I here by post the code why I came across while exploring one technique.
Y = repmat((1:m)', [1 n]);
X = repmat(1:n, [m 1]) - labels_left;
X(X<1) = 1;
indices = sub2ind([m,n],Y,X);
final_labels = labels_left;
final_labels(abs(labels_left - labels_right(indices))>=1) = -1;
In above code labels left is single channel image.[m n] is the size of that image. I want to know how this sub2ind works in above code.And Iam also facing problem in the last statement which contains
labels_right(indices)
what the above expression evaluates to.Here labels right is also an image
Maybe a smaller example could help understand:
%# image matrix
M = rand(4,3)
[m n] = size(M)
%# meshgrid, and convert to linear indices
[X,Y] = meshgrid(1:n,1:m)
indices = sub2ind([m,n],Y,X)
%# extract those elements
M(indices)
The matrix M:
>> M
M =
0.95717 0.42176 0.65574
0.48538 0.91574 0.035712
0.80028 0.79221 0.84913
0.14189 0.95949 0.93399
the grid of (x,y) coordinates of all points:
>> X,Y
X =
1 2 3
1 2 3
1 2 3
1 2 3
Y =
1 1 1
2 2 2
3 3 3
4 4 4
converted to linear indices:
>> indices
indices =
1 5 9
2 6 10
3 7 11
4 8 12
then we index into the matrix using those indices.
>> M(indices)
ans =
0.95717 0.42176 0.65574
0.48538 0.91574 0.035712
0.80028 0.79221 0.84913
0.14189 0.95949 0.93399
Note that: M(indices(i,j)) = M(Y(i,j)),X(i,j)).
Related
I have got the following code. I need to rewrite it without looping. How should I do it?
l1 = [1 2 3 2 1];
l2 = [3 4 4 5 4];
A = zeros(5,5);
for i=1:5
A(i, l1(i):l2(i)) = 1;
end
A
You can use bsxfun -
I = 1:5 % Array corresponding to iterator : "for i=1:5"
out = bsxfun(#le,l1(:),I) & bsxfun(#ge,l2(:),I)
If you need a double datatype array, convert to double, like so -
out_double = double(out)
Add one more into the mix then! This one simply uses a cumsum to generate all the 1s - so it does not use the : operator at all - It's also fully parallel :D
l1 = [1 2 3 2 1];
l2 = [3 4 4 5 4];
A = zeros(5,5);
L1 = l1+(1:5)*5-5; %Convert to matrix location index
L2 = l2+(1:5)*5-5; %Convert to matrix location index
A(L1) = 1; %Place 1 in that location
A(L2) = 1; %Place 1 in that location
B = cumsum(A,1) ==1 ; %So fast
Answer = (A|B)'; %Lightning fast
Answer =
1 1 1 0 0
0 1 1 1 0
0 0 1 1 0
0 1 1 1 1
1 1 1 1 0
Here is how you could build the matrix without using a loop.
% Our starting values
l1 = [1 2 3 2 1];
l2 = [3 4 4 5 4];
% Coordinate grid of the right size (we don't need r, but I keep it there for illustration)
[r,c] = ndgrid(1:5);
% Build the logical index based on our lower and upper bounds on the column indices
idx_l1=bsxfun(#ge,c,l1');
idx_l2=bsxfun(#le,c,l2');
% The result
A = zeros(size(idx_l1));
A(idx_l1&idx_l2)=1
You may need something like [r,c] = ndgrid(1:numel(l1),1:10).
Also if your matrix size is truly huge and memory becomes an issue, you may want to stick to a loop anyway, but for 'normal size' this could be faster.
There should be some skepticism in every vectorization. If you measure the time actually your loop is faster than the given answers, mostly because you only perform in place write.
Here is another one that would probably get faster for larger sizes but I haven't tested:
tic
myind = [];
for i = 1:5
myind = [myind (5*(i-1))+[l1(i):l2(i)]];
end
A(myind) = 1;
toc
gives the transposed A because of the linear indexing order.
