I have a 2D array and I want to create a 1D by MATLAB, satisfying the requirement that each element of the 1D output was created by the value of a given index into the 2D array. Example 2D array is
A=[2 4 6; 1 9 7.3 4 5]
And indexes for the 1D array
X=[1;2;3]
Y=[1;2;3]
I want to store the 1D array with elements determined by
B=A(x,y) % x,y are index in X and Y matrix
Example of building the 1D array:
X=[1;2;3]
Y=[1;2;3]
B=[A(1,1);A(2,2);A(3,3)]=[2; 9; 5]
This is my code
B=zeros(1,length(A));
B=A(...) %I don't know it
How can I implement it?
Thanks all.
You are looking for sub2ind:
A=[2 4 6; 1 9 7; 3 4 5]
X=[1;2;3]; Y=[1;2;3];
B = A(sub2ind(size(A),X,Y))
B =
2
9
5
You can use cellfun to do it. You convert A into cell by column, and execute f for each element of the cell.
A=[2 4 6; 1 2 7];
% some example f funcion that just adds the col_index_A and row_index_A
f = #(col_index_A, row_index_A) col_index_A + row_index_A;
% execute f with parameters that come from each column of A
B = cellfun(#(c) f(c(1), c(2)), num2cell(A, 1));
B =
3 6 13
I am not sure I understand your question but i think you want to apply functions on a 2-by-n matrix
Try
for pos=1:size(a,2)
b(pos) = f(a(:,pos));
end
Related
I have a 3d matrix H(i,j,k) with dimensions (i=1:m,j=1:n,k=1:o). I will use a simple case with m=n=o = 2:
H(:,:,1) =[1 2; 3 4];
H(:,:,2) =[5 6; 7 8];
I want to filter this matrix and project it to an (m,n) matrix by selecting for each j in 1:n a different k in 1:0.
For instance, I would like to retrieve (j,k) = {(1,2), (2,1)}, resulting in matrix G:
G = [5 2; 7 4];
This can be achieved with a for loop:
filter = [2 1]; % meaning filter (j,k) = {(1,2), (2,1)}
for i = 1:length(filter)
G(:,i) = squeeze(H(:,i,filter(i)));
end
But I'm wondering if it is possible to avoid the for loop via some smart indexing.
You can create all the linear indices to get such an output with the expansion needed for the first dimension with bsxfun. The implementation would look like this -
szH = size(H)
offset = (filter-1)*szH(1)*szH(2) + (0:numel(filter)-1)*szH(1)
out = H(bsxfun(#plus,[1:szH(1)].',offset))
How does it work
(filter-1)*szH(1)*szH(2) and (0:numel(filter)-1)*szH(1) gets the linear indices considering only the third and second dimension elements respectively. Adding these two gives us the offset linear indices.
Add the first dimenion linear indices 1:szH(1) to offset array in elementwise fashion with bsxfun to give us the actual linear indices, which when indexed into H would be the output.
Sample run -
H(:,:,1) =
1 2
3 4
H(:,:,2) =
5 6
7 8
filter =
2 1
out =
5 2
7 4
I want to show 2dim. Surface plots for different combinations of 2 parameters of a 3- or higher-dimensional array in matlab. The data for the non-shown dimensions are integrated (i.e. summed in the remaining dimensions). I am using surf(), and for parameter combinations other than (1,2) (eg. (1,3), (2,3) ...) I have to rearrange the data matrices in order to make it work.
I am looking for an alternative command (or shorter code) which does this work.
Here's the code:
a=zeros(3,3,2);
a(:,:,1) = [1 2 3 ;4 5 6; 7 8 9; 10 11 12]; % // data matrix
a(:,:,2) = -[1 2 3 ;4 5 6; 7 8 9; 10 11 12]*2; % // data matrix
ai=[[1 2 3 4]' [5 6 7 0]' [8 9 0 0]']; % // parameter vector
mat12 = sum(a,3);
surf(ai(1:3,2),ai(1:4,1),mat12)
aux13 = sum(a,2);
for i = 1:2; mat13(:,i) = aux13(:,:,i);
surf(ai(1:2,3),ai(1:4,1),mat13)
aux23 = sum(a,1);
for i = 1:2; mat23(i,:) = aux23(:,:,i);
surf(ai(1:3,2),ai(1:2,3),mat23)
In other words, I am looking for a way to use surf for matrices mat13 and mat23 without the aux13, aux23 variables and the for loop.
First your example doesn't run because you declare a=zeros(3,3,2); as a matrix [3x3x2] but you immediately try to populate it as a [4x3x2] matrix, so I had to adjust your first line to: a=zeros(4,3,2);
If I run your code with that adjustment, your auxiliary variable and for loops are to reform/reshape a matrix stripped of it's singleton dimension. Matlab provide a handy function for that : squeeze.
For example, your variable aux13 is of dimension [4x1x2], then mat13=squeeze(aux13); achieve the same thing than your for loop. Your matrix mat13 is now of dimension [4x2].
