsxfun in this regard, it performs automatic broadcasting on two inputs where each dimension in the output array will match whichever of the two inputs had the largest value. The condition is that for each dimension independently, they should both match or one of them is a singleton dimension (i.e. 1).
Therefore for your particular example, condition will be 106 x 116 x 111 while the output of the permute operation will be 1 x 1 x 1 x N. condition is also technically 106 x 116 x 111 x 1 and using bsxfun, we would thus get an output array of size 106 x 116 x 111 x N. Performing the element-wise #times operation, the permuted vector a will thus broadcast itself over all dimensions where each 3D slice i will contain the value of i. The condition matrix will then duplicate itself over the fourth dimension so we have N copies of the condition matrix, and performing the element-wise multiply thus completes what you need. This is doable as the logical mask you created contains only 0s and 1s. By multiplying element-wise with this mask, only the values that are 1 will register a change. Specifically, if you multiply the values at these locations by any non-zero value, they will change to these non-zero values. Using this logic, you'd want to make these values a. It is important to note that I had to cast condition to double as bsxfun only allows two inputs of the same class / data type to be used.
To visually see that this is correct, I'll show scatter plots of each blob where the colour of each blob would denote what label in a it belongs to:
close all;
N = 4;
clrs = 'rgbm';
figure;
for ii = 1 : N
blob = output(:,:,:,ii);
subplot(2,2,ii);
plot3(X(blob == ii), Y(blob == ii), Z(blob == ii), [clrs(ii) '.']);
end
We get:
Notice that the spatial extent of each ellipsoid is the same but what is different are the colours assigned to each blob. I've made it such that the values for the first blob, or those assigned to a = 1 are red, those that are assigned to a = 2 are assigned to green, a = 3 to blue and finally a = 4 to magenta.
If you absolutely want to be sure that the coordinates of each blob are equal across all blobs, you can use find in each 3D blob individually and ensure that the non-zero indices are all equal:
inds = arrayfun(#(x) find(output(:,:,:,x)), a, 'un', 0);
all_equal = isequal(inds{:});
The code finds in each blob the column major indices of all non-zero locations in 3D. To know whether or not we got it right, each 3D blob should all contain the same non-zero column major indices. The arrayfun call goes through each 3D blob and returns a cell array where each cell element is the column major indices for a particular blob. We then pipe this into isequal to ensure that all of these column major indices arrays are equal. We get:
>> all_equal
all_equal =
1
I have x,y pixel coordinates of multiple objects that have been tracked from an image (3744x5616). The coordinates are stored in a structure called objects, e.g.
objects(1).centre = [1868 1236]
The objects are each uniquely identified by a numerical code, e.g.
objects(i).code = 33
I want to be able to record each time any two objects come within a radius of 300 pixels each other. What would be the best way to check through if any objects are touching and then record the identity of both objects involved in the interaction, like object 33 interacts with object 34.
Thanks!
Best thing I can think of right now is a brute force approach. Simply check the distances from one object's centre with the rest of the other objects and manually check if the distances are < 300 pixels.
If you want this fast, we should probably do this without any toolboxes. You can intelligently do this with vanilla MATLAB using bsxfun. First, create separate arrays for the X and Y coordinates of each object:
points = reshape([objects.centre], 2, []);
X = points(1,:);
Y = points(2,:);
[objects.centre] accesses the individual coordinates of each centre field in your structure and unpacks them into a comma-separated list. I reshape this array so that it is 2 rows where the first row is the X coordinate and the second row is the Y coordinate. I extract out the rows and place them into separate arrays.
Next, create two difference matrices for each X and Y where the rows denote one unique coordinate and the columns denote another unique coordinate. The values inside this matrix are the differences between the point i at row i and point j at column j:
Xdiff = bsxfun(#minus, X.', X);
Ydiff = bsxfun(#minus, Y.', Y);
bsxfun stands for Binary Singleton EXpansion FUNction. If you're familiar with the repmat function, it essentially replicates matrices and vectors under the hood so that both inputs you're operating on have the same size. In this case, what I'm doing is specifying X or Y as both of the inputs. One is the transposed version of the other. By doing this bsxfun automatically broadcasts each input so that the inputs match in dimension. Specifically, the first input is a column vector of X and so this gets repeated and stacked horizontally for as many times as there are values in X.
