I have a for loop nested thrice in a matlab program. Can any of you help me optimize it.
w=5;
a = rand(m*n,10); b=rand(m,n);
for i = 1 : m
for j = 1 : n
for k = 1 : l
if (i-w >= 1 && i+w <= m)
featureL = a(((i-1)*n)+j,:); featureR = a(((i-1)*n)+j-d,:);
D1(i,j,k) = sqrt( sum( (featureL - featureR) .* (featureL - featureR) ) );
D2(i,j,k) = mean2( b(i-w:i+w, j-w:j+w) );
end
end
end
end
I know the performance could be heavily improved by using meshgrid, but I am not sure how to do it.
Thanks in anticipation.
Can it be done something like this..
[X Y Z] = meshgrid(1:m,1:n,1:l);
D1(something containing X,Y,Z) = sqrt( sum( ( a(something cont. X,Y) - a(something cont. X,Y)).*(a(something cont. X,Y) - a(something cont. X,Y)) ) );
% similarly D2
Thanks a lot!.
I've found that a good way to attack these things is incrementally. Start by examining everything in the innermost loop, and see if it can be done at a higher level. This will reduce repeated computations.
For example, you can perform your if (i-w >= 1 && i+w <= m) two levels higher (since it only depends on i,w, and m), reducing if checks and skipping loop iterations.
Once that is done, your featureL and featureR calculations can be moved up one level; they are performed inside the k loop but only depend on j. Similarly, sqrt( sum( (featureL - featureR) .* (featureL - featureR) ) ) can be computed outside of the k loop, put into a variable, and assigned later.
In fact, as far as I can see you can get rid of the entire k loop since k is never used. Here's your code with some of this applied:
w=5;
a = rand(m*n,10);
b=rand(m,n);
for i = 1 : m
if (i-w >= 1 && i+w <= m)
for j = 1 : n
featureL = a(((i-1)*n)+j,:);
featureR = a(((i-1)*n)+j-d,:);
x = sqrt( sum( (featureL - featureR) .* (featureL - featureR) ) );
y = mean2( b(i-w:i+w, j-w:j+w) )
D1(i,j,:) = x;
D2(i,j,:) = y;
end
end
end
Related
I am given a set of points (p1,q1) (p2,q2) ... (p20,q20) which satisfy the function q = 1/(ap + b)^2 except that one of these does not satisfy the given relation. The values of a and b are not given to me. All I have with me is two inputs p and q as arrays. I need to find the index of the point which does not satisfy the given relation.
The way I proceeded to solve is to find the values of a and b using the first two pairs (p1,q1) and (p2,q2) and check if the remaining points satisfy the function for the solved values of a and b. The results will be stored in a logical matrix. I wish to make use of the logical matrix to pick out the odd pair, but unable to proceed further.
Specifically, the challenge is to make use of vectorization in MATLAB to find the odd point, instead of resorting to for-loops. I think that I will have to first search for the only logical zero in any of the row. In that case, the column index of that zero will fetch me the odd point. But, if there are more than one zeros in all 4 rows, then the odd point is either of the first two pairs. I need help in translating this to efficient code in MATLAB.
Please note that vectors p and q have been named as x and y in the below code.
function [res, sol] = findThePair(x, y)
N = length(x);
syms a b
vars = [a,b];
eqns = [y(1) - 1/(a*x(1) + b)^2 == 0; y(2) - 1/(a*x(2) + b)^2];
[solA, solB] = solve(eqns,vars);
sol = [double(solA) double(solB)]; %solution of a & b (total 4 possibilites)
xTest = x(3:end); % performing check on remaining points
yTest = y(3:end);
res = zeros(4, N-2); % logical matrix to store the results of equality check
for i = 1:4
A = sol(i,1); B = sol(i, 2);
res(i, :) = [yTest == 1./(A*xTest + B).^2]; % perform equality check on remaining points
end
Let's do some maths up front, to avoid needing loops or vectorisation. At most this leaves us with half a dozen function evaluations, and we only need 5 points.
