A MATLAB array has an infinite number of trailing singleton dimension (theoretically).
So a 1x10 array is in fact a 1x10x1x1x1... array. I suppose that MATLAB just hides the trailing singleton dimension.
You can observe this behaviour with this code snippet:
permute([1:10],[2,1,4,3])
MATLAB deals with permuting the 3rd and 4th dimension without any problem even if it's a 1D array.
But why does MATLAB not hide the trailing singleton dimension in the 2nd dimension?
size(rand(10,1))
gives:
ans =
10 1
size(rand(10,1,1))
also gives:
ans =
10 1
Is that because MATLAB is optimized for linear algebra? So, is it simply a convention?
Related
If I have a square matrix of arrays such as:
[1,2], [2,3]
[5,9], [1,4]
And I want to get the mean of the first values in the arrays of each row such:
1.5
3
Is this possible in Matlab?
I've used the mean(matrix, 2) command to do this with a matrix of single values, but I'm not sure how to extend this to deal with the arrays.
Get the first elements in all arrays of matrix, then call mean function
mean(matrix(:,:,1))
maybe you need to reshape before call mean
a = matrix(:,:,1);
mean(a(:))
You can apply mean function inside mean function to get the total mean value of the 2D array at index 1. You can do similary with array at index 2. Consider the following snapshot.
After staring at your problem for a long time, it looks like your input is a 3D matrix where each row of your formatting corresponds to a 2D matrix slice. Therefore, in proper MATLAB syntax, your matrix is actually:
M = cat(3, [1,2; 2,3], [5,9; 1,4]);
We thus get:
>> M = cat(3, [1,2; 2,3], [5,9; 1,4])
M(:,:,1) =
1 2
2 3
M(:,:,2) =
5 9
1 4
The first slice is the matrix [1,2; 2,3] and the second slice is [5,9; 1,4]. From what it looks like, you would like the mean of only the first column of every slice and return this as a single vector of values. Therefore, use the mean function and index into the first column for all rows and slices. This will unfortunately become a singleton 3D array so you'll need to squeeze out the singleton dimensions.
Without further ado:
O = squeeze(mean(M(:,1,:)))
We thus get:
>> O = squeeze(mean(M(:,1,:)))
O =
1.5000
3.0000
I have two vectors
A = [...] %size 1x320
B = [...] %size 1x192
I would like to combine the two vectors in one but the way I want to combine them is the following:
Take the first 5 elements of vector A then add 3 elements from vector B add the next 5 elements from vector A then add the next element from vector B and so on until the both vectors are combined in one. I think the process should be repeated 64 times since 320/5=64 and 192/3=64.
Is there any built-in Matlab function to do that?
I don't think that there is a built-in function that does exactly that, but the following will do what you want:
A=randi(10,1,320);
B=randi(10,1,192);
C=zeros(1,length(A)+length(B));
for i=1:5
C(i:8:end)=A(i:5:end);
end
for i=6:8
C(i:8:end)=B(i-5:3:end);
end
Then the array C is the combined array.
Edit: Another way to do that, without for loops:
A=randi(10,1,320);
B=randi(10,1,192);
A_new=reshape(A,5,[]);
B_new=reshape(B,3,[]);
C=[A_new;B_new];
C=reshape(C,[1,numel(C)]);
In this solution, by specifying the third parameter in reshape(A,5,[]) to be [], we allow it to adjust the number of columns according to the length of A, given that the number of rows in the reshaped array is 5. In addition, numel(C) is the total number of elements in the array C. So this solution can be easily generalized to higher number of arrays as well.
I have a cell array A Mx3 in size where each entry contains a further cell-array Nx1 in size, for example when M=9 and N=5:
All data contained within in any given cell array is in vector format and of equal length. For example, A{1,1} contains 5 vectors 1x93 in size whilst A{1,2} contains 5 vectors 1x100 in size:
I wish to carry out this procedure on each of the 27 cells:
B = transpose(cell2mat(A{1,1}));
B = sort(B);
C = std(B,0,2); %Calculate standard deviation
Ultimately, the desired outcome would be, for the above example, 27 columns (9x3) containing the standard deviation results (padded with 0 or NaNs to handle differing lengths) printed in the order A{1,1}, A{1,2}, A{1,3}, A{2,1}, A{2,2}, A{2,3} and so forth.
I can do this by wrapping the above code into a loop to iterate over each one of the 27 cells in the correct order however, I was wondering if there was a clever cellfun or more succinct method to accomplish this particularly without the use of a loop?
