I have 5 columns x, y, r, g, b with values of line number, column number, red, green and blue. The lines of this n by 5 matrix are not in a particular order, however they are consistent with image(x,y) and the r,g,b.
I would like to do something like I=uint8(zeros(480,640,3) and just change those rgb values based on the n by 5 mat.
Something along the lines of I(mat(:,1), mat(:,2), 1)=mat(:,3) for red etc
The following uses the concept of linear indexing and the versatile bsxfun function:
m = 640; %// number of rows
n = 480; %// number of columns
I = zeros(m, n, 3, 'uint8'); %// initiallize directly as uint8
I(bsxfun(#plus, x(:)+(y(:)-1)*m, (0:2)*m*n)) = [r(:) g(:) b(:)]; %// fill values
Small example: for
m = 2;
n = 3;
x = [1 2 1];
y = [1 1 2];
r = [ 1 2 3];
g = [11 12 13];
b = [21 22 23];
the code produces
I(:,:,1) =
1 3 0
2 0 0
I(:,:,2) =
11 13 0
12 0 0
I(:,:,3) =
21 23 0
22 0 0
An alternative:
INDr = sub2ind([480, 640, 3], mat(:, 1), mat(:,2), ones([numel(mat(:,3)), 1]));
INDg = sub2ind([480, 640, 3], mat(:, 1), mat(:,2), 2*ones([numel(mat(:,3)), 1]));
INDb = sub2ind([480, 640, 3], mat(:, 1), mat(:,2), 3*ones([numel(mat(:,3)), 1]));
I=uint8(zeros(480,640, 3));
I(INDr)=mat(:,3);
I(INDg)=mat(:,4);
I(INDb)=mat(:,5);
Note that in Matlab, the convention between axes is different between images and arrays.
Related
I have two vectors, and I'm trying to find ALL coincidences of one on the other within a certain tolerance without using a for loop.
By tolerance I mean for example if I have the number 3, with tolerance 2, I will want to keep values within 3±2, so (1,2,3,4,5).
A = [5 3 4 2]; B = [2 4 4 4 6 8];
I want to obtain a cell array containing on each cell the numbers of all the coincidences with a tolerance of 1 (or more) units. (A = B +- 1)
I have a solution with zero units (A = B), which would look something like this:
tol = 0;
[tf, ia] = ismembertol(B,A,tol,'DataScale',1); % For tol = 0, this is equivalent to using ismember
idx = 1:numel(B);
ib = accumarray(nonzeros(ia), idx(tf), [], #(x){x}) % This gives the cell array
The output is:
ib =
[]
[]
[2 3 4]
[1]
Which is as desired.
If I change the tolerance to 1, the code doesn't work as intended. It outputs instead:
tol = 1
[tf, ia] = ismembertol(B,A,tol,'DataScale',1); % For tolerance = 1, this is equivalent to using ismember
idx = 1:numel(B);
ib = accumarray(nonzeros(ia), idx(tf), [], #(x){x}) % This gives the cell array
ib =
[5]
[2 3 4]
[]
[1]
When I would expect to obtain:
ib =
[2 3 4 5]
[1 2 3 4]
[2 3 4]
[1]
What am I doing wrong? Is there an alternative solution?
Your problem is that, in the current state of your code, ismembertol only outputs 1 index per element of B found in A, so you lose information in the cases where an element can be found several times within tolerance.
As per the documentation You can use the 'OutputAllIndices',true value pair argument syntax, to output what you want in ia with just a call to ismembertol:
A = [5 3 4 2]; B = [2 4 4 4 6 8];
tol = 0;
[tf, ia] = ismembertol(A,B,tol,'DataScale',1,'OutputAllIndices',true);
celldisp(ia) % tol = 0
ia{1} =
0
ia{2} =
0
ia{3} =
2
3
4
ia{4} =
1
celldisp(ia) % tol = 1
ia{1} =
2
3
4
5
ia{2} =
1
2
3
4
ia{3} =
2
3
4
ia{4} =
1
Here is a manual approach, just to provide another method. It computes an intermediate matrix of all absolute differences (using implicit expansion), and from the row and column indices of the entries that are less than the tolerance it builds the result:
A = [5 3 4 2];
B = [2 4 4 4 6 8];
tol = 1;
[ii, jj] = find(abs(A(:).'-B(:))<=tol);
ib = accumarray(jj, ii, [numel(A) 1], #(x){x});
Note that this approach
may be memory-intensive, because of the intermediate matrix;
can be made to work in old Matlab versions, because it doesn't use ismembertol; but then implicit expansion has to be replaced by explicitly calling bsxfun:
[ii, jj] = find(abs(bsxfun(#minus, A(:).', B(:)))<=tol);
ib = accumarray(jj, ii, [numel(A) 1], #(x){x});
I have an array of length n. The array has braking energy values, and the index number represents time in seconds.
