Hi I have an three dimensional octave array A of size [x y z]
Now I have another array B of dimensions n * 3
say B(0) gives [3 3 1]
I need to access that location in A ie A(3, 3, 1) = say 15
something like A(B(0))
How do I go about it?
See the help for sub2ind (and ind2sub).
However, nowadays people recommend to use loops.
Well, first, B(0) is invalid index, as addressing in MATLAB and Octave begins from 1. Other issue is that you want that B(0) would contain a vector [3 3 1 ]. Matrices in MATLAB can not contain other matrices, only scalars. So you need to use a 3x3 cell array, a 3x3 struct or a 4-dimensional array. I'll choose here the cell array option, because I find it easiest and most convenient.
% Set random seed (used only for example data generation).
rng(123456789);
% Let's generate some pseudo-random example data.
A = rand(3,3,3);
A(:,:,1) =
0.5328 0.7136 0.8839
0.5341 0.2570 0.1549
0.5096 0.7527 0.6705
A(:,:,2) =
0.6434 0.8185 0.2308
0.7236 0.0979 0.0123
0.7487 0.0036 0.3535
A(:,:,3) =
0.1853 0.8994 0.9803
0.7928 0.3154 0.5421
0.6122 0.4067 0.2423
% Generate an example 3x3x3 cell array of indices, filled with pseudo-random 1x3 index vectors.
CellArrayOfIndicesB = cellfun(#(x) randi(3,1,3), num2cell(zeros(3,3,3)), 'UniformOutput', false);
% Example #1. Coordinates (1,2,3).
Dim1 = 1;
Dim2 = 2;
Dim3 = 3;
% The code to get the corresponding value of A directly.
ValueOfA = A(CellArrayOfIndicesB{Dim1,Dim2,Dim3}(1), CellArrayOfIndicesB{Dim1,Dim2,Dim3}(2), CellArrayOfIndicesB{Dim1,Dim2,Dim3}(3));
ValueOfA =
0.8839
% Let's confirm that by first checking where CellArrayOfIndicesB{1,2,3} points to.
CellArrayOfIndicesB{1,2,3}
ans =
[ 1 3 1 ]
% CellArrayOfIndicesB{1,2,3} points to A(1,3,1).
% So let's see what is the value of A(1,3,1).
A(1,3,1)
ans =
0.8839
% Example #2. Coordinates (3,1,2).
Dim1 = 3;
Dim2 = 1;
Dim3 = 2;
ValueOfA = A(CellArrayOfIndicesB{Dim1,Dim2,Dim3}(1), CellArrayOfIndicesB{Dim1,Dim2,Dim3}(2), CellArrayOfIndicesB{Dim1,Dim2,Dim3}(3));
ValueOfA =
0.4067
CellArrayOfIndicesB{3,1,2}
ans =
[ 3 2 3 ]
A(3,2,3)
ans =
0.4067
Related
A is a given N x R xT array. I must split it horizontally to N sub-arrays of size L x M and then group each z together in an array K and take a mean.
For Example: A is the array rand(N,R,T)= rand( 16, 3 ,3); Now I am going to split it:
A=rand( 16, 3 ,3) : A(1,:,:), A(2,:,:), A(3,:,:), A(4,:,:), ... , A(16,:,:).
I have 16 slices.
B_1=A(1,:,:); B_2=A(2,:,:); B_3=A(3,:,:); ... ; B_16=A(16,:,:);
The next step is grouping together every 3 ( for example).
Now I am going create K_i as :
K_1(1,:,:)=B_1;
K_1(2,:,:)=B_2;
K_1(3,:,:)=B_3;
...
K_8(1,:,:)=B_14;
K_8(2,:,:)=B_15;
K_8(3,:,:)=B_16;
The average array is found as:
C_1=[B_1 + B_2 + B_3]/3
...
C_8= [ B_14 + B_15 + B_16] /3
I have implemented it as:
A_reshape = reshape(squeeze(A), size(A,2), size(A,3),2, []);
mean_of_all_slices = permute(mean(A_reshape , 3), [1 2 4 3]);
Question 1 I have checked by hand. It gives me a wrong result. How to fix it? [SOLVED]
EDIT 2 I need to simulate the following computation:
take a product each slice of the array K_i with another array P_p: It means:
for `K_1` is given `P_1`): `B_1 * P_1` , `B_2 * P_1`, `B_3 * P_1`
...
for `K_8` is given `P_8`): `B_14 * P_8` , `B_15 * P_8`, `B_16 * P_8`
I have solved!!!
