I am trying to optimize my code and am not sure how and if I would be able to vectorize this particular section??
for base_num = 1:base_length
for sub_num = 1:base_length
dist{base_num}(sub_num) = sqrt((x(base_num) - x(sub_num))^2 + (y(base_num) - y(sub_num))^2);
end
end
The following example provides one method of vectorization:
%# Set example parameters
N = 10;
X = randn(N, 1);
Y = randn(N, 1);
%# Your loop based solution
Dist1 = cell(N, 1);
for n = 1:N
for m = 1:N
Dist1{n}(m) = sqrt((X(n) - X(m))^2 + (Y(n) - Y(m))^2);
end
end
%# My vectorized solution
Dist2 = sqrt(bsxfun(#minus, X, X').^2 + bsxfun(#minus, Y, Y').^2);
Dist2Cell = num2cell(Dist2, 2);
A quick speed test at N = 1000 has the vectorized solution running two orders of magnitude faster than the loop solution.
Note: I've used a second line in my vectorized solution to mimic your cell array output structure. Up to you whether you want to include it or two combine it into one line etc.
By the way, +1 for posting code in the question. However, two small suggestions for the future: 1) When posting to SO, use simple variable names - especially for loop subscripts - such as I have in my answer. 2) It is nice when we can copy and paste example code straight into a script and run it without having to do any changes or additions (again such as in my answer). This allows us to converge on a solution more rapidly.
Related
How can I simulate this question using MATLAB?
Out of 100 apples, 10 are rotten. We randomly choose 5 apples without
replacement. What is the probability that there is at least one
rotten?
The Expected Answer
0.4162476
My Attempt:
r=0
for i=1:10000
for c=1:5
a = randi(1,100);
if a < 11
r=r+1;
end
end
end
r/10000
but it didn't work, so what would be a better way of doing it?
Use randperm to choose randomly without replacement:
A = false(1, 100);
A(1:10) = true;
r = 0;
for k = 1:10000
a = randperm(100, 5);
r = r + any(A(a));
end
result = r/10000;
Short answer:
Your problem follow an hypergeometric distribution (similar to a binomial distribution but without replacement), if you have the necessary toolbox you can simply use the probability density function of the hypergeometric distribution:
r = 1-hygepdf(0,100,10,5) % r = 0.4162
Since P(x>=1) = P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) = 1-P(x=0)
Of course, here I calculate the exact probability, this is not an experimental result.
To get further:
Noticed that if you do not have access to hygepdf, you can easily write the function yourself by using binomial coefficient:
N = 100; K = 10;
n = 5; k = 0;
r = 1-(nchoosek(K,k)*nchoosek(N-K,n-k))/nchoosek(N,n) % r = 0.4162
You can also use the binomial probability density function, it is a bit more tricky (but also more intuitive):
r = 1-prod(binopdf(0,1,10./(100-[0:4])))
Here we compute the probability to obtain 0 rotten apple five time in a row, the probabily increase at every step since we remove 1 good juicy apple each time. And then, according to the above explaination, we take 1-P(x=0).
There are a couple of issues with your code. First of all, implicitly in what you wrote, you replace the apple after you look at it. When you generate the random number, you need to eliminate the possibility of choosing that number again.
I've rewritten your code to include better practices:
clear
n_runs = 1000;
success = zeros(n_runs, 1);
failure = zeros(n_runs, 1);
approach = zeros(n_runs, 1);
for ii = 1:n_runs
apples = 1:100;
a = randperm(100, 5);
if any(a < 11)
success(ii) = 1;
elseif a >= 11
failure(ii) = 1;
end
approach(ii) = sum(success)/(sum(success)+sum(failure));
end
figure; hold on
plot(approach)
title("r = "+ approach(end))
hold off
The results are stored in an array (called approach), rather than a single number being updated every time, which means you can see how quickly you approach the end value of r.
Another good habit is including clear at the beginning of any script, which reduces the possibility of an error occurring due to variables stored in the workspace.