The structure of array is as follows:
Index 1 to 140, array has zero values. (Vehicle not braking)
Index 141 to 200, array has random energy values. (Vehicle was braking and regenerating energy)
Index 201 to 325, array has zero values. (Vehicle not braking)
Index 326 to 405, array has random energy values. (Vehicle was braking and regenerating energy)
...and so on for an array of length n.
What I want to do is to get starting and ending index number of each set of energy values.
For example the above sequence gives this result:
141 - 200
326 - 405
...
Can someone please suggest what method or technique can I use to get this result?
Using diff is a quick way to do this.
Here is a demo (see the comments for details):
% Junk data for demo. Indices shown above for reference
% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
x = [0, 0, 0, 2, 3, 4, 0, 0, 1, 1, 7, 9, 3, 4, 0, 0, 0];
% Logical converts all non-zero values to 1
% diff is x(2:end)-x(1:end-1), so picks up on changes to/from zeros
% Instead of 'logical', you could have a condition here,
% e.g. bChange = diff( x > 0.5 );
bChange = diff( logical( x ) );
% bChange is one of the following for each consecutive pair:
% 1 for [0 1] pairs
% 0 for [0 0] or [1 1] pairs
% -1 for [1 0] pairs
% We inflate startIdx by 1 to index the non-zero value
startIdx = find( bChange > 0 ) + 1; % Indices of [0 1] pairs
endIdx = find( bChange < 0 ); % Indices of [1 0] pairs
I'll leave it as an exercise to capture the edge cases where you add a start or end index if the array starts or ends with a non-zero value. Hint: you could handle each case separately or pad the initial x with additional end values.
Output of the above:
startIdx
>> [4, 9]
endIdx
>> [6, 14]
So you can format this however you like to get the spans 4-6, 9-14.
This task is performed by two methods Both works perfectly.
Wolfie Method:
bChange = diff( EnergyB > 0 );
startIdx = find( bChange > 0 ) + 1; % Indices of [0 1] pairs
endIdx = find( bChange < 0 ); % Indices of [1 0] pairs
Result:
startIdx =
141
370
608
843
endIdx =
212
426
642
912
Second Method:
startends = find(diff([0; EnergyB > 0; 0]));
startends = reshape(startends, 2, [])';
startends(:, 2) = startends(:, 2) - 1
Result:
startends =
141 212
370 426
608 642
843 912
Suppose I have an m x n matrix A .
Is there a way to create B, a (n x m) x n matrix whose "diagonal" is formed by A's columns ?
Example:
A = [1 2;
3 4]
B = [1 0;
3 0;
0 2;
0 4]
Here is a way:
Convert A to a cell array of its columns, using mat2cell;
From that cell array generate a comma-separated list, and use it as an input to blkdiag.
Code:
A = [1 2; 3 4]; %// example data
C = mat2cell(A, size(A,1), ones(1,size(A,2))); %// step 1
B = blkdiag(C{:}); %// step 2
This produces
B =
1 0
3 0
0 2
0 4
Here is a short script to accomplish this. It works for any dimensions of A.
A=[1 2; 3 4];
[R C] = size(A);
for i=1:C
B( 1+R*(i-1) : R*i , i ) = A(:,i);
end
I have a 3D-cell array designated as A{s,i,h}, serving as a store for large amounts of numerical data during a nested-loop portion of my script. Some of the cell entries will be blank [ ], whilst the rest consist of numbers - either singular or in arrays (1 x 10 double etc.):
I want to convert this cell array to a set of 2D matrices.
Specifically, one separate matrix for each value of h (h is always equal 1:3) and one column in each matrix for every value of s. Each column will contain all the numerical data combined - it does not need to be separated by i.
How can I go about this? I ordinarily deal with 3D-cell arrays in this form to produce separate matrices (one for each value of h) using something like this:
lens = sum(cellfun('length',reshape(A,[],size(A,3))),1);
max_length = max(lens);
mat = zeros(max_length,numel(lens));
mask = bsxfun(#le,[1:max_length]',lens);
mat(mask) = [A{:}];
mat(mat==0) = NaN;
mat = sort(mat*100);
Matrix1 = mat(~isnan(mat(:,1)),1);
Matrix2 = mat(~isnan(mat(:,2)),2);
Matrix3 = mat(~isnan(mat(:,3)),3);
However in this instance, each matrix had only a single column. I'm have trouble adding multiple columns to each output matrix.
1. Result in the form of a cell array of matrices (as requested)
Here's one possible approach. I had to use one for loop. However, the loop can be easily avoided if you accept a 3D-array result instead of a cell array of 2D-arrays. See second part of the answer.
If you follow the comments in the code and inspect the result of each step, it's straightforward to see how it works.