I have a problem with sorting some finance data based on firmnumbers. So given is a matrix that looks like:
[1 3 4 7;
1 2 7 8;
2 3 7 8;]
On Matlab i would like the matrix to be sorted as follows:
[1 0 3 4 7 0;
1 2 0 0 7 8;
0 2 3 0 7 8;]
So basically every column needs to consist of 1 type of number.
I have tried many things but i cant get the matrix sorted properly.
A = [1 3 4 7;
1 2 7 8;
2 3 7 8;]
%// Get a unique list of numbers in the order that you want them to appear as the new columns
U = unique(A(:))'
%'//For each column (of your output, same as columns of U), find which rows have that number. Do this by making A 3D so that bsxfun compares each element with each element
temp1 = bsxfun(#eq,permute(A,[1,3,2]),U)
%// Consolidate this into a boolean matrix with the right dimensions and 1 where you'll have a number in your final answer
temp2 = any(temp1,3)
%// Finally multiply each line with U
bsxfun(#times, temp2, U)
So you can do that all in one line but I broke it up to make it easier to understand. I suggest you run each line and look at the output to see how it works. It might seem complicated but it's worthwhile getting to understand bsxfun as it's a really useful function. The first use which also uses permute is a bit more tricky so I suggest you first make sure you understand that last line and then work backwards.
What you are asking can also be seen as an histogram
A = [1 3 4 7;
1 2 7 8;
2 3 7 8;]
uniquevalues = unique(A(:))
N = histc(A,uniquevalues' ,2) %//'
B = bsxfun(#times,N,uniquevalues') %//'
%// bsxfun can replace the following instructions:
%//(the instructions are equivalent only when each value appears only once per row )
%// B = repmat(uniquevalues', size(A,1),1)
%// B(N==0) = 0
Answer without assumptions - Simplified
I did not feel comfortable with my old answer that makes the assumption of everything being an integer and removed the possibility of duplicates, so I came up with a different solution based on #lib's suggestion of using a histogram and counting method.
The only case I can see this not working for is if a 0 is entered. you will end up with a column of all zeros, which one might interpret as all rows initially containing a zero, but that would be incorrect. you could uses nan instead of zeros in that case, but not sure what this data is being put into, and if it that processing would freak out.
EDITED
Includes sorting of secondary matrix, B, along with A.
A = [-1 3 4 7 9; 0 2 2 7 8.2; 2 3 5 9 8];
B = [5 4 3 2 1; 1 2 3 4 5; 10 9 8 7 6];
keys = unique(A);
[counts,bin] = histc(A,transpose(unique(A)),2);
A_sorted = cell(size(A,1),1);
for ii = 1:size(A,1)
for jj = 1:numel(keys)
temp = zeros(1,max(counts(:,jj)));
temp(1:counts(ii,jj)) = keys(jj);
A_sorted{ii} = [A_sorted{ii},temp];
end
end
A_sorted = cell2mat(A_sorted);
B_sorted = nan(size(A_sorted));
for ii = 1:size(bin,1)
for jj = 1:size(bin,2)
idx = bin(ii,jj);
while ~isnan(B_sorted(ii,idx))
idx = idx+1;
end
B_sorted(ii,idx) = B(ii,jj);
end
end
B_sorted(isnan(B_sorted)) = 0
You can create at the beginning a matrix with 9 columns , and treat the values in your original matrix as column indexes.
A = [1 3 4 7;
1 2 7 8;
2 3 7 8;]
B = zeros(3,max(A(:)))
for i = 1:size(A,1)
B(i,A(i,:)) = A(i,:)
end
B(:,~any(B,1)) = []
Suppose I have an array X of size n by p by q. I would like to reshape it as a matrix with p rows, and in each row put the concatenation of the n rows of size q, resulting in a matrix of size p by nq.
I managed to do it with a loop but it takes a while say if n=1000, p=300, q=300.
