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?
Related
This question already has an answer here:
How do I generate the following matrix and vector from the given input data in MATLAB?
(1 answer)
Closed 5 years ago.
I have a 7*1 vector a = (1:7).'. I want to form a matrix A of size 4*4 from vector a such that the elements of a form the anti-diagonals of matrix A as follows:
A = [1 2 3 4;
2 3 4 5;
3 4 5 6;
4 5 6 7]
I would like this to work for a general a, not just when the elements are consecutive integers.
I appreciate any help.
Setting up the indexing
Adding the two outputs of meshgrid can give the indices:
[x, y] = meshgrid(1:4, 0:3);
x + y;
% ans = [1 2 3 4
% 2 3 4 5
% 3 4 5 6
% 4 5 6 7];
If a was just as in your example you could stop there. Alternatively, use this to index a general vector a. For comparison, I'll use the same example input as rahnema1 did for their method:
a = [4 6 2 7 3 5 1];
[x, y] = meshgrid(1:4, 0:3);
A = a(x + y);
% A = [4 6 2 7
% 6 2 7 3
% 2 7 3 5
% 7 3 5 1]
There are many ways you could create the indices instead of using meshgrid, see the benchmarking functions below for some exampels!
Benchmarking and seven different methods.
Here are some timings for running different methods, including methods using cumsum, repmat, hankel and a simple for loop. This benchmark was done in Matlab 2015b, so takes advantage of Matlab optimisations etc. which the Octave benchmarks in rahnema1's answer might not do. I also use the timeit function which is more robust than tic/toc because it does multiple trials etc.
function benchie()
n = 10000; % (large) square matrix size
a = 1:2*n-1; % array of correct size, could be anything this long
f1 = #() m1(a,n); disp(['bsxfun: ', num2str(timeit(f1))]);
f2 = #() m2(a,n); disp(['cumsum: ', num2str(timeit(f2))]);
f3 = #() m3(a,n); disp(['meshgrid: ', num2str(timeit(f3))]);
f4 = #() m4(a,n); disp(['repmat: ', num2str(timeit(f4))]);
f5 = #() m5(a,n); disp(['for loop: ', num2str(timeit(f5))]);
f6 = #() m6(a,n); disp(['hankel1: ', num2str(timeit(f6))]);
f7 = #() m7(a,n); disp(['hankel2: ', num2str(timeit(f7))]);
end
% Use bsxfun to do broadcasting of addition
function m1(a,n); A = a(bsxfun(#plus, (1:n), (0:n-1).')); end
% Use cumsum to do cumulative vertical addition to create indices
function m2(a,n); A = a(cumsum([(1:n); ones(n-1,n)])); end
% Add the two meshgrid outputs to get indices
function m3(a,n); [x, y] = meshgrid(1:n, 0:n-1); A = a(x + y); end
% Use repmat twice to replicate the meshgrid results, for equivalent one liner
function m4(a,n); A = a(repmat((1:n)',1,n) + repmat(0:n-1,n,1)); end
% Use a simple for loop. Initialise A and assign values to each row in turn
function m5(a,n); A = zeros(n); for ii = 1:n; A(:,ii) = a(ii:ii+n-1); end; end
% Create a Hankel matrix (constant along anti-diagonals) for indexing
function m6(a,n); A = a(hankel(1:n,n:2*n-1)); end
% Create a Hankel matrix directly from elements
function m7(a,n); A = hankel(a(1:n),a(n:2*n-1)); end
Output:
bsxfun: 1.4397 sec
cumsum: 2.0563 sec
meshgrid: 2.0169 sec
repmat: 1.8598 sec
for loop: 0.4953 sec % MUCH quicker!
hankel1: 2.6154 sec
hankel2: 1.4235 sec
So you are best off using rahnema1's suggestion of bsxfun or direct generation of a hankel matrix if you want a one liner, here is a brilliant StackOverflow answer which explains some of bsxfun's advantages: In Matlab, when is it optimal to use bsxfun?