%// Example data
A(:,:,1) = { 1:2, 3:5, 6:9; 10 11:12 13:15 };
A(:,:,2) = { 16:18, 19:22, 23; 24:28, [], 29:30 };
%// Let's go
[S, I, H] = size(A);
B = permute(A, [2 1 3]); %// permute rows and columns
B = squeeze(mat2cell(B, I, ones(1, S), ones(1, H))); %// group each col of B into a cell...
B = cellfun(#(x) [x{:}], B, 'uniformoutput', false); %// ...containing a single vector
t = cellfun(#numel, B); %// lengths of all columns of result
result = cell(1,H); %// preallocate
for h = 1:H
mask = bsxfun(#le, (1:max(t(:,h))), t(:,h)).'; %'// values of result{h} to be used
result{h} = NaN(size(mask)); %// unused values will be NaN
result{h}(mask) = [B{:,h}]; %// fill values for matrix result{h}
end
Result in this example:
A{1,1,1} =
1 2
A{2,1,1} =
10
A{1,2,1} =
3 4 5
A{2,2,1} =
11 12
A{1,3,1} =
6 7 8 9
A{2,3,1} =
13 14 15
A{1,1,2} =
16 17 18
A{2,1,2} =
24 25 26 27 28
A{1,2,2} =
19 20 21 22
A{2,2,2} =
[]
A{1,3,2} =
23
A{2,3,2} =
29 30
result{1} =
1 10
2 11
3 12
4 13
5 14
6 15
7 NaN
8 NaN
9 NaN
result{2} =
16 24
17 25
18 26
19 27
20 28
21 29
22 30
23 NaN
2. Result in the form of 3D array
As indicated above, using a 3D array to store the result permits avoiding loops. In the code below, the last three lines replace the loop used in the first part of the answer. The rest of the code is the same.
%// Example data
A(:,:,1) = { 1:2, 3:5, 6:9; 10 11:12 13:15 };
A(:,:,2) = { 16:18, 19:22, 23; 24:28, [], 29:30 };
%// Let's go
[S, I, H] = size(A);
B = permute(A, [2 1 3]); %// permute rows and columns
B = squeeze(mat2cell(B, I, ones(1, S), ones(1, H))); %// group each col of B into a cell...
B = cellfun(#(x) [x{:}], B, 'uniformoutput', false); %// ...containing a single vector
t = cellfun(#numel, B); %// lengths of all columns of result
mask = bsxfun(#le, (1:max(t(:))).', permute(t, [3 1 2])); %'// values of result to be used
result = NaN(size(mask)); %// unused values will be NaN
result(mask) = [B{:}]; %// fill values
This gives (compare with result of the first part):
>> result
result(:,:,1) =
1 10
2 11
3 12
4 13
5 14
6 15
7 NaN
8 NaN
9 NaN
result(:,:,2) =
16 24
17 25
18 26
19 27
20 28
21 29
22 30
23 NaN
NaN NaN
Brute force approach:
[num_s, num_i, num_h] = size(A);
cellofmat = cell(num_h,1);
for matrix = 1:num_h
sizemat = max(cellfun(#numel, A(:,1,matrix)));
cellofmat{matrix} = nan(sizemat, num_s);
for column = 1:num_s
lengthcol = length(A{column, 1, matrix});
cellofmat{matrix}(1:lengthcol, column) = A{column, 1,matrix};
end
end
Matrix1 = cellofmat{1};
Matrix2 = cellofmat{2};
Matrix3 = cellofmat{3};
I don't know what your actual structure looks like but this works for A that is setup using the following steps.
A = cell(20,1,3);
for x = 1:3
for y = 1:20
len = ceil(rand(1,1) * 10);
A{y,1,x} = rand(len, 1);
end
end
I have two 3d arrays:
A(:,:,1) = [1 2 3; 4 5 6; 7 8 9];
A(:,:,2) = [1 0 0; 0 1 0; 0 0 1];
A(:,:,3) = [3 2 1; 6 5 4; 9 8 7];
...
and
B(:,:,1) = [1 1 1; 2 2 2; 3 3 3];
B(:,:,2) = [1 0 0; 0 1 0; 0 0 1];
B(:,:,3) = [3 3 3; 2 2 2; 1 1 1];
...
They both consist of 3x3 matrices and their third dimensions are very large. I want to obtain the array of matrices that exist in both arrays. I am doing it in a for loop by comparing element-wise (matrix-wise). It takes a very long time, so I am looking for an easier way (or an existing function) to do the same.
Thanks!
Collapse the first two dimensions into one and transpose, so that matrices of the 3D array become rows of a matrix. That way you can use intersect(...,'rows'). Finally, transpose back and reshape back:
[m, n, p] = size(A);
result = intersect(reshape(A, [], p).', reshape(B, [], p).', 'rows');
result = reshape(result.', m, n, []);