Similarly this is done for the Y value. After you do this, you perform an element-wise subtraction for both outputs and you get the component wise subtraction between one point and another point for X and Y where the row gives you the first point, and the column gives you the second point. As a toy example, imagine we had X = [1 2 3]. Doing a bsxfun call using the above code gives:
>> Xdiff = bsxfun(#minus, [1 2 3].', [1 2 3])
Xdiff =
## | 1 2 3
----------------------
1 | 0 -1 -2
2 | 1 0 -1
3 | 2 1 0
There are some additional characters I placed in the output, but these are used solely for illustration and to give you a point of reference. By taking a row value from the ## column and subtracting from a column value from the ## row gives you the desired subtract. For example, the first row second column illustrates 1 - 2 = -1. The second row, third column illustrates 2 - 3 = -1. If you do this for both the X and Y points, you get the component-wise distances for one point against all of the other points in a symmetric matrix.
You'll notice that this is an anti-symmetric matrix where the diagonal is all 0 ... makes sense since the distance of one dimension of one point with respect to itself should be 0. The bottom left triangular portion of the matrix is the opposite sign of the right... because of the order of subtraction. If you subtracted point 1 with point 2, doing the opposite subtraction gives you the opposite sign. However, let's assume that the rows denote the first object and the columns denote the second object, so you'd want to concentrate on the lower half.
Now, compute the distance, and make sure you set either the upper or lower triangular half to NaN because when computing the distance, the sign gets ignored. If you don't ignore this, we'd find duplicate objects that interact, so object 3 and object 1 would be a different interaction than object 1 and object 3. You obviously don't care about the order, so set either the upper or lower triangular half to NaN for the next step. Assuming Euclidean distance:
dists = sqrt(Xdiff.^2 + Ydiff.^2);
dists(tril(ones(numel(objects))==1)) = NaN;
The first line computes the Euclidean distance of all pairs of points and we use tril to extract the lower triangular portion of a matrix that consists of all logical 1. Extracting this matrix, we use this to set the lower half of the matrix to NaN. This allows us to skip entries we're not interested in. Note that I also set the diagonal to 0, because we're not interested in distances of one object to itself.
Now that you're finally here, search for those objects that are < 300 pixels:
[I,J] = find(dists < 300);
I and J are row/column pairs that determine which rows and columns in the matrix have values < 300, so in our case, each pair of I and J in the array gives you the object locations that are close to each other.
To finally figure out the right object codes, you can do:
codes = [[objects(I).code].' [objects(J).code].'];
This uses I and J to access the corresponding codes of those objects that were similar in a comma-separated list and places them side by side into a N x 2 matrix. As such, each row of codes gives you unique pairs of objects that satisfied the distance requirements.
For copying and pasting:
points = reshape([objects.centre], 2, []);
X = points(1,:);
Y = points(2,:);
Xdiff = bsxfun(#minus, X.', X);
Ydiff = bsxfun(#minus, Y.', Y);
dists = sqrt(Xdiff.^2 + Ydiff.^2);
dists(tril(ones(numel(objects))==1)) = NaN;
[I,J] = find(dists < 300);
codes = [[objects(I).code].' [objects(J).code].'];
Toy Example
Here's an example that we can use to verify if what we have is correct:
objects(1).centre = [1868 1236];
objects(2).centre = [2000 1000];
objects(3).centre = [1900 1300];
objects(4).centre = [3000 2000];
objects(1).code = 33;
objects(2).code = 34;
objects(3).code = 35;
objects(4).code = 99;
I initialized 4 objects with different centroids and different codes. Let's see what the dists array gives us after we compute it:
>> format long g
>> dists
dists =
NaN 270.407100498489 71.5541752799933 1365.69396278961
NaN NaN 316.227766016838 1414.2135623731
NaN NaN NaN 1303.84048104053
NaN NaN NaN NaN
I intentionally made the last point farther than any of the other three points to ensure that we can show cases where there are points not near other ones.