F0=[];
for k=1:size(F,1)
F0=[F0,squeeze(X(k,:,:))];
end
Is there a faster way?
I think this is what you want:
Y = reshape(permute(X, [2 1 3]), size(X,2), []);
Example with n=2, p=3, q=4:
>> X
X(:,:,1) =
0 6 9
8 3 0
X(:,:,2) =
4 7 1
3 7 4
X(:,:,3) =
4 7 2
6 7 6
X(:,:,4) =
6 1 9
1 4 3
>> Y = reshape(permute(X, [2 1 3]), size(X,2), [])
Y =
0 8 4 3 4 6 6 1
6 3 7 7 7 7 1 4
9 0 1 4 2 6 9 3
Try this -
reshape(permute(X,[2 3 1]),p,[])
Thus, for code verification, one can look into a sample case run -
n = 2;
p = 3;
q = 4;
X = rand(n,p,q)
F0=[];
for k=1:n
F0=[F0,squeeze(X(k,:,:))];
end
F0
F0_noloop = reshape(permute(X,[2 3 1]),p,[])
Output is -
F0 =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
F0_noloop =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
Rather than using vectorization to solve the problem, you could look at the code to try and figure out what may improve performance. In this case, since you know the size of your output matrix F0 should be px(n*q), you could pre-allocate memory to F0 and avoid the constant resizing of the matrix at each iteration of the for loop
n=1000;
p=300;
q=300;
F0=zeros(p,n*q);
for k=1:size(F,1)
F0(:,(k-1)*q+1:k*q) = squeeze(F(k,:,:));
end
While probably not as efficient as the other two solutions, it is an alternative. Try the above and see what happens!
I have a 3D image, divided into contiguous regions where each voxel has the same value. The value assigned to this region is unique to the region and serves as a label. The example image below describes the 2D case:
1 1 1 1 2 2 2
1 1 1 2 2 2 3
Im = 1 4 1 2 2 3 3
4 4 4 4 3 3 3
4 4 4 4 3 3 3
I want to create a graph describing adjaciency between these regions. In the above case, this would be:
0 1 0 1
A = 1 0 1 1
0 1 0 1
1 1 1 0
I'm looking for a speedy solution to do this for large 3D images in MATLAB. I came up with a solution that iterates over all regions, which takes 0.05s per iteration - unfortunately, this will take over half an hour for an image with 32'000 regions. Does anybody now a more elegant way of doing this? I'm posting the current algorithm below:
labels = unique(Im); % assuming labels go continuously from 1 to N
A = zeros(labels);
for ii=labels
% border mask to find neighbourhood
dil = imdilate( Im==ii, ones(3,3,3) );
border = dil - (Im==ii);
neighLabels = unique( Im(border>0) );
A(ii,neighLabels) = 1;
end
imdilate is the bottleneck I would like to avoid.
Thank you for your help!
I came up with a solution which is a combination of Divakar's and teng's answers, as well as my own modifications and I generalised it to the 2D or 3D case.
To make it more efficient, I should probably pre-allocate the r and c, but in the meantime, this is the runtime:
For a 3D image of dimension 117x159x126 and 32000 separate regions: 0.79s
For the above 2D example: 0.004671s with this solution, 0.002136s with Divakar's solution, 0.03995s with teng's solution.
I haven't tried extending the winner (Divakar) to the 3D case, though!
noDims = length(size(Im));
validim = ones(size(Im))>0;
labels = unique(Im);
if noDims == 3
Im = padarray(Im,[1 1 1],'replicate', 'post');
shifts = {[-1 0 0] [0 -1 0] [0 0 -1]};
elseif noDims == 2
Im = padarray(Im,[1 1],'replicate', 'post');
shifts = {[-1 0] [0 -1]};
end
% get value of the neighbors for each pixel
% by shifting the image in each direction
r=[]; c=[];
for i = 1:numel(shifts)
tmp = circshift(Im,shifts{i});
r = [r ; Im(validim)];
c = [c ; tmp(validim)];
end
A = sparse(r,c,ones(size(r)), numel(labels), numel(labels) );
% make symmetric, delete diagonal
A = (A+A')>0;
A(1:size(A,1)+1:end)=0;
Thanks for the help!
