I have a 3 dimensional array (10x3x3) in Matlab and I want to change any value greater than 999 to Inf. However, I only want this to apply to (:,:,2:3) of this array.
All the help I have found online seems to only apply to the whole array, or 1 column of a 2D array. I can't work out how to apply this to a 3D array.
I have tried the following code, but it becomes a 69x3x3 array after I run it, and I don't really get why. I tried to copy the code from someone using a 2D array, so I just think I don't really understand what the code is doing.
A(A(:,:,2)>999,2)=Inf;
A(A(:,:,3)>999,3)=Inf;
One approach with logical indexing -
mask = A>999; %// get the 3D mask
mask(:,:,1) = 0; %// set all elements in the first 3D slice to zeros,
%// to neglect their effect when we mask the entire input array with it
A(mask) = Inf %// finally mask and set them to Infs
Another with linear indexing -
idx = find(A>999); %// Find linear indices that match the criteria
valid_idx = idx(idx>size(A,1)*size(A,2)) %// Select indices from 2nd 3D slice onwards
A(valid_idx)=Inf %// Set to Infs
Or yet another with linear indexing, almost same as the previous one with the valid index being calculated in one step and thus enabling us a one-liner -
A(find(A(:,:,2:3)>999) + size(A,1)*size(A,2))=Inf
Related
I have a 2D array PointAndTangent of dimension 8500 x 5. The data is row-wise with 8500 data rows and 5 data values for each row. I need to extract the row index of an element in 4th column when this condition is met, for any s:
abs(PointAndTangent[:,3] - s) <= 0.005
I just need the row index of the first match for the above condition. I tried using the following:
index = np.all([[abs(s - PointAndTangent[:, 3])<= 0.005], [abs(s - PointAndTangent[:, 3]) <= 0.005]], axis=0)
i = int(np.where(np.squeeze(index))[0])
which doesn't work. I get the follwing error:
i = int(np.where(np.squeeze(index))[0])
TypeError: only size-1 arrays can be converted to Python scalars
I am not so proficient with NumPy in Python. Any suggestions would be great. I am trying to avoid using for loop as this is small part of a huge simulation that I am trying.
Thanks!
Possible Solution
I used the following
idx = (np.abs(PointAndTangent[:,3] - s)).argmin()
It seems to work. It returns the row index of the nearest value to s in the 4th column.
You were almost there. np.where is one of the most abused functions in numpy. Half the time, you really want np.nonzero, and the other half, you want to use the boolean mask directly. In your case, you want np.flatnonzero or np.argmax:
mask = abs(PointAndTangent[:,3] - s) <= 0.005
mask is a 1D array with ones where the condition is met, and zeros elsewhere. You can get the indices of all the ones with flatnonzero and select the first one:
index = np.flatnonzero(mask)[0]
Alternatively, you can select the first one directly with argmax:
index = np.argmax(mask)
The solutions behave differently in the case when there are no rows meeting your condition. Three former does indexing, so will raise an error. The latter will return zero, which can also be a real result.
Both can be written as a one-liner by replacing mask with the expression that was assigned to it.
Say A is a 3x4x5 array. I am given a vector a, say of dimension 2 and b of dimension 2. If I do A(a,b,:) it will give 5 matrices of dimensions 2x2. I instead want the piecewise vectors (without writing a for loop).
So, I want the two vectors of A which are given by (a's first element and b's first element) and (a's second element and b's second element)
How do I do this without a for loop? If A were two dimensions I could do this using sub2ind. I don't know how to access the entire vectors.
You can use sub2ind to find the linear index to the first element of each output vector: ind = sub2ind(size(A),a,b). To get the whole vectors, you can't do A(ind,:), because the : has to be the 3rd dimension. However, what you can do is reshape A to be 2D, collapsing the first two dimensions into one. We have a linear index to the vectors we want, that will correctly index the first dimension of this reshaped A:
% input:
A = rand(3,4,5);
a = [2,3];
b = [1,2];
% expected:
B = [squeeze(A(a(1),b(1),:)).';squeeze(A(a(2),b(2),:)).']
