I have two 3D matrices of the same size. Say, A contain values ranging from 0 to 1 and B contains certain discrete values.
I want to extract values from B matrix which are above certain threshold values in A?
Can anyone help me out?
Like this? Have a look at logical indexing.
% Input.
B = reshape(1:18, 2, 3, 3)
A = rand(size(B))
% Threshold.
thr = 0.75
% Output.
output = B(A > thr)
B =
ans(:,:,1) =
1 3 5
2 4 6
ans(:,:,2) =
7 9 11
8 10 12
ans(:,:,3) =
13 15 17
14 16 18
A =
ans(:,:,1) =
0.80533 0.24370 0.89180
0.90358 0.22422 0.69243
ans(:,:,2) =
0.119366 0.168337 0.771999
0.206004 0.065481 0.979772
ans(:,:,3) =
0.0057303 0.1469925 0.0556628
0.0454038 0.4122576 0.9847027
thr = 0.75000
output =
1
2
5
11
12
18
Related
Consider a matrix X. I have to update a submatrix of X, X(row1:row2, col1:col2), with a matrix Z (of size row2-row1+1, col2-col1+1) but only on those positions where a logical matrix L (of size row2-row1+1, col2-col1+1) is true.
E.g. if
X=[ 1 2 3 4 5 6
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 34 36]
Z=[31 41
32 42]
L=[ 1 0
0 1]
row1 = 2; row2 = 3; col1 = 3; col2 = 4
then after the update I should get:
X=[ 1 2 3 4 5 6
11 12 31 14 15 16
21 22 23 42 25 26
31 32 33 34 34 36]
Currently I do the following:
Y = X(row1:row2, col1:col2);
Y(L) = Z(L);
X(row1:row2, col1:col2) = Y;
This code is in a tight loop and according to Matlab's (v2019a) profiler is the main bottleneck of my program. In the real code X is a 2000x1500x3 cube; row1, row2, col1, col2, Z and L change in the loop.
The question is whether it can be rewritten into a single / faster assignment.
Thanks.
Honestly, without seeing your actual code, I get the sense that your solution may be as fast as you can get. The reason I say that is because I tested a few different solutions by creating some random sample data closer to your actual problem. I assumed X is an image of type uint8 with size 2000-by-1500-by-3, Z is size N-by-N (i.e. we will only be modifying the first page of X), L is an N-by-N logical array, and the row and column indices are randomly chosen:
X = randi([0 255], 2000, 1500, 3, 'uint8');
N = 20; % Submatrix size
Z = randi([0 255], N, N, 'uint8');
L = (rand(N, N) > 0.5);
row1 = randi([1 2000-N]);
row2 = row1+N-1
col1 = randi([1 1500-N]);
col2 = col1+N-1;
I then tested 3 different solutions: your original solution, a solution using find and sub2ind to create a linear index for X, and a solution that creates a logical index for X:
% Original solution:
Y = X(row1:row2, col1:col2, 1);
Y(L) = Z(L);
X(row1:row2, col1:col2, 1) = Y;
% Linear index solution:
[rIndex, cIndex] = find(L);
X(sub2ind(size(X), rIndex+row1-1, cIndex+col1-1)) = Z(L);
% Logical index solution
[R, C, ~] = size(X);
fullL = false(R, C);
fullL(row1:row2, col1:col2) = L;
X(fullL) = Z(L);
I tested these repeatedly with randomly-generated sample data using timeit and found that your original solution is consistently the fastest. The linear index solution is very close, but slightly slower. The logical index solution takes more than twice as long.
