y = 0;
for m = 0:variable
for n = 0:m
y = y + f(n,m);
end
end
I vectorized the inner loop this way,
y = 0;
for m = 0:variable
n = 0:m
y = y + f(n,m);
end
This resulted in around 60% speed increase for my code. How do I also vectorize the outer loop?
You are probably looking for the meshgrid function. It is designed to fill in the sort of m by n combinations that it looks like you need. For example:
>> m = 1:4;
>> n = 1:3;
>> [mGridValues, nGridValues] = meshgrid(m,n)
mGridValues =
1 2 3 4
1 2 3 4
1 2 3 4
nGridValues =
1 1 1 1
2 2 2 2
3 3 3 3
This is a little more complicated since your inner loop depends on the value of your outer loop. So you will need to mask out the undesired [n, m] pairs (see below).
Modifying the prototype code that you have provided, you would end up with something like this:
[mValues, nValues] = meshgrid(0:variable, 0:variable); %Start with a full combination of values
mask = mValues >= nValues; %Identify all values where m >= n
mValues = mValues(mask); % And then remove pairs which do not
nValues = nValues(mask); % meet this criteria
y = f(nValues, mValues ); %Perform whatever work you are performing here
Related
I am trying to implement the paper detection of copy move forgery using histogram of oriented gradients.
The algorithm is:
Divide the image into overlapping blocks.
Calculate feature vectors for each block and store them in a matrix.
Sorting the matrix lexicographically
Using block matching to identify forged regions.
https://www.researchgate.net/publication/276518650_Detection_of_copy-move_image_forgery_using_histogram_of_orientated_gradients
I am stuck with the 3rd step and can't proceed.
The code I have implemented is:
clc;
clear all;
close all;
%read image
img = imread('006_F.png');
img=rgb2gray(img);
img=imresize(img, 1/4);
figure(1);
imshow(img);
b=16; %block size
nrc=5; %no. of rows to check
td=416; %threshold
[r, c]=size(img);%Rows and columns;
column=(r-b+1)*(c-b+1);
M= zeros(column,4);
Mi = zeros(1,2);
i=1;
disp('starting extraction of features');
for r1 = 1:r-b+1
for c1 = 1:c-b+1
% Extract each block
B = img(r1:r1+b-1,c1:c1+b-1);
features = extractHOGFeatures(B);%extracting features
M(i, :) = features;
Mi(i,:) = [r1 c1];
i=i+1;
end
end
[S, index] = sortrows(M , [ 1 2 3 4]);
P= zeros(1,6);
b2=r-b+1;
disp('Finding Duplicates');
for i = 1:column
iv = index(i);
xi=mod(iv,b2) + 1;
yi=ceil(iv/b2);
j = i+1;
while j < column && abs(i - j) < 5
jv=index(j);
xj=mod(jv,b2) + 1;
yj=ceil(jv/b2);
z=sqrt(power(xi-xj,2) + power(yi-yj,2));
% only process those whose size is above Nd
if z > 16
offset = [xi-xj yi-yj];
P = [P;[xi yi xj yj xi-xj yi-yj]];
end
j = j + 1;
end
end
rows = size(P,1);
P(:,6) = P(:,6) - min(P(:,6));
P(:,5) = P(:,5) - min(P(:,5));
maxValP = max(P(:,6)) + 1;
P(:,5) = maxValP .* P(:,5) + P(:,6);
mostfrequentval = mode(P(:,5));
disp('Creating Image');
idx = 2;
% Create a copy of the image and mask it
RI = img;
while idx < rows
x1 = P(idx,1);
y1 = P(idx,2);
x2 = P(idx,3);
y2 = P(idx,4);
if (P(idx,5) == mostfrequentval)
RI(y1:y1,x1:x1) = 0;
RI(y2:y2,x2:x2) = 0;
end
idx = idx + 1;
end;
After going through some references indicated in the paper you are working on (ref. [8] and [20]):
The lexicographic sorting is the equivalent of the alphabetical one, for numbers i.e., [1 1 1 1] < [1 1 2 1] < [2 3 4 5] < [2 4 4 5]
So, in your case, you case use the function sortrows() in the following way:
A = [1 1 1 1;1 1 1 2;1 1 1 4;1 2 2 2; 1 2 2 1; 1 4 6 3; 2 3 4 5; 2 3 6 6]; % sample matrix
[B,idx] = sortrows(A,[1 2 3 4]); % Explicit notation but it is the Matlab default setting so equivalent to sortrows(A)
It means: Sort the rows of A by first looking at the first column and, in case of equality, looking at the second one, and so on.
If your are looking for a reverse order, you specify '-' before the number of the column.
So in the end, your code is good and if the results are not as expected it has to come from another step of the implementation...
Edit: the parameter idx records the original index of the sorted rows.
I have a matrix D of distances between 3 places and 4 persons
example D(2,3) = 10 means person 3 is far away from place 2 of 10 units.
D=[23 54 67 32
32 5 10 2
3 11 13 5]
another matrix A with the same number of rows (3 places) and where A(i,:) correspond to the persons that picked place i
example for place 1, persons 1 and 3 picked it
no one picked place 2
and persons 2 and 4 picked place 3
A=[1 3 0
0 0 0
2 4 0]
I want to reorder each row of A by the persons who are closest to the place it represents.
In this example, for place 1, person 1 is closer to it than person 3 based on D so nothing to do.
nothing to do for place 2
and there is a change for place 3 since person 4 is closer than 2 to place 3 D(3,2)>D(3,4)
The result should be
A=[1 3
0 0
4 2 ]
each row(place) in A can have 0 or many non zeros elements in it (persons that picked it)
Basically, I want to reorder elements in each row of A based on the rows of D (the closest to the location comes first), something like this but here A and D are not of the same size (number of columns).
[SortedD,Ind] = sort(D,2)
for r = 1:size(A,1)
A(r,:) = A(r,Ind(r,:));
end
There is another Matlab function sortrows(C,colummn_index) that can do the trick. It can sort rows based on the elements in a particular column. So if you transpose your matrix A (C = A') and extend the result by adding to the end the proper column, according to which you want to sort a required row, then you will get what you want.
To be more specific, you can do something like this:
clear all
D=[23 54 67 32;
32 5 10 2;
3 11 13 5];
A=[1 0;
3 0;
4 2 ];
% Sort elements in each row of the matrix A,
% because indices of elements in each row of the matrix D are always
% ascending.
A_sorted = sort(A,2);
% shifting all zeros in each row to the end
for i = 1:length(A_sorted(:,1))
num_zeros = sum(A_sorted(i,:)==0);
if num_zeros < length(A_sorted(i,:))
z = zeros(1,num_zeros);
A_sorted(i,:) = [A_sorted(i,num_zeros+1:length(A_sorted(i,:))) z];
end;
end;
% Prelocate in memory an associated array of the corresponding elements in
% D. The matrix Dr is just a reduced derivation from the matrix D.
Dr = zeros(length(A_sorted(:,1)),length(A_sorted(1,:)));
% Create a matrix Dr of elements in D corresponding to the matrix A_sorted.
for i = 1:length(A_sorted(:,1)) % i = 1:3
for j = 1:length(A_sorted(1,:)) % j = 1:2
if A_sorted(i,j) == 0
Dr(i,j) = 0;
else
Dr(i,j) = D(i,A_sorted(i,j));
end;
end;
end;
% We don't need the matrix A_sorted anymore
clear A_sorted
% In order to use the function SORTROWS, we need to transpose matrices
A = A';
Dr = Dr';
% The actual sorting procedure starts here.
for i = 1:length(A(1,:)) % i = 1:3
C = zeros(length(A(:,1)),2); % buffer matrix
C(:,1) = A(:,i);
C(:,2) = Dr(:,i);
C = sortrows(C,2);
A(:,i) = C(:,1);
% shifting all zeros in each column to the end
num_zeros = sum(A(:,i)==0);
if num_zeros < length(A(:,i))
z = zeros(1,num_zeros);
A(:,i) = [A(num_zeros+1:length(A(:,i)),i) z]';
end;
end;
% Transpose the matrix A back
A = A';
clear C Dr z
I am trying to solve this problem:
Write a function called cancel_middle that takes A, an n-by-m
matrix, as an input where both n and m are odd numbers and k, a positive
odd integer that is smaller than both m and n (the function does not have to
check the input). The function returns the input matrix with its center k-by-k
matrix zeroed out.
Check out the following run:
>> cancel_middle(ones(5),3)
ans =
1 1 1 1 1
1 0 0 0 1
1 0 0 0 1
1 0 0 0 1
1 1 1 1 1
My code works only when k=3. How can I generalize it for all odd values of k? Here's what I have so far:
function test(n,m,k)
A = ones(n,m);
B = zeros(k);
A((end+1)/2,(end+1)/2)=B((end+1)/2,(end+1)/2);
A(((end+1)/2)-1,((end+1)/2)-1)= B(1,1);
A(((end+1)/2)-1,((end+1)/2))= B(1,2);
A(((end+1)/2)-1,((end+1)/2)+1)= B(1,3);
A(((end+1)/2),((end+1)/2)-1)= B(2,1);
A(((end+1)/2),((end+1)/2)+1)= B(2,3);
A(((end+1)/2)+1,((end+1)/2)-1)= B(3,1);
A(((end+1)/2)+1,((end+1)/2))= B(3,2);
A((end+1)/2+1,(end+1)/2+1)=B(3,3)
end
You can simplify your code. Please have a look at
Matrix Indexing in MATLAB. "one or both of the row and column subscripts can be vectors", i.e. you can define a submatrix. Then you simply need to do the indexing correct: as you have odd numbers just subtract m-k and n-k and you have the number of elements left from your old matrix A. If you divide it by 2 you get the padding on the left/right, top/bottom. And another +1/-1 because of Matlab indexing.
% Generate test data
n = 13;
m = 11;
A = reshape( 1:m*n, n, m )
k = 3;
% Do the calculations
start_row = (n-k)/2 + 1
start_col = (m-k)/2 + 1
A( start_row:start_row+k-1, start_col:start_col+k-1 ) = zeros( k )
function b = cancel_middle(a,k)
[n,m] = size(a);
start_row = (n-k)/2 + 1;
start_column = (m-k)/2 + 1;
end_row = (n-k)/2 + k;
end_column = (m-k)/2 + k;
a(start_row:end_row,start_column:end_column) = 0;
b = a;
end
I have made a function in an m file called cancel_middle and it basically converts the central k by k matrix as a zero matrix with the same dimensions i.e. k by k.
the rest of the matrix remains the same. It is a general function and you'll need to give 2 inputs i.e the matrix you want to convert and the order of submatrix, which is k.
Given some multidimensional matrix A in Octave / Matlab,
What's the easiest way to get a matrix of the same size as A where all elements are replaced by their index along the k'th dimension
ie for the matrix
A =
ans(:,:,1) =
0.095287 0.191905
0.226278 0.749100
ans(:,:,2) =
0.076826 0.131639
0.862747 0.699016
I want a function f such that
f(A,1) =
ans(:,:,1) =
1 1
2 2
ans(:,:,2) =
1 1
2 2
f(A,2) =
ans(:,:,1) =
1 2
1 2
ans(:,:,2) =
1 2
1 2
and
f(A, 3) =
ans(:,:,1) =
1 1
1 1
ans(:,:,2) =
2 2
2 2
Also, given a sparse matrix B
What's the easiest way to get another sparse matrix of the same size where the nonzero elements are replaced by their index along the k'th dimension? (so same problem as above, but for only the nonzero elements)
Ideally I'm looking for a way which is well-vectorized for octave (meaning it doesn't explicitly loop over anything)
CLARIFICATION: For the sparse matrix one, I'm looking for a solution which does not involve creating a full size(B) matrix at any point
ndgrid() does what you want, although not in the format you are looking for. If you know the dims of the input A beforehand, you can use the following line to create the N-dimentional mesh grid:
% for matrix a where ndims(a) == 3
[x, y, z] = ndgrid (1:size(a,1), 1:size(a,2), 1:size(a,3));
% x is like f(a, 1)
% y is like f(a, 2)
% z is like f(a, 3)
You may be able to write a custom wrapper around ndgrid() to convert it to the function format you are looking for.
In case anyone's curious, since I didn't know about ndgrid, here's the answer I came up with:
function [y] = indices(a,k)
s = size(a);
n = s(k);
D = length(s);
x = permute(a,[k,1:(k-1),(k+1):D]);
y = reshape(x,n,[]);
y = diag(1:n) * ones(size(y));
y = reshape(y,size(x));
y = permute(y,[(2:k),1,(k+1):D]);
endfunction
function [y] = spindices(a,k)
s = size(a);
n = s(k);
D = length(s);
x = permute(a,[k,1:(k-1),(k+1):D]);
y = reshape(x,n,[]);
y = spdiag(1:n) * spones(y);
y = reshape(y,size(x));
y = permute(y,[(2:k),1,(k+1):D]);
endfunction
I have written a function that is supposed to do the following:
Take as an input two sets
Take the distance between the two sets using pdist2 which code is shown here.
It will take the distance between the two sets at the beginning. Then, for the second set, at each iteration, it will set the (i,j) location to 0 and calculate the distance with this change. An, when it goes to the next iteration, it should change the next location value to '0' while at the same time return the preceding value which was set to '0' to its original value.
Note that the result from pdist2 originally return as a matrix, but for the comparison, I sum up the matrix values to use them for comparison.
Based on that, I have written the following function (note that you can use the pdist2.m function from the link here):
function m = pixel_minimize_distance(x,y)
sum1=0;
sum2=0;
[r c] = size(y);
d1 = pdist2(x,y);
[r1 c1] = size(d1);
for i=1:r1
for j=1:c1
sum1=sum1+d1(i,j);
end
end
maximum = sum1;
for i=1:r
for j=1:c
o = y(i,j)
y(i,j) = 0;
d2 = pdist2(x,y);
[r2 c2] = size(d2);
for i=1:r2
for j=1:c2
sum2=sum2+d2(i,j);
end
end
if sum2 >= maximum
if o ~= 0
maximum = sum2;
m = o;
end
end
if sum2 <= maximum
maximum = maximum;
end
y(i,j)=o;
end
end
end
Now, this is what I have run as a test:
>> A=[1 2 3; 6 5 4];
>> B=[4 5 3; 7 8 1];
>> pixel_minimize_distance(A,B)
o =
4
o =
4
o =
1
o =
7
o =
7
o =
0
ans =
7
See the the answer here is 7 (scroll down if you cannot see it), while the expected value when I calculate this manually should be 3, as since when we set it to 0 the sum of the distance will be 142.
Any idea what could be wrong in the code? I think it would be in the location in the code of setting o = y(i,j) where o denotes original value, but really couldn't figure a way of solving that.
Thanks.
I think you have many redundant commands in your code. I just removed them, nothing else. I am getting value of m as 3. I used MATLAB's pdist2 function with squared euclidean distance (since that is the default in the function you provided). I did not get 142 as the distance.
Here is the code:
function m = pixel_minimize_distance(x,y)
[r c] = size(y);
maximum = (sum(sum(pdist2(x,y)))).^2; %explained below
for i=1:r
for j=1:c
o = y(i,j);
y(i,j) = 0
sum2 = (sum(sum(pdist2(x,y)))).^2;
if sum2 >= maximum
if o ~= 0
maximum = sum2;
m = o;
end
end
y(i,j)=o;
end
end
end
and output is:
y =
0 5 3
7 8 1
y =
4 0 3
7 8 1
y =
4 5 0
7 8 1
y =
4 5 3
0 8 1
y =
4 5 3
7 0 1
y =
4 5 3
7 8 0
m =
3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Explanation:
You have written the following snippet of code:
d2 = pdist2(x,y);
[r2 c2] = size(d2);
for i=1:r2
for j=1:c2
sum2=sum2+d2(i,j);
end
end
what this simply does is calculates the distance between two sets using pdist2 and sums up the entire distance matrix to come up with one value stored in sum2 in your case.
Lets look at the my code:
sum2 = (sum(sum(pdist2(x,y)))).^2;
pdist2 will give the distance. First sum command will sum along the rows and then the second one will sum along the columns to give you a total of all values in the matrix (This is what you did with two for loops). Now, the reason behind .^2 is:
In the original pdist2 function in the link which you have provided, you can see from the following snippet of code:
if( nargin<3 || isempty(metric) ); metric=0; end;
switch metric
case {0,'sqeuclidean'}
that squared Euclidean is the default distance, whereas in MATLAB, Euclidean distance is default. Therefore, I have squared the term.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%