As you can see, points (1,2) and (1,3) are all near each other, which is what we get when we complete the rest of the code. This corresponds to objects 33, 34 and 35 with pairings of (33,34) and (33,35). Points with codes 34 and 35 I made slightly smaller, but they are still greater than the 300 pixel threshold, so they don't count either:
>> codes
codes =
33 34
33 35
Now, if you want to display this in a prettified format, perhaps use a for loop:
for vec = codes.'
fprintf('Object with code %d interacted with object with code %d\n', vec(1), vec(2));
end
This for loop is a bit tricky. It's a little known fact that for loops can also accept matrices and the index variable gives you one column of each matrix at a time from left to right. Therefore, I transposed the codes array so that each pair of unique codes becomes a column. I just access the first and second element of each column and print it out.
We get:
Object with code 33 interacted with object with code 34
Object with code 33 interacted with object with code 35
I have a problem with MATLAB bar3 plots: Here is what I have:
m x n Array Values containing values of a measurement.
Another m x n Array Angles Represents the angle at which a value was measured (e.g. the 3rd value was measured at an angle of 90°). The angular values for each measurement value are stored in another variable.
I need a range for my x-axis from -180° to +180°. This alone is no problem. But how do I hand over my measurement values? I have to somehow link them to the angular values. So that each value in Values is somehow linked to it's angular value in Angles. For my y-axis, I can simply count from 0 to the amount of rows of my Values Array.
EXAMPLE:
Valueslooks like:
3 5 6
2 1 7
5 8 2
Angles looks like:
37° 38° 39°
36° 37° 38°
34° 35° 36°
Values(1,1) = 3 was measured at Angles(1,1) = 37° for example.
At each angle, the number of bars varies depending on how many measurements exist for that angle. bar3 needs a matrix input. In order to build a matrix, missing values are filled with NaN.
Warning: NaNs are usually ignored by plotting commands, but bar3 apparently breaks this convention. It seems to replace NaNs by zeros! So at missing values you'll get a zero-height bar (instead of no bar at all).
[uAngles, ~, uAngleLabels] = unique(Angles); %// get unique values and
%// corresponding labels
valuesPerAngle = accumarray(uAngleLabels(:), Values(:), [], #(v) {v});
%// cell array where each cell contains all values corresponding to an angle
N = max(cellfun(#numel, valuesPerAngle));
valuesPerAngle = cellfun(#(c) {[c; NaN(N-numel(c),1)]}, valuesPerAngle);
%// fill with NaNs to make all cells of equal lenght, so that they can be
%// concatenated into a matrix
valuesPerAngle = cat(2, valuesPerAngle{:}); %// matrix of values for each angle,
%// filled with NaNs where needed
bar3(uAngles, valuesPerAngle.'); %'// finally, the matrix can be plotted
ylabel('Angles')
xlabel('Measurement')
With your example Values and Angles this gives:
I would like to plot two matrices both of (4*36 double) size. The first contains rho and the second contains depths for the 36 locations
well I looked into surf but it reads two arrays and one matrix rather than two matrices and yes I would like to plot them as column graph
here is an example
rho= magic(36);
rho(5:1:end,:)=[];
D= magic(36);
D(5:1:end,:)=[];
D=sort(depth);
So right now the matrix rho contains the densities for the 36 location at four different depths. The matrix D contains the four different depths at which the reading at rho is found. The first element in the first matrix corresponds to the first element in the second matrix and so on
in the end what I would like to have is the 36 column with the different reading from (rho) plotted against appropriate depth in (D)
I hope I helped make it clearer somehow
Simple example of plotting four sets of X and Y data:
X = repmat(1:36, [4 1]);
Y(1,:) = rand(1,36);
Y(2,:) = 0.2 * (1:36);
Y(3,:) = 5 * sin(linspace(-pi,pi,36));
Y(4,:) = 0.1 * (1:36).^2;
figure
plot(X', Y')
This results in
Note - in order to get four series to plot like this, the data has to be in COLUMNS. The original data was in 4x36 matrix, so it was in ROWS. I used the transpose operator (apostrophe - X' rather than just X) to get the data organized in columns.
Maybe this helps...