Let's define some example data:
X = randi(9,5,6);
Y = 10+X;
row1 = 2;
row2 = 4;
col1 = 3;
col2 = 4;
L = logical([0 1; 0 0; 1 1]);
Then:
ind_subm = bsxfun(#plus, (row1:row2).',size(X,1)*((col1:col2)-1));
% linear index for submatrix
ind_subm_masked = ind_subm(L);
% linear index for masked submatrix
X(ind_subm_masked) = Y(ind_subm_masked);
Example results:
X before:
X =
6 2 1 7 9 6
3 3 3 5 5 7
6 3 8 6 5 4
7 4 1 3 3 4
2 5 9 5 5 9
L:
L =
3×2 logical array
0 1
0 0
1 1
X after:
X =
6 2 1 7 9 6
3 3 3 15 5 7
6 3 8 6 5 4
7 4 11 13 3 4
2 5 9 5 5 9
I want to create matrix B from matrix A, with the following rules:
Non-diagonal elements A are like non-diagonal elements of B.
The main diagonal of A is the antidiagonal of B
The antidiagonal of A is the main diagonal of B.
For example:
A = [ 1 2 3 4;
7 8 9 10;
13 14 15 16;
19 20 21 22 ];
B = [ 4 2 3 1;
7 9 8 10;
13 15 14 16;
22 20 21 19 ];
How can I create B given A?
You can create all of the indices, then it's a single assignment.
% Get size of square matrix A
n = size(A,1);
% Indicies are 1:n^2 by default
idx = 1:n^2;
% Swap diagonal and antidiagonal indices
idx( [1:(n+1):n^2, n^2-n+1:1-n:n] ) = [n^2-n+1:1-n:n, 1:(n+1):n^2];
% Use the indexing array to create B from A, reshape to be n*n
B = reshape( A( idx ), n, n );
Output for your example A:
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
There are so many ways to reach that results, it is just an indexing exercise. Here is one (of the many) way to reach that result for any square matrix of size n:
%% input
A=[ 1 2 3 4 ;
7 8 9 10 ;
13 14 15 16 ;
19 20 21 22 ];
%% Calculate linear indices for the diagonal and antidiagonal
n=size(A,1) ;
idxdiag = 1:(n+1):n^2 ; % => idxdiag = [1 6 11 16]
idxantidiag = n:(n-1):n^2-1 ; % => idxantidiag = [4 7 10 13]
%% Generate B
B = A ; % start with a simple copy (for the non-diagonal elements)
% Method 1: direct indice assignment
B(idxdiag) = diag(fliplr(A)) ; % Assign diagonal elements of B
B(idxantidiag) = flipud(diag(A)) ; % Assign antidiagonal elements of B
% Method 2: summation
B([idxdiag idxantidiag]) = 0 ;
B = B + diag(diag(fliplr(A))) + fliplr(diag(diag(A))) ;
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
Both methods return exactly the same matrix B.
I suggest you familiarise yourself with the MATLAB function used to understand what is going on behind the scene:
fliplr
flipud
diag
and may be have a read at:
Matrix Indexing in MATLAB
I thought a little differently and came to a conclusion
A=[1 2 3 4;7 8 9 10;13 14 15 16; 19 20 21 22];;
n=size(A,1) ;
B=zeros(n,n) ;
for i=1:n
for j=1:n
if i==j
B(i,j)=A(i,n-i+1);
elseif j==n-i+1
B(i,j)=A(i,i);
else
B(i,j)=A(i,j);
end
end
end
B
Here's a variant using eye, find, and flip to generate linear indices:
ind1 = find(eye(size(A)));
ind2 = flip(find(flip(eye(size(A)))));
B = A;
B([ind1 ind2]) = B([ind2 ind1]);
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
And here's a variant of the above that uses just eye and flip to generate logical indices:
ind1 = eye(size(A), 'logical');
ind2 = flip(ind1);
B = A;
B(ind1) = flip(A(ind2));
B(ind2) = flip(A(ind1));
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
I want to create a square matrix that takes input n and create a matrix incrementing from 0 up to n^2
eg.