However, the for loop is more than twice as quick! Conclusion: Matlab has lots of neat ways to achieve things like this, sometimes a simple for loop with some appropriate pre-allocation and Matlab's internal optimisations can be the quickest.
You can use hankel:
n= 4;
A= hankel(a(1:n),a(n:2*n-1))
Other solution(expansion/bsxfun):
In MATLAB r2016b /Octave It can be created as:
A = a((1:4)+(0:3).')
In pre r2016b you can use bsxfun:
A = a(bsxfun(#plus,1:4, (0:3).'))
Example input/output
a = [4 6 2 7 3 5 1]
A =
4 6 2 7
6 2 7 3
2 7 3 5
7 3 5 1
Using the benchmark provided by #Wolfie tested in Octave:
_____________________________________
|Method |memory peak(MB)|timing(Sec)|
|=========|===============|===========|
|bsxfun |2030 |1.50 |
|meshgrid |3556 |2.43 |
|repmat |2411 |2.64 |
|hankel |886 |0.43 |
|for loop |886 |0.82 |
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
I'm a total beginner to matlab and I'm currently writing a script for extracting data from a thermographic video.
Firstly the video is cut in separate frames. The first frame is opened as a sample picture to define the coordinates of sampling points. The goal is then to select the rgb values of those defined coordinates from a set of frames and save them into a matrix.
Now I have a problem separating the matrix to n smaller matrices.
e.g I'm defining the number of points to be selected to n=2 , with a picture count of 31. Now it returns a matrix stating the rgb codes for 31 pictures, each at 2 points, in a 62x3 double matrix...
Now I want to extract the 1st, 3rd, 5th....etc... row to a new matrix...this should be done in a loop, according to the number of n points...e.g 5 points on each picture equals 5 matrices, containing values of 31 pictures....
this is an extract of my code to analyse the pictures, it returns the matrix 'values'
files = dir('*.jpg');
num_files = numel(files);
images = cell(1, num_files);
cal=imread(files(1).name);
n = input('number of selection points?: ');
imshow(cal);
[x,y] = ginput(n);
eval(get(1,'CloseRequestFcn'))
%# x = input('x-value?: '); manual x selection
%# y = input('y-value?: '); manual y selection
for k = 1:num_files
images{k} = imread(files(k).name);
end
matrix=cell2mat(images);
count=(0:size(matrix,1):size(matrix,1)*num_files);
for k = 1:num_files
a(k)= mat2cell(impixel(matrix,x+count(k),y));
end
values = cat(1,a{:})
Easy fix
Do you mean that if you have:
n = 2;
k = 2; % for example
matrix = [1 2 3;
4 5 6;
7 8 9;
8 7 6];
you want it to become
b{1} = [1 2 3;
7 8 9];
b{2} = [4 5 6;
8 7 6];
This can be easily done with:
for ii = 1:n
b{ii} = matrix(1:n:end,:);
end
Better fix
Of course it's also possible to just reshape your data matrix and use that instead of the smaller matrices: (continuing with my sample data ^^)
>> d=reshape(matrix',3,2,[]);
>> squeeze(d(:,1,:))
ans =
1 7
2 8
3 9
>> squeeze(d(:,2,:))
ans =
4 8
5 7
6 6
Good practice
Or, my preferred choice: save the data immediately in an easy to access way. Here I think it will be an matrix of size: [num_files x num_points x 3]
If you want all the first points:
rgb_data(:,1,:)
only the red channel of those points:
rgb_data(:,1,1)
and so on.
I think this is possible with this:
rgb_data = zeros(num_files, num_points, 3);
for kk = 1:num_files
rgb_data(kk,:,:) = impixel(images{kk},x+count(k),y);
end
But I don't understand the complete meaning of your code (eg: why matrix=cell2mat(images) ??? and then of course:
count=(0:size(matrix,1):size(matrix,1)*num_files);
is just count=0:num_files;
so I'm not sure what would come out of impixel(matrix,x+count(k),y) and I used images{k} :)
The expression is:
for i=1:n
X(:,i) = [P{i}(:)];
end
where X is a DxN matrix and P is a cell-array.
reshape(cat(3,P{:}),[numel(P{1}) n])
Of course, the above solution is just for fun. I would recommend profiling both solutions and only using this one if it has a significant performance advantage.