q = 1 / (a*p + b)^2
% ->
sqrt(q) * ( a*p + b ) = 1
% ->
a = ( 1 - b*sqrt(q) ) / ( p * sqrt(q) )
% Sub in some points (1 and 2) ->
a1 = ( 1 - b*sqrt(q1) ) / ( p1 * sqrt(q1) )
a2 = ( 1 - b*sqrt(q2) ) / ( p2 * sqrt(q2) )
% a1 and a2 should be the same ->
( 1 - b*sqrt(q1) ) * ( p2 * sqrt(q2) ) = ( 1 - b*sqrt(q2) ) * ( p1 * sqrt(q1) )
% Rearrange ->
b = ( p2*sqrt(q2) - p1*sqrt(q1) ) / ( (p2-p1)*sqrt(q1)*sqrt(q2) )
We have two unknowns, a and b. All we need are two points to create simultaneous equations. I'll use the following logic
Choose (pm, qm) and (pn, qn) with any m ~= n.
Calculate a and b using the above equation.
test whether (pr, qr) fits with the calculated a and b.
If it fits, we know all three of these must be on the curve, and we have a and b.
If it doesn't fit, we know either point m, n, or r is the outlier. Return to step (1) with two other points, the calculated a and b must be correct, as we've not fitted to the outlier.
Here is some code to implement this:
% Random coeffs, keep things unknown
a = rand*10;
b = rand*10;
% Set up our data
p = 1:20;
q = 1 ./ (a*p + b).^2;
% Create an outlier
q( 3 ) = q( 3 ) + 1;
% Steps as described
% 1.
p1 = p(1); p2 = p(2);
q1 = q(1); q2 = q(2);
% 2.
bGuess = ( p2*sqrt(q2) - p1*sqrt(q1) ) / ( (p2-p1)*sqrt(q1)*sqrt(q2) );
aGuess = ( 1 - bGuess*sqrt(q1) ) / ( p1 * sqrt(q1) );
% 3.
p3 = p(3);
q3Guess = 1 / ( aGuess*p3 + bGuess )^2;
tol = 1e-7; % Use tolerance rather than == comparison to avoid float issues
if abs( q3Guess - q(3) ) < tol
% success
aFit = aGuess;
bFit = bGuess;
else
% p1, p2 or p3 is an outlier! Repeat using other points
% If there's known to be only one outlier, this should give the result
p1 = p(4); p2 = p(5);
q1 = q(4); q2 = q(5);
bFit = ( p2*sqrt(q2) - p1*sqrt(q1) ) / ( (p2-p1)*sqrt(q1)*sqrt(q2) );
aFit = ( 1 - bFit*sqrt(q1) ) / ( p1 * sqrt(q1) );
end
% Validate
fprintf( 'a is valid: %d, b is valid: %d\n', abs(a-aFit)<tol, abs(b-bFit)<tol )
I don't really understand how you were trying to solve this and what do syms (i.e. symbolic variables) have to do with this, so I'll show you how I would solve this problem.
Since we're essentially looking for an outlier, we might as well convert the problem to something that's easier to work with. For this reason, instead of using q as-is, I'm going to invert it: this way, we'd be dealing with an equation of a parabola - which is easy.
Next, knowing that our points should lie on a parabola, we can fit the equation of the parabola (or equivalently - find the coefficients of the polynomial that describes the relation of the input to the output). The polynomial is a^2*x^2+2*a*b*x+b^2, and so the coefficients are {a^2, 2*a*b, b^2}.
Since the majority of the points (19 out of 20) lie on the same parabola, the outlier will always have a larger error, which would make it stand out, no matter how close it is to the parabola (within the limitations of machine precision) - you can see an extreme example of this in the code below.
Fitting of a parabola is performed using polynomial interpolation (see also: Vandermonde matrix).
function I = q55241683()
%% Generate the ground truth:
TRUE_A = 2.3;
TRUE_B = -pi;
IDX_BAD = 5;
p = 1:0.04:1.76;
q = (TRUE_A * p + TRUE_B).^-2;
q(IDX_BAD) = (1-1E-10)*q(IDX_BAD); % notice just how close this is to being valid
%% Visualize dataset:
% figure(); plot(p,q.^-1);
%% Solve
I = findThePair(p, q.^-1);
%% Test
if IDX_BAD == I
disp('Great success!');
else
disp('Complete failure!');
end
end
function I = findThePair(x,y)
% Fit a parabola to {x vs. y^-1}
P = x(:).^(2:-1:0)\y(:); %alternatively: P = polyfit(x,y.^-1,2)
% Estimate {a,b} (or {-a,-b})
est_A = sqrt(P(1));
est_B = P(2)/(2*est_A);
% Compute the distances of the points from the fit (residuals), find the biggest:
[~,I] = max( abs(y - (est_A*x + est_B).^2) );
end
Can you help me please translate this equation into a code?