input: n = 2
output = [1 2
4 3]
input = 4
output = [1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13]
n = 4;
output = reshape(1:n^2,n,n)';
output(2:2:end,:) = fliplr(output(2:2:end,:))
Example:
input = [1 255 0 0 0 9 9 9 1 6 6 6 6 6 6 1]; % array of numbers (uint8)
output = [1 255 0 0 0 255 9 9 9 255 1 6 6 6 255 6 6 6 255 1];
% output must have 255 inserted at positions 6, 10, 15, 19
% because 0, 9, 6, 6 have occurred three times respectively
outputIndex = [6 10 15 19];
% outputIndex must indicate the positions where 255 was inserted
This could be one vectorized approach to get things done efficiently -
%// Input
A = [1 255 0 0 0 9 9 9 1 6 6 6 6 6 6 1]
%// Input paramter (how many times a value must be repeated for detection)
search_count = 3;
%// Find difference between consecutive elemnts and set all non zero
%// differences as ones, otherwise as zeros in a binary array
diffA = diff(A)~=0
%// Find start and end indices of "islands" of same value
starts = strfind([1 diffA],[1 zeros(1,search_count-1)])
ends = strfind([diffA 1],[zeros(1,search_count-1) 1])+search_count
%// For each island of same valued elements, find out where first group ends
firstgrp = starts + search_count
%// Find how many times a group of that same value of search_count times repeats
%// within each "island" of same valued elements. Also get the max repeats.
pattern_repeats = floor((ends - starts)./search_count)
max_repeat = max(pattern_repeats)
%// All possible repeat indices within all islands
all_repeats = bsxfun(#plus,firstgrp,[0:max_repeat-1]'*(search_count)) %//'
%// Use a binary mask to select only those repeats allowed with pattern_repeat
out_idx = all_repeats(bsxfun(#lt,[0:max_repeat-1]',pattern_repeats)) %//'
out_idx = out_idx + [0:numel(out_idx)-1]' %//'
%// Create output arary, insert 255 at out_idx locations and put values
%// from input array into rest of the locations
out = zeros(1,numel(A)+numel(out_idx));
out(out_idx) = 255
out(out==0) = A
Code run -
>> A
A =
Columns 1 through 13
1 255 0 0 0 9 9 9 1 6 6 6 6
Columns 14 through 16
6 6 1
>> out_idx
out_idx =
6
10
15
19
>> out
out =
Columns 1 through 13
1 255 0 0 0 255 9 9 9 255 1 6 6
Columns 14 through 20
6 255 6 6 6 255 1
I don't understand the downvotes, it's actually an interesting question.
Here the long answer:
n = 3;
subst = 255;
input = [1 255 0 0 0 9 9 9 1 6 6 6 6 6 6 61];
%// mask
X = NaN(1,numel(input));
%// something complicated (see below)
X(filter(ones(1,n-1),1,~([0. diff(input)])) == n-1) = 1;
%// loop to split multiple occurences of n-tuples
for ii = 1:numel(input)
if X(ii) == 1 && ii < numel(X)-n+1
X(ii+1:ii+n-1) = NaN(1,n-1);
end
end
%// output vector
D = [input; X.*subst];
E = D(:);
output = E(isfinite(E))
%// indices of inserted 255
D = [input.*0; X.*subst];
E = D(:);
outputIndex = find(E(isfinite(E)))
Explanation of the complicated part:
%// finite differences of input
A = [0 diff(input)];
%// conversion to logical
B = ~A;
%// mask consecutive values
mask = filter(ones(1,n-1),1,B) == n-1;
%// set masked values to 1
X(mask) = 1;
If you have the image processing toolbox you can save the loop with this fancy oneliner for getting the mask:
mask = accumarray(bwlabel(filter(ones(1,n-1),1,~([0. diff(input)])) == n-1).'+1,1:numel(input),[],#(x) {getfield(sort(x),{find(mod(cumsum(1:numel(x)),n) == 1)})});
X = NaN(1,numel(input));
X(vertcat(mask{2:end})) = subst;
%// output vector
D = [input; X];
E = D(:);
output = E(isfinite(E))
%// indices of inserted 255
D = [input.*0; X];
E = D(:);
outputIndex = find(E(isfinite(E)))
I have a vector that contains repeated numbers like so:
[1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6] and so on. What I want to do is to group the similar values (1's, 5's, etc.). I would like to have each of the unique values in a row of a big matrix, like:
[ 1 1 1 1 0 0
5 5 5 5 0 0
93 93 93 0 0 0
6 6 6 6 6 6]
I don't know the maximum number of occurrence of a unique value, so it is ok to create an initial zero matrix with a large number of columns (that I am sure is bigger than the maximum number of occurrence of a unique value).