Try this out -
Im = padarray(Im,[1 1],'replicate');
labels = unique(Im);
box1 = [-size(Im,1)-1 -size(Im,1) -size(Im,1)+1 -1 1 size(Im,1)-1 size(Im,1) size(Im,1)+1];
mat1 = NaN(numel(labels),numel(labels));
for k2=1:numel(labels)
a1 = find(Im==k2);
for k1=1:numel(labels)
a2 = find(Im==k1);
t1 = bsxfun(#plus,a1,box1);
t2 = bsxfun(#eq,t1,permute(a2,[3 2 1]));
mat1(k2,k1) = any(t2(:));
end
end
mat1(1:size(mat1,1)+1:end)=0;
If it works for you, share with us the runtimes as comparison? Would love to see if the coffee brews any faster than half an hour!
Below is my attempt.
Im = [1 1 1 1 2 2 2;
1 1 1 2 2 2 3;
1 4 1 2 2 3 3;
4 4 4 4 3 3 3;
4 4 4 4 3 3 3];
% mark the borders
validim = zeros(size(Im));
validim(2:end-1,2:end-1) = 1;
% get value of the 4-neighbors for each pixel
% by shifting the images 4 times in each direction
numNeighbors = 4;
adj = zeros([prod(size(Im)),numNeighbors]);
shifts = {[0 1] [0 -1] [1 0] [-1 0]};
for i = 1:numNeighbors
tmp = circshift(Im,shifts{i});
tmp(validim == 0) = nan;
adj(:,i) = tmp(:);
end
% mark neighbors where it does not eq Im
imDuplicates = repmat(Im(:),[1 numNeighbors]);
nonequals = adj ~= imDuplicates;
% neglect the border
nonequals(isnan(adj)) = 0;
% get these neighbor values and the corresponding Im value
compared = [imDuplicates(nonequals == 1) adj(nonequals == 1)];
% construct your 'A' % possibly could be more optimized here.
labels = unique(Im);
A = zeros(numel(labels));
for i = 1:size(compared,1)
A(compared(i,1),compared(i,2)) = 1;
end
#Lisa
Yours reasoning is elegant, though it obviously gives wrong answers for labels on the edges.
Try this simple label matrix:
Im =
1 2 2
3 3 3
3 4 4
The resulting adjacency matrix , according to your code is:
A =
0 1 1 0
1 0 1 1
1 1 0 1
0 1 1 0
which claims an adjacency between labels "2" and "4": obviously wrong. This happens simply because you are reading padded Im labels based on "validim" indices, which now doesn't match the new Im and goes all the way down to the lower borders.
I'm trying to increase the efficiency of my MATLAB code. What it does is, it replaces the nonzero elements of a matrix with the multiplication of the rest of the nonzero elements in the same row. For instance,
X = [2 3 6 0; 0 3 4 2]
transforms into
X = [18 12 6 0; 0 8 6 12]
It's an easy task to implement within a for loop. It checks every row, finds nonzero values and do its replacements. I want to get rid of the for loop though. Is there a way to implement this without a loop?
Code
X = [2 3 6 0; 0 3 4 2];
X1 = X;
X1(~X) = 1;
out = bsxfun(#rdivide,prod(X1,2),X1).*(X~=0)
Output
out =
18 12 6 0
0 8 6 12
Probably getting the row product first once and then divide by the element you don't want is the simplest way:
X = [2 3 6 0; 0 3 4 2]
Y=X
%get the product of all elements in a row
Y(Y==0)=1
Y=prod(Y,2)
%repeat Y to match the size of X
Y=repmat(Y,1,size(X,2))
%For all but the zero elements, divide Y by X, which is the product of all other elements.
X(X~=0)=Y(X~=0)./X(X~=0)