I tried this code but its only the first part
theSum = sum(M(:, y) .* S(:, y) ./ (1 + K(:, y)))
EDIT: sorry, I was having a brainfart. The answer below makes no assumptions on the nature of M, K, etc, which is why I recommended functions like that. But they're clearly matrices. I'll make another answer, I'll leave this here for reference though in case it's useful
I would start by making the M, K, X, L, and O expressions into simple functions, so that you can easily call them as M(z,y), X(z,y) (or X(z,j) depending on the input you need ) etc.
Then you will convert each summation into a for loop and collect the result (you can think about vectorisation later, right now focus on translating the problem). The double summation is essentially a nested for loop, where the result of the inner loop is used in the outer one at each outer iteration.
So your end result should look something like:
Summation1 = 0;
for z = 1 : Z
tmp = M(z,y) / K(z,y) * (X(z,y) / (1 + L(z,y));
Summation1 = Summation1 + tmp;
end
Summation2 = 0;
for j = 1 : Y
if j ~= y
for z = 1 : Z
tmp = (M(z,j) * X(z,j) * O(j)) / (K(z,j)^2 * (1 + L(z,j)) * X(z,y);
Summation2 = Summation2 + tmp;
end
end
end
Result = Summation1 - Summation2;
(Btw, this assumes that all operations are on scalars. If M(z,y) outputs a vector, adjust for elementwise operations appropriately)
IF M, K, etc are all matrices, and all operations are expected to be element-wise, then this is a vectorised approach for this equation.
Left summation is
S1 = M(1:Z,y) ./ K(1:Z,y) .* X(1:Z,y) ./ (1 + L(1:Z,y));
S1 = sum(S1);
Right summation is (assuming (O is a horizontal vector)
S2 = M(1:Z, 1:Y) .* X(1:X, 1:Y) .* repmat(O(1:Y), [Z,1]) ./ ...
(K(1:Z, 1:Y) .^ 2 .* (1 + L(1:Z, 1:Y))) .* X(1:Z, 1:Y);
S2(:,y) = []; % remove the 'y' column from the matrix
S2 = sum(S2(:)); % add all elements
End result: S1 - S2
this is lambda version vectorized:
equation = #(y,M,K,X,L,O) ...
sum(M(:,y)./K(:,y).*X(:,y)./(1+L(:,y))) ...
-sum(sum( ...
bsxfun( ...
#times ...
,M(:,[1:y-1,y+1:end]) ...
.* X(:,[1:y-1,y+1:end]) ...
.* O(:,[1:y-1,y+1:end]) ...
./ (K(:,[1:y-1,y+1:end]) .^ 2 ...
.*(1+ L(:,[1:y-1,y+1:end]))) ...
,X(:,y) ...
) ...
));
%%% example:
y = 3;
Y = 5;
Z = 10;
M = rand(Y, Z);K = rand(Y, Z);X = rand(Y, Z);L = rand(Y, Z);O = rand(Y, Z);
equation(y,M,K,X,L,O)
I have a linear system Ax = b , which is created by natural splines and looks like this:
where
The code in matlab which is supposed to solve the system is the following:
clear;
clc;
x = [...] ;
a = [...];
x0 = ...;
n = length(x) - 1 ;
for i = 0 : (n-1)
h(i+1) = x(i+2) - x(i+1) ;
end
b= zeros( n+1 , 1 ) ;
for i =2: n
b(i,1) = 3 *(a(i+1)-a(i))/h(i) - 3/h(i-1)*(a(i) - a(i-1) ) ;
end
%linear system solution.
l(1) =0 ; m(1) = 0 ; z(1) = 0 ;
for i =1:(n-1)
l(i+1) = 2*( x(i+2) - x(i) ) - h(i)* m(i) ;
m(i+1) = h(i+1)/l(i+1);
z(i+1) = ( b(i+1) - h(i)*z(i) ) / l ( i+1) ;
end
l(n+1) =1;
z(n+1) = 0 ;
c(n+1) = 0 ;
for j = ( n-1) : (-1) : 0
c(j+1) = z(j+1) - m(j+1)*c(j+2) ;
end
but I can't understand which method is being used for solving the linear system.