Any help is highly appreciated.
How about this?
A = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6];
[a,b] = hist(A,unique(A))
f = #(x) [ones(1,a(x)) zeros(1,max(a)-a(x))]
X = cell2mat( arrayfun(#(x) {f(x)*b(x)}, 1:numel(b) )' )
returns:
X =
1 1 1 1 0 0
5 5 5 5 0 0
6 6 6 6 6 6
93 93 93 0 0 0
I know the order is different, is that important? Otherwise:
n = hist(A,1:max(A)) % counts how often every number apperas
[a b] = unique(A,'stable') % gets all unique numbers
n = n(a) % correlates count and numbers
f = #(x) [ones(1,n(x)) zeros(1,max(n)-n(x))] % creates the logical index
% vector for every single row
X = cell2mat( arrayfun(#(x) {f(x)*b(x)}, 1:numel(b) )' ) %fills the rows
or inspired by Luis Mendo's Answer a little shorter:
n = hist(A,1:max(A));
a = unique(A,'stable')
n = n(a)
Y = repmat(a',1,max(n)).*bsxfun(#le, cumsum(ones(max(n),numel(n))), n)'
returns:
X =
1 1 1 1 0 0
5 5 5 5 0 0
93 93 93 0 0 0
6 6 6 6 6 6
For the bored people out there, there is a one-line solution:
X = getfield(cell2mat(arrayfun(#(x,y) padarray( padarray(x,[0 y],'replicate','pre'),[0 max(hist(A,1:max(A)))-y],'post'),1:max(A),hist(A,1:max(A)),'uni',0)'),{unique(A,'stable'),2:1+max(hist(A,1:max(A)))})
Or an almost lovely two-liner:
n = hist(A,1:max(A))
X = getfield(cell2mat(arrayfun(#(x,y) padarray( padarray(x,[0 y],'replicate',...
'pre'),[0 max(n)-y],'post'),1:max(A),n,'uni',0)'),...
{unique(A,'stable'),2:1+max(n)})
just for fun ;)
Vectorized solution (no loops):
x = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6]; %// data
ind = [find(diff(x)) numel(x)]; %// end of each run of equal values
values = x(ind); %// unique values (maintaining order)
count = diff([0 ind]); %// count of each value
result = bsxfun(#le, meshgrid(1:max(count),1:numel(values)), count.'); %'// mask
result = bsxfun(#times, result, values.'); %'// fill with the values
EDIT:
Alternative procedure that avoids the second bsxfun:
x = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6]; %// data
ind = [find(diff(x)) numel(x)];
values = x(ind); %// unique values (maintaining order)
count = diff([0 ind]); %// count of each value
mask = bsxfun(#le, ndgrid(1:max(count),1:numel(values)), count);
result = zeros(size(mask)); %// pre-allocate and pre-shape (transposed) result
result(mask) = x; %// fill in values
result = result.';
This could be one approach -
%%// Input
array1 = [1 1 1 1 5 5 5 5 93 93 93 6 6 6 6 6 6];
%// Main Processing
id = unique(array1,'stable'); %//Find the unique numbers/IDs
mat1 = zeros(numel(id),nnz(array1==mode(array1))); %%// Create a matrix to hold the final result
for k=1:numel(id)
extent_each_id = nnz(array1==id(k)); %%// Count of no. of occurances for each ID
mat1(k,1:extent_each_id)=id(k); %%// Starting from the left to the extent for each ID store that ID
end
Gives -
mat1 =
1 1 1 1 0 0
5 5 5 5 0 0
93 93 93 0 0 0
6 6 6 6 6 6