Disclaimer: this answers a previous version of the question.
In cases such as this I would suggest relying on built-ins, which have a predictable behavior. In your case, this would be movmean (introduced in R2016a):
WIN_SZ = 2; % Window size for averaging
AVG_DIM = 1; % Dimension for averaging
tmp = movmean(A, WIN_SZ , AVG_DIM ,'Endpoints', 'discard');
C = tmp(1:WINDOW_SZ:end, :, :); % This only selects A1+A2, A3+A4 etc.
If your MATLAB is a bit older, this can also be done using convolution (convn, introduced before R2006):
WIN_SZ = 3;
tmp = convn(A, ones(WIN_SZ ,1)./WIN_SZ, 'valid'); % Shorter than A in dim1 by (WIN_SZ-1)
C = tmp(1:WINDOW_SZ:end, :, :); % dim1 size is: ceil((size(A,1)-(WIN_SZ-1))/3)
BTW, the step where you create B from slices of A can be done using
B = num2cell(A,[2,3]); % yields a 16x1 cell array of 1x3x3 double arrays
I have an application with an array of matrices. I have to manipulate the diagonals several times. The other elements are unchanged. I want to do things like:
for j=1:nj
for i=1:n
g(i,i,j) = gd(i,j)
end
end
I have seen how to do this with a single matrix using logical(eye(n)) as a single index, but this does not work with an array of matrices. Surely there is a way around this problem. Thanks
Use a linear index as follows:
g = rand(3,3,2); % example data
gd = [1 4; 2 5; 3 6]; % example data. Each column will go to a diagonal
s = size(g); % size of g
ind = bsxfun(#plus, 1:s(1)+1:s(1)*s(2), (0:s(3)-1).'*s(1)*s(2)); % linear index
g(ind) = gd.'; % write values
Result:
>> g
g(:,:,1) =
1.000000000000000 0.483437118939645 0.814179952862505
0.154841697368116 2.000000000000000 0.989922194103104
0.195709075365218 0.356349047562417 3.000000000000000
g(:,:,2) =
4.000000000000000 0.585604389346560 0.279862618046844
0.802492555607293 5.000000000000000 0.610960767605581
0.272602365429990 0.551583664885735 6.000000000000000
Based on Luis Mendo's answer, a version that may perhaps be more easy to modify depending on one's specific purposes. No doubt his version will be more computationally efficient though.
g = rand(3,3,2); % example data
gd = [1 4; 2 5; 3 6]; % example data. Each column will go to a diagonal
sz = size(g); % Get size of data
sub = find(eye(sz(1))); % Find indices for 2d matrix
% Add number depending on location in third dimension.
sub = repmat(sub,sz(3),1); %
dim3 = repmat(0:sz(1)^2:prod(sz)-1, sz(1),1);
idx = sub + dim3(:);
% Replace elements.
g(idx) = gd;
Are we already playing code golf yet? Another slightly smaller and more readable solution
g = rand(3,3,2);
gd = [1 4; 2 5; 3 6];
s = size(g);
g(find(repmat(eye(s(1)),1,1,s(3))))=gd(:)
g =
ans(:,:,1) =
1.00000 0.35565 0.69742
0.85690 2.00000 0.71275
0.87536 0.13130 3.00000
ans(:,:,2) =
4.00000 0.63031 0.32666
0.33063 5.00000 0.28597
0.80829 0.52401 6.00000
I have matrices:
a= 0.8147 0.1270 0.6324
0.9058 0.9134 0.0975
b= 0.2785 0.9649 0.9572
0.5469 0.1576 0.4854
0.9575 0.9706 0.8003
c = 0.1419 0.7922
0.4218 0.9595
0.9157 0.6557
and also I have another matrix
I= 1 3 1 1
2 1 3 2
I want to get d matrix such that
d= a(1,3) b(3,1) c(1,1)
a(2,1) b(1,3) c(3,2)
where indices come as two consecutive entries of I matrix.
This is one example I get. However, I get different size matrices for a,b,c,.. and I.