Since no for loop is needed, you can completely bypass your auxiliary variable by calling squeeze directly on the result of your summation: mat13=squeeze( sum(a,2) );
Full example, the code below does exactly the same than your code sample:
mat12 = sum(a,3);
surf(ai(1:3,2),ai(1:4,1),mat12)
mat13 = squeeze( sum(a,2) ) ;
surf(ai(1:2,3),ai(1:4,1),mat13)
mat23 = squeeze( sum(a,1) ) ;
mat23 = mat23.' ; %'// <= note the "transpose" operation here
surf(ai(1:3,2),ai(1:2,3),mat23)
Note that I had to transpose mat23 to make it match the one in your example.
sum(a,1) is [1x3x2] => squeeze that and you obtain a [3x2] matrix but your code arrange the same values in a [2x3] matrix, so the use of the transpose. The transpose operator has a shorthand notation .'.
I used it in the example in a separate line just to highlight it. Once understood you can simply write the full operation in one line:
mat23 = squeeze(sum(a,1)).' ;
The way you write your loops isn't exactly MATLAB syntax. Below is the correct loop syntax shown.
On line 2 and 3, you are trying to load (4x3)-matrices into (3x3)-matrices. That is why you get a subscript error. You could resolve it by making the zeros-matrix bigger. Here's some Syntax fixed:
a=zeros(4,3,2);
a(:,:,1) = [1 2 3 ;4 5 6; 7 8 9; 10 11 12]; % // data matrix
a(:,:,2) = -[1 2 3 ;4 5 6; 7 8 9; 10 11 12]*2; % // data matrix
ai=[[1 2 3 4]' [5 6 7 0]' [8 9 0 0]']; % // parameter vector
mat12 = sum(a,3);
surf(ai(1:3,2),ai(1:4,1),mat12)
aux13 = sum(a,2);
for i = 1:2 mat13(:,i) = aux13(:,:,i);
surf(ai(1:2,3),ai(1:4,1),mat13)
end
aux23 = sum(a,1);
for i = 1:2 mat23(i,:) = aux23(:,:,i);
surf(ai(1:3,2),ai(1:2,3),mat23)
end
Now, what are you exactly trying to do inside those loops?
I wanna delete all the subsets of cell c, suppose I have 6 cell vectors: c{1}=[1 2 3]; c{2}=[2 3 4];
c{3}=[1 2 3 4 5 6]; c{4}=[2 3 4 7]; c{5}=[2 3 7]; c{6}=[4 5 6]; then I wanna delete [1 2 3], [2 3 4] and [4 5 6]. I used two for loops to find all these subsets, but it's too slow for large datasets, is there any simple way can do this?
The following code removes a vector if it's a subset of any other vector. The approach is very similar to my answer to this other question:
n = numel(c);
[i1 i2] = meshgrid(1:n); %// generate all pairs of cells (their indices, really)
issubset = arrayfun(#(k) all(ismember(c{i1(k)},c{i2(k)})), 1:n^2); %// subset?
issubset = reshape(issubset,n,n) - eye(n); %// remove diagonal
c = c(~any(issubset)); %// remove subsets
Note that, in your example, [2 3 7] should also be removed.
You could find the cells that are exact matches for a particular input vector s1 using the following approach:
indx = find(cell2mat(cellfun(#(x)strcmp(num2str(x),num2str(s1)),c,'un', 0)));
You can then loop over matches (which should now be a much smaller set), and remove them by setting their contents to an empty set:
for ii=1:length(indx)
c{:,ii} = [];
end
I have an array (300x6) and I want to use it for this block in simulink. But I have a problem. How can I use "for iterator" block to take the q(i,:), q(i+1,:) , q(i+2,:) ... elements respectively? Or is there any other way to do this idea? Thanks.
In Matlab, for iterates over matrix columns. So you can just transpose to iterate over rows:
A = [1 2 3; 4 5 6; 7 8 9]; %// example matrix
for v = A.'
v = v.'; %// v will be [1 2 3], then [4 5 6] etc
%// Do stuff with v
end
I'm not into Simulink but I guess you can adapt this to that environment.
I have the array
a=1:20
and a series of indices which indicate where I want to start pulling data out:
i=[4,12]
For each index i, I want that index and the next four (well, x, really) elements in a column or row. I'll avoid getting to close to the end of the array, so that special case can be disregarded.
If I was hard-coding this, I could use:
a([4:8;12:16])
and this would achieve my result.
But i may have many different values.
Any thoughts on how I can transform a list of indices into a matrix of ranges, or other ways to solve this problem?
EDIT
I am using Matlab 2007; it would be preferable if the solution relied solely on Matlab's internals and toolboxes. bsxfun is not present until 2007a.
Let i be your indicesx and x the number of elements you want in addition to the elements in i, then you can use
i = [4 6 8];
x = 4;
bsxfun(#plus, 0:x, i(:))
to get a matrix of indices:
ans =
4 5 6 7 8
6 7 8 9 10
8 9 10 11 12
If you do not have access to bsxfun you can use repmat instead:
i = [4 6 8];
x = 4;
repmat(i(:), 1, x+1) + repmat(0:x, length(i), 1)
Here is a solution without bsxfun but with repmat inspired by the previous answer.
i = [4 6 8];
x = 4;
p = repmat(1:x,length(i),1);
q = repmat(i',1,x);
p+q