Maintenance and readability are also very important factors to consider when writing code.
If you obtained the cell array via mat2cell, you may be wanting to arrange blocks of an image into the columns of an array X. This can be achieved in a single step using the command IM2COL
%# rearrange the large array so that each column of X
%# corresponds to the 4 pixels of each 2-by-2 block
X = im2col(largeArray,[2 2],'distinct');
You might be able to get away with:
P{1} = [ 1 2; 3 4];
P{2} = [ 7 8; 9 10];
P{3} = [ 11 12; 13 14];
X = [P{:}]
X =
1 2 7 8 11 12
3 4 9 10 13 14
Then some sort of reshape() to get to where you want to be.
Short Version
How can I do concatMap in MATLAB? I'm trying to build a single vector from a series of smaller, differently sized vectors. I know I can do:
result = [];
for i=1:N
result = [result nextPart(i)];
end
but that has a serious speed impact and there must be a smarter way to do concatMap.
Long Version
I'm trying to write a MATLAB function that returns the counterdiagonals of a block. For example, if you have the block:
1 2 4
3 5 7
6 8 9
then counterDiagonals(block) should return [1 2 3 4 5 6 7 8 9].
I have a function that will find a single counter diagonal of a block. i.e. counterDiagonal(x, 3) will return [4 5 6].
Therefore, counterDiagonals should be as simple as concatMap counterDiagonal(x, i) (1:N) where N is (2*length(block)-1). How can I do this in MATLAB in an efficient way?
One problem with the accepted answer: if the matrix A had zeros, they will be incorrectly removed from the result.. Instead you should work on the indices of the elements:
A = [0 2 4; 3 5 7; 6 8 9]; %# Sample matrix (contains zeros)
ind = reshape(1:numel(A), size(A)); %# indices of elements
ind = fliplr( spdiags( fliplr(ind) ) ); %# get the anti-diagonals (or use ROT90)
ind(ind==0) = []; %# keep non-zero indices
result = A(ind); %# get elements in desired order
This is very similar to this answer I gave in a previous question (the difference was that the anti-digaonals were in reverse order).
I believe what you want to do can be accomplished using the functions ROT90 and SPDIAGS:
A = [1 2 4; 3 5 7; 6 8 9]; %# Sample matrix
result = rot90(A); %# Rotate the matrix counter-clockwise
result = spdiags(result); %# Find all the diagonals
result = result(result ~= 0).'; %'# Remove zero padding and format the results
%# into a row vector
And you should end up with result = [1 2 3 4 5 6 7 8 9].
EDIT: As Amro mentions in a comment, the above code assumes that there are no zeroes in the original matrix A. If there are zeroes in the original matrix, one solution is to replace them with a non-zero flag value that you know doesn't appear in the original matrix (like, for example, NaN), run the above code, then replace the flag values in the result:
A = [0 2 4; 3 0 7; 6 8 0]; %# Sample matrix
result = rot90(A); %# Rotate the matrix counter-clockwise
result(result == 0) = nan; %# Replace zeroes with NaN
result = spdiags(result); %# Find all the diagonals
result = result(result ~= 0).'; %'# Remove zero padding and format the results
%# into a row vector
result(isnan(result)) = 0; %# Put the original zeroes back
Short version:
If you preassign your result array, everything will be a lot faster.
result = zeros(1,knownLengthOfResultsArray); %# such as "numel(block)"
ct = 1;
for i=1:N
tmp = nextPart(i);
nTmp = length(tmp);
result(ct:ct+nTmp-1) = tmp;
ct = ct + nTmp;
end
Long version:
However, it may be even more efficient to rewrite your algorithm. See e.g. the answers to this question (use fliplr on your array first), or #gnovice's answer.