Added: I is m x (n+3) which includes indices, and other (n+2) matrices which have corresponding entries are X,A1,A2,...,An,Y. When n is given, A1,A2,...,An matrices are generated.
Can someone please help me to write Matlab code for this task?
You can do it with varargin. Assuming that your matrices are constructed such that you can form your desired output in the way you want (Updated according to Carmine's answer):
function out = IDcombiner(I, varargin)
out = zeros(size(I, 1), nargin-1);
idx = #(m, I, ii) (sub2ind(size(m), I(:, ii), I(:, ii+1)));
for ii = 1:1:nargin-1
out(:, ii) = varargin{ii}(idx(varargin{ii}, I, ii));
end
Now using this function you can make your selection on a flexible number of inputs:
out = IDcombiner(I, a, b, c)
out =
0.6324 0.9575 0.1419
0.9058 0.9572 0.6557
There is also a one-liner solution, which I do not recommend, since it dramatically decreases the readability of the code and doesn't help you gain much:
IDcombiner = #(I,varargin) ...
cell2mat(arrayfun(#(x) varargin{x}(sub2ind(size(varargin{x}), ...
I(:,x), I(:,x+1))), 1:nargin-1, 'UniformOutput', false));
Normally a matrix is not interpreted as a list of indices, but you can have this if you use sub2ind. To use it you need the size of the matrix you are addressing. Let's make an example starting with a:
a(sub2ind(size(a), I(:,1), I(:,2)))
The code does not change if you first assign the newly generated matrices to a variable name.
will use the column I(:,1) as rows and I(:,2) as columns.
To make the code more readable you can define an anonymous function that does this, let's call it idx:
idx = #(m,I,i)(sub2ind(size(m), I(:,i), I(:,i+1)))
So finally the code will be
d = [a(idx(a,I,1)), b(idx(b,I,2)), c(idx(c,I,3))]
The code does not change if you first assign the newly generated matrices to a variable name.
Other details
Let's make an example with 2 central matrices:
a = rand(3,1) % 3 rows, 1 column
b = rand(3,3) % 3 rows, 3 columns
c = rand(3,3) % another squared matrix
d = rand(3,1) % 3 rows, 1 column
The definition of the anonymous function is the same, you just change the definition of the output vector:
output = [a(idx(a,I,1)), b(idx(b,I,2)), c(idx(c,I,3)), d(idx(d,I,3))]
Keep in mind that following that pattern you always need a I matrix with (n_matrices + 1) columns.
Generalization
Let's generalize this code for a number n of central matrices of size rxr and for "side matrices" of size rxc. I will use some values of those parameters for this example, but you can use what you want.
Let me generate an example to use:
r = 3;
c = 4;
n = 3;
a = rand(r,c); % 2D array
b = rand(r,r,n); % 3D array, along z = 1:n you have 2D matrices of size rxr
c = rand(r,c);
I = [1 3 1 2 1 3; 2 1 3 1 1 1];
The code I wrote can easily be extended using cat to append matrices (note the 2 in the function tells MATLAB to append on the direction of the columns) and a for cycle:
idx = #(m,I,i)(sub2ind(size(m), I(:,i), I(:,i+1)))
d = a(idx(a,I,1));
for i = 1:n
temp = b(:,:,i);
d = cat(2,d,temp(idx(tmp,I,i+1)));
end
d = cat(2,d,c(idx(c,I,n+1)));
If you really don't want to address anything "by hand", you can use cell arrays to put all the matrices together and then cyclically apply the anonymous function to each matrix in the cell array.
How can I reshape a 2d array to a 3d array with the last column being used as pages?
All data found in array2d should be in pages
example:
array2d=[7,.5,12; ...
1,1,1; ...
1,1,1; ...
4,2,4; ...
2,2,2; ...
2,2,2; ...
3,3,3; ...
3,3,3; ...
3,3,3];
The first page in the array would be
7,.5,12;
1,1,1;
1,1,1;
The second page in the array would be
4,2,4;
2,2,2;
2,2,2;
The third page in the array would be
3,3,3;
3,3,3;
3,3,3;
This is a 9x3 array how can I get it to be a 9x3x? (not sure what this number should be so I placed a question mark as a place holder) multidimensional array?