You should probably realize that cellfun is essentially a glorified for loop over cells. There's simply extra error checking and all that to ensure that the whole thing works. In any case, yes it's possible to do what you're asking in a single cellfun call. Note that I am simply going to apply the same logic as you would have in a for loop with cellfun. Also note that because you're using cell arrays, you have no choice but to iterate over the entire master cell array. However, what you'll want to do is pad each resulting column vector in each output in the final cell array so that they all share the same length. We can do that with another two cellfun calls - one to determine the largest vector length and another to perform the padding operation.
Something like this could work:
% Step #1 - Sort the vectors in each cell array, then find row-wise std
B = cellfun(#(x) std(sort(cell2mat(x).'), 0, 2), A, 'un', 0);
% Step #2 - Determine the largest length vector and pad
sizes = cellfun(#numel, B);
B = cellfun(#(x) [x; nan(max(sizes(:)) - numel(x), 1)], B, 'un', 0);
The first line of code takes each element in A, converts each cell element into a N x 5 column matrix (i.e. cell2mat(x).'), we then sort each column individually with sort, then take the standard deviation row-wise. Because the output is ultimately a vector, we must make sure that the 'UniformOutput' flag is 0, or 'un=0'. Once we complete the standard deviation calculation, we determine the total number of elements for each resulting column vector for all cell elements, determine the largest size then use another cellfun call to pad these vectors so they all match the same size.
To finally get your desired output, you need to transpose the cell array, then unroll the elements in column major order. Remember that MATLAB accesses things in column major, so a common trick to get things in row-major (what you want) as opposed to column major is to first transpose, then unroll in column-major fashion to perform a row-major readout. Doing this in one line is tricky, so you'll need to not only transpose the cell array, you must use reshape to ensure that the elements are read out in row major format, but then ensuring that the result is placed in a row of cells, then call cell2mat so you can piece these vectors together. The final result should be a 27 column matrix where we have pieced all of these vectors together in a single row-wise fashion:
C = cell2mat(reshape(B.', 1, []));
I currently have a 3 dimensional matrix and I want to extract a single row (into the third dimension) from it by index (say matrix(2,1,:)). I initially anticipated that the result of this would be a 1 dimensional matrix however what I got was a 1 by 1 by n matrix. Usually this wouldn't be a problem but some of the functions I'm using don't like 3D matrices. For example see the problem replicated below:
threeDeeMatrix=rand(3,3,3);
oneDeeAttempt=threeDeeMatrix(1,1,:);
norm(oneDeeAttempt)
Which returns the error message:
Error using norm
Input must be 2-D.
This is because oneDeeAttempt is
oneDeeAttempt(:,:,1) =
0.8400
oneDeeAttempt(:,:,2) =
0.0700
oneDeeAttempt(:,:,3) =
0.7663
rather than [0.8400 0.0700 0.7663]
How can I strip these extra dimensions? The only solution I can come up with is to use a loop to manually copy the values but that seems a little excessive.
Using permute to rearrange the matrix
The solution (which I found in the final stages of asking this) is to use permute which rearranges the order of the dimensions (similar to a=a' for 2D matrices). Once the unit dimensions are last they are stripped from the matrix and it becomes 1 dimensional.
oneDee=permute(oneDeeAttempt,[3 1 2]) %rearrange so the previous third dimension is now the first
%the matrix is now 3 by 1 by 1 which becomes 3
Using squeeze to remove leading singleton dimensions
As pointed out by Luis Mendo squeeze will very simply remove these leading singleton dimensions without having to worry about which dimensions are non singleton
oneDee=squeeze(oneDeeAttempt);
I am not 100% what the role of the 1: is here. At which index to start the copy? But then why not two such parameters for the rank 2 array?
To be a little more explicit:
In Fortran90 and up you access a single array value by giving a single index, and access a subarray (a window of the array) by giving a range of indices separated by a colon.
a(1) = 0.
a(2:5) = (/3.14,2.71,1.62,0./)
You can also give a step size.
a(1:5:2) = (/0,2.71,0./)
And finally, you can leave out values and a default will be inserted. So if a runs from index 1 to 5 then I could write the above as
a(::2) = (/0,2.71,0./)
and the 1 and 5 are implied. Obviously, you shouldn't leave these out if it makes the code unclear.
With a multidimensional array, you can mix and match these on each dimension, as in your example.
You're taking a slice of array2, namely the elements in the D'th column from row 1 to C and putting them in the slice of array1 which is elements 1 through A
So both slices are 1-dimensional arrays
Slice may not be the correct term in Fortran