I have tried searching for an answer on this, but can't find one that specifically addresses my issue. Although vectorizing in MATLAB must draw many questions in, the problem I am having is less general than typical examples I have found on the web. My background is more in C++ than MATLAB, so this is an odd concept for me to get my head around.
I am trying to evolve a Hamiltonian matrix from it's initial state (being a column vector where all elements but the last is a 0, and the last is a 1) to a final state as time increases. This is achieved by sequentially applying a time evolution operator U to the state. I also want to use the new state at each time interval to calculate an observable property.
I have achieved this, as can be seen in the code below. However, I need to make this code as efficient as possible, and so I was hoping to vectorize, rather than rely on for loops. However, I am unsure of how to vectorize this code. The problem I have is that on each iteration of the for loop, the column vector psi should change its values. Each new psi is then used to calculate my observable M for each interval of time. I am unsure of how to track the evolution of psi such that I can end up with a row vector for M, giving the outcome of the application of each new psi.
time = tmin:dt:tmax;
H = magic(2^N)
X = [0,1;1,0]
%%% INITIALISE COLUMN VECTOR
init = sparse(2^N,1);
init(2^N) = 1;
%%% UNITARY TIME EVOLUTION OPERATOR
U = expm(-1i*H*dt);
%%% TIME EVOLVUTION
for num = 1:length(time)
psi = U*init;
init = psi;
%%% CALCULATE OBSERVABLE
M(num) = psi' * kron(X,speye(2^(N-1))) * psi
end
Any help would be greatly appreciated.
I have quickly come up with the following partially vectorized code:
time = tmin:dt:tmax;
H = magic(2^N);
X = [0,1;1,0];
%%% INITIALISE COLUMN VECTOR
init = sparse(2^N,1);
init(2^N) = 1;
%%% UNITARY TIME EVOLUTION OPERATOR
U = expm(-1i*H*dt);
%%% TIME EVOLVUTION
% preallocate psi
psi = complex(zeros(2^N, length(time)));
% compute psi for all timesteps
psi(:,1) = U*init;
for num = 2:length(time)
psi(:,num) = U*psi(:, num-1);
end
% precompute kronecker product (if X is constant through time)
F = kron(X,speye(2^(N-1)));
%%% CALCULATE OBSERVABLE
M = sum((psi' * F) .* psi.', 2);
However, it seems that the most computationally intensive part of your problem is computation of the psi. For that I can't see any obvious way to vectorize as it depends on the value computed in the previous step.
This line:
M = sum((psi' * F) .* psi.', 2);
is a little Matlab trick to compute psi(:,i)'*F*psi(:,i) in a vectorized way.
I'm trying to code a loop in Matlab that iteratively solves for an optimal vector s of zeros and ones. This is my code
N = 150;
s = ones(N,1);
for i = 1:N
if s(i) == 0
i = i + 1;
else
i = i;
end
select = s;
HI = (item_c' * (weights.*s)) * (1/(weights'*s));
s(i) = 0;
CI = (item_c' * (weights.*s)) * (1/(weights'*s));
standarderror_afterex = sqrt(var(CI - CM));
standarderror_priorex = sqrt(var(HI - CM));
ratio = (standarderror_afterex - standarderror_priorex)/(abs(mean(weights.*s) - weights'*select));
ratios(i) = ratio;
s(i) = 1;
end
[M,I] = min(ratios);
s(I) = 0;
This code sets the element to zero in s, which has the lowest ratio. But I need this procedure to start all over again, using the new s with one zero, to find the ratios and exclude the element in s that has the lowest ratio. I need that over and over until no ratios are negative.
Do I need another loop, or do I miss something?
I hope that my question is clear enough, just tell me if you need me to explain more.
Thank you in advance, for helping out a newbie programmer.