What I'm trying to get is to have
All the ones would be on one dimension/page all the two's would be another dimension/page etc... –
I tried reshape(array2d,[9,3,1]) and it's still a 9x3
Use permute with reshape -
N = 3; %// Cut after every N rows to form a "new page"
array3d = permute(reshape(array2d,N,size(array2d,1)/N,[]),[1 3 2]) %// output
Assuming that each slice of your matrix is the same in dimensions, we can do this very easily. Let's call the number of rows and columns that each slice would have to be M and N respectively. In your example, this would be M = 3 and N = 3. As such, assuming array2d is of the above form, we can do the following:
M = 3;
N = 3; %// This is also simply the total number of columns we have,
%// so you can do size(array2d, 2);
outMatrix = []; %// Make this empty. We will populate as we go.
%// Figure out how many slices we need
numRows = size(array2d,1) / M;
for k = 1 : numRows
%// Extract the k'th slice
%// Reshape so that it has the proper dimensions
%// of one slice
sliceK = reshape(array2d(array2d == k), M, N);
%// Concatenate in the third dimension
outMatrix = cat(3,outMatrix,sliceK);
end
With your example, we thus get:
>> outMatrix
outMatrix(:,:,1) =
1 1 1
1 1 1
1 1 1
outMatrix(:,:,2) =
2 2 2
2 2 2
2 2 2
outMatrix(:,:,3) =
3 3 3
3 3 3
3 3 3
This method should generalize for any number of rows and columns for each slice, provided that each slice shares the same dimensions.
Your array is already of size 1 in the 3rd dimension (in other words, it is already 9x3x1, to prove this try entering array2d(1,1,1)). If you want to concatenate 2d matrices along the 3rd dimension you can use cat.
For example:
a = [1,2;3,4];
b = [5,6;7,8];
c = cat(3,a,b);
c will be a 2x2x2 matrix.
This piece of code is specific for this example, I hope you will be able to understand how to go for other data samples.
out2 = [];
col = size(array2d,2);
for i = 1:3
temp2 = reshape(array2d(array2d == i),[],col);
out2 = cat(3,out2,temp2);
end
I have a M x N sized structure array with fields var and val which are vectors.
What I would like to do is to get an M x N sized matrix A where each element A(i, j) contains the sum value of vector var (or val) from structure array
For example:
myStructure(1,5)
ans =
var: 1
val: [0.0100 0.1800 0.8100]
sum(myStructure(1,5).val)
ans =
1
myStructure(7,8)
ans =
var: [1 3]
val: [1x9 double]
myStructure(7,8).val
ans =
Columns 1 through 6
0.1111 0.1111 0.1111 0.1111 0.1111 0.1111
Columns 7 through 9
0.1111 0.1111 0.1111
Therefore A(1,5) should be 1 and the same way all elements A(i,j) should be equal to sum(myStructure(i,j).val).
Does anyone know how this could be done in Matlab without using for loops?
I've tried to use sum function in a number of ways (sum(messages.val) and sum(messages(:,:).val) ...) but couldn't get the desired result.
You can get the field elements into one matrix using:
svals = [myStructure.val];
If val is always the same length (let's name it P), this'll be a numel(myStructure)*P x 1 vector containing all values of all fields in sequence. You can reshape it of course back:
[N,M]=size(myStructure);
P = numel(myStructure(1,1).val);
svals = reshape(svals,[P M N]);
and now just sum the first dimension, which leaves you the MxN sized A matrix:
A = squeeze(sum(svals,1));
squeeze is applied in this last step to remove the resulting singleton dimension (otherwise A would be of size 1xMxN).
If the vallength can vary, I see no other way than looping it, or using arrayfun, which is essentially the same as looping:
A = arrayfun(#(x) sum(x.val),myStructure);
Here is a slightly different solution. First lets create an array structure for testing:
s = struct();
for i=1:5
for j=1:3
s(i,j).var = i+j;
s(i,j).val = rand(1,randi(10)); %# different lengths vectors
end
end
Now we do the sum:
A = cellfun(#sum, reshape({s.val}, size(s)))
A =
1.9278 3.0719 5.8731
3.2377 0.43874 2.2374
3.0661 2.8892 4.1455
1.9093 1.4758 1.441
4.8731 0.5308 3.4076