% solution:
ind = sub2ind(size(A),a,b);
C = reshape(A,[],size(A,3));
C = C(ind,:)
assert(isequal(B,C))
You can change a and b to be 3d arrays just like A and then the sub2ind should be able to index the whole matrix. Like this:
Edit: Someone pointed out a bug. I have changed it so that a correction gets added. The problem was that ind1, which should have had the index number for each desired element of A was only indexing the first "plane" of A. The fix is that for each additional "plane" in the z direction, the total number of elements in A in the previous "planes" must be added to the index.
A=rand(3,4,5);
a=[2,3];
b=[1,2];
a=repmat(a,1,1,size(A,3));
b=repmat(b,1,1,size(A,3));
ind1=sub2ind(size(A),a,b);
correction=(size(A,1)*size(A,2))*(0:size(A,3)-1);
correction=permute(correction,[3 1 2]);
ind1=ind1+repmat(correction,1,2,1);
out=A(ind1)
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 two arrays within a <1x2 cell>. I want to permute those arrays. Of course, I could use a loop to permute each one, but is there any way to do that task at once, without using loops?
Example:
>> whos('M')
Name Size Bytes Class Attributes
M 1x2 9624 cell
>> permute(M,p_matrix)
This does not permute the contents of the two arrays within M.
I could use something like:
>> for k=1:size(M,2), M{k} = permute(M{k},p_matrix); end
but I'd prefer not to use loops.
Thanks.
This seems to work -
num_cells = numel(M) %// Number of cells in input cell array
size_cell = size(M{1}) %// Get sizes
%// Get size of the numeric array that will hold all of the data from the
%// input cell array with the second dimension representing the index of
%// each cell from the input cell array
size_num_arr = [size_cell(1) num_cells size_cell(2:end)]
%// Dimensions array for permuting with the numeric array holding all data
perm_dim = [1 3:numel(size_cell)+1 2]
%// Store data from input M into a vertically concatenated numeric array
num_array = vertcat(M{:})
%// Reshape and permute the numeric array such that the index to be used
%// for indexing data from different cells ends up as the final dimension
num_array = permute(reshape(num_array,size_num_arr),perm_dim)
num_array = permute(num_array,[p_matrix numel(size_cell)+1])
%// Save the numeric array as a cell array with each block from
%// thus obtained numeric array from its first to the second last dimension
%// forming each cell
size_num_arr2 = size(num_array)
size_num_arr2c = num2cell(size_num_arr2(1:end-1))
M = squeeze(mat2cell(num_array,size_num_arr2c{:},ones(1,num_cells)))
Some quick tests show that mat2cell would prove to be the bottleneck, so if you don't mind indexing into the intermediate numeric array variable num_array and use it's last dimension for an equivalent indexing into M, then this approach could be useful.
Now, another approach if you would like to preserve the cell format would be with arrayfun, assuming each cell of M to be a 4D numeric array -
M = arrayfun(#(x) num_array(:,:,:,:,x),1:N,'Uniform',0)
This seems to perform much better than with mat2cell in terms of performance.
Please note that arrayfun isn't a vectorized solution as most certainly it uses loops behind-the-scenes and seems like mat2cell is using for loops inside its source code, so please do keep all these issues in mind.
I'm trying to iterate trough a fixed size 3d array in order to plot the 3rd vector dimension like this:
%respo is a 3D array of fixed size defined above
for ii = 1:size(respo,1)
for jj = 1:size(respo,2)
plot(squeeze(respo(ii,jj,1:8)))
end
end
Is there a better way to do this than by 2 level for loop with pointing exactly to the vector plotted at each iteration?
I get there is a linear indexing for each array in MATLAB, but I struggle to come up with a way that saves from the double-looping.
Well I guess you could reshape it to only need one loop:
respo_2D = reshape(respo, [], size(respo,3))
So now
for ii = 1:size(respo_2D, 1)
plot(respo(ii,1:8));
end
(or potentially even plot(respo_2D(:,1:8)') depending on what you're trying to do)
plot applied to a matrix plots the columns of that matrix. So: rearrange dimensions such that the third becomes the new first and the others get combined into the new second, and then just call plot
plot(reshape(permute(respo, [3 1 2]), size(respo,3), []))