Edit
I think that I need to add some form of while loop as well. But I can't see how to structure this. This is the flow that I want
With all items included (s(i) = 1 for all i), calculate HI, CI and the standard errors and list the ratios, exclude item i (s(I) = 0) which corresponds to the lowest negative ratio.
With the new s, including all ones but one zero, calculate HI, CI and the standard errors and list the ratios, exclude item i, which corresponds to the lowest negative ratio.
With the new s, now including all ones but two zeros, repeat the process.
Do this until there is no negative element in ratios to exclude.
Hope that it got more clear now.
Ok. I want to go through a few things before I list my code. These are just how I would try to do it. Not necessarily the best way, or fastest way even (though I'd think it'd be pretty quick). I tried to keep the structure as you had in your code, so you could follow it nicely (even though I'd probably meld all the calculations down into a single function or line).
Some features that I'm using in my code:
bsxfun: Learn this! It is amazing how it works and can speed up code, and makes some things easier.
v = rand(n,1);
A = rand(n,4);
% The two lines below compute the same value:
W = bsxfun(#(x,y)x.*y,v,A);
W_= repmat(v,1,4).*A;
bsxfun dot multiplies the v vector with each column of A.
Both W and W_ are matrices the same size as A, but the first will be much faster (usually).
Precalculating dropouts: I made select a matrix, where before it was a vector. This allows me to then form a variable included using logical constructs. The ~(eye(N)) produces an identity matrix and negates it. By logically "and"ing it with select, then the $i$th column is now select, with the $i$th element dropped out.
You were explicitly calculating weights'*s as the denominator in each for-loop. By using the above matrix to calculate this, we can now do a sum(W), where the W is essentially weights.*s in each column.
Take advantage of column-wise operations: the var() and the sqrt() functions are both coded to work along the columns of a matrix, outputting the action for a matrix in the form of a row vector.
Ok. the full thing. Any questions let me know:
% Start with everything selected:
select = true(N);
stop = false; % Stopping flag:
while (~stop)
% Each column leaves a variable out...
included = ~eye(N) & select;
% This calculates the weights with leave-one-out:
W = bsxfun(#(x,y)x.*y,weights,included);
% You can comment out the line below, if you'd like...
W_= repmat(weights,1,N).*included; % This is the same as previous line.
% This calculates the weights before dropping the variables:
V = bsxfun(#(x,y)x.*y,weights,select);
% There's different syntax, depending on whether item_c is a
% vector or a matrix...
if(isvector(item_c))
HI = (item_c' * V)./(sum(V));
CI = (item_c' * W)./(sum(W));
else
% For example: item_c is a matrix...
% We have to use bsxfun() again
HI = bsxfun(#rdivide, (item_c' * V),sum(V));
CI = bsxfun(#rdivide, (item_c' * W),sum(W));
end
standarderror_afterex = sqrt(var(bsxfun(#minus,HI,CM)));
standarderror_priorex = sqrt(var(bsxfun(#minus,CI,CM)));
% or:
%
% standarderror_afterex = sqrt(var(HI - repmat(CM,1,size(HI,2))));
% standarderror_priorex = sqrt(var(CI - repmat(CM,1,size(CI,2))));
ratios = (standarderror_afterex - standarderror_priorex)./(abs(mean(W) - sum(V)));
% Identify the negative ratios:
negratios = ratios < 0;
if ~any(negratios)
% Drop out of the while-loop:
stop = true;
else
% Find the most negative ratio:
neginds = find(negratios);
[mn, mnind] = min(ratios(negratios));
% Drop out the most negative one...
select(neginds(mnind),:) = false;
end
end % end while(~stop)
% Your output:
s = select(:,1);
If for some reason it doesn't work, please let me know.
In Matlab I need to accumulate overlapping diagonal blocks of a large matrix. The sample code is given below.
Since this piece of code needs to run several times, it consumes a lot of resources. The process is used in array signal processing for a so-called subarray smoothing or spatial smoothing. Is there any way to do this faster?