If I had to guess I would say that the LU method is used, adjusted for tridiagonal matrices, but I still can't find the connection with the code...
Any help would be appreciated!!!
The coefficients look a little odd (particularly that 2 in the l equation), but it looks like a specialized Thomas Algorithm where:
The second-to-last loop performs a forward elimination of the subdiagonal to bring the matrix into upper triangular form.
The last loop performs the back substitution for the solution.
The code doesn't seem to match one-to-one with the general algorithm since the solution is using the vectors that compose the diagonals instead of the diagonals themselves with no apparent preallocation of memory. So I can't say if this method is "better" than the general one off the bat.
I would like to use the LAPACK routines for factorisation and inversion of matrices using the fully packed rectangular format, as this requires only n(n+1)/2 elements to be stored for a symmetric nxn matrix. So far, I am setting up the matrix in 'packed' format and transform it calling routine DTPTTF. However, this requires a second array. I would like to build my matrix directly in fully packed rectangular format (to save on space) - is there an 'addressing' function which will give me the position of the i,j-th element? or could somebody point me to the relevant formula?
to partly answer my own question: inspecting the source code of DTPTTF and the example given therein, I've worked out the adress for one of the four possible constellations (the only one I need), namely uplo ='L' and trans ='N'. below is my fortran function:
! ==================================== ! returns address for RFP format
integer function ijfprf( ii, jj, n ) ! for row jj and column ii
! ==================================== ! for UPLO = 'L' and TRANSR = 'N' only!
implicit none
integer, intent(in) :: ii, jj, n
integer :: i, j, k, n1, k1
if( ii <= jj ) then
i = ii; j = jj
else
i = jj; j = ii
end if
k = n/2
if( mod(n,2) == 0 ) then ! n even
n1 = n + 1
if( i <= k ) then
ijfprf = 1 + (i - 1) * n1 + j
else
ijfprf = ( j - k - 1 ) * n1 + i - k
end if
else ! n odd
k1 = k + 1
if( i > k1 ) then
ijfprf = ( j - k1 ) * n + i - k1
else
ijfprf = ( i - 1 ) * n + j
end if
end if
return
end function ijfprf
I have an issue with a code performing some array operations. It is getting too slow, because I am using loops. I am trying for some time to optimize this code and to re-write it with less or without loops. Until now unsuccessfully. Can you please help me solve this:
YVal = 1:1:100000;
M_MAX = 1000;
N_MAX = 2000;
clear YTemp
tic
for M=1:1:M_MAX
for N = 1:1:N_MAX
YTemp(M,N) = sum(YVal (N+1:N+M) ) - sum(YVal (1:M) );
end
end
For large N_MAX and M_MAX the execution time of these two loops is very high. How can I optimize this?
Thank you!
Assuming YVal is larger than N_MAX+M_MAX
sum1 = cumsum( YVal(1:(M_MAX+N_MAX)) ); % sum1(M) = sum(YVal(1:M))
If I'm not mistaken, then
sum( YVal( N+1:N+M ) ) = sum1( N + M ) - sum1( N )
And therefore
YTemp( M, N ) = sum1( N + M ) - sum1( N ) - sum1( M )
Using ndgrid
[M N] = ndgrid( 1:M_MAX, 1:N_MAX );
YTemp = sum1( N + M ) - sum1( N ) - sum1( M );
Have I got it right?
EDIT:
Another go without ndgrid
sum1 = cumsum( YVal( 1 : (N_MAX+M_MAX) ) );
YTemp = bsxfun( #minus, ...
bsxfun( #minus, ...
sum1( bsxfun( #plus, 1:N_MAX, (1:M_MAX)' ) ) , sum1 ),...
sum1' );
You should be able to speed it up a little by hoisting the invariant term out of the inner loop, e.g.
for M=1:1:M_MAX
sum2 = sum(YVal(1:M));
for N = 1:1:N_MAX
YTemp(M,N) = sum(YVal(N+1:N+M)) - sum2;
end
end