% some values for parameters
M = 1000; % size of array
m = 400; % size of subarray
n = M-m+1; % number of subarrays
R = randn(M)+1i*rand(M);
% main code
S = R(1:m,1:m);
for i = 2:n
S = S + R(i:m+i-1,i:m+i-1);
end
ATTEMPTS:
1) I tried the following alternative vectorized version, but unfortunately it became much slower!
[X,Y] = meshgrid(1:m);
inds1 = sub2ind([M,M],Y(:),X(:));
steps = (0:n-1)*(M+1);
inds = repmat(inds1,1,n) + repmat(steps,m^2,1);
RR = sum(R(inds),2);
S = reshape(RR,m,m);
2) I used Matlab coder to create a MEX file and it became much slower!
I've personally had to fasten up some portions of my code lately. Being not an expert at all, I would recommend trying the following:
1) Vectorize:
Getting rid of the for-loop
S = R(1:m,1:m);
for i = 2:n
S = S + R(i:m+i-1,i:m+i-1)
end
and replacing it for an alternative based on cumsum should be the way to go here.
Note: will try and work on this approach on a future Edit
2) Generating a MEX-file:
In some instances, you could simply fire up the Matlab Coder app (given that you have it in your current Matlab version).
This should generate a .mex file for you, that you can call as it was the function that you are trying to replace.
Regardless of your choice (1) or 2)), you should profile your current implementation with tic; my_function(); toc; for a fair number of function calls, and compare it with your current implementation:
my_time = zeros(1,10000);
for count = 1:10000
tic;
my_function();
my_time(count) = toc;
end
mean(my_time)
I am writing MATLAB scripts since some time and, still, I do not understand how it works "under the hood". Consider the following script, that do some computation using (big) vectors in three different ways:
MATLAB vector operations;
Simple for cycle that do the same computation component-wise;
An optimized cycle that is supposed to be faster than 2. since avoid some allocation and some assignment.
Here is the code:
N = 10000000;
A = linspace(0,100,N);
B = linspace(-100,100,N);
C = linspace(0,200,N);
D = linspace(100,200,N);
% 1. MATLAB Operations
tic
C_ = C./A;
D_ = D./B;
G_ = (A+B)/2;
H_ = (C_+D_)/2;
I_ = (C_.^2+D_.^2)/2;
X = G_ .* H_;
Y = G_ .* H_.^2 + I_;
toc
tic
X;
Y;
toc
% 2. Simple cycle
tic
C_ = zeros(1,N);
D_ = zeros(1,N);
G_ = zeros(1,N);
H_ = zeros(1,N);
I_ = zeros(1,N);
X = zeros(1,N);
Y = zeros(1,N);
for i = 1:N,
C_(i) = C(i)/A(i);
D_(i) = D(i)/B(i);
G_(i) = (A(i)+B(i))/2;
H_(i) = (C_(i)+D_(i))/2;
I_(i) = (C_(i)^2+D_(i)^2)/2;
X(i) = G_(i) * H_(i);
Y(i) = G_(i) * H_(i)^2 + I_(i);
end
toc
tic
X;
Y;
toc
% 3. Opzimized cycle
tic
X = zeros(1,N);
Y = zeros(1,N);
for i = 1:N,
X(i) = (A(i)+B(i))/2 * (( C(i)/A(i) + D(i)/B(i) ) /2);
Y(i) = (A(i)+B(i))/2 * (( C(i)/A(i) + D(i)/B(i) ) /2)^2 + ( (C(i)/A(i))^2 + (D(i)/B(i))^2 ) / 2;
end
toc
tic
X;
Y;
toc
I know that one shall always try to vectorize computations, being MATLAB build over matrices/vectors (thus, nowadays, it is not always the best choice), so I am expecting that something like:
C = A .* B;
is faster than:
for i in 1:N,
C(i) = A(i) * B(i);
end
What I am not expecting is that it is actually faster even in the above script, despite the second and the third methods I am using go through only one cycle, whereas the first method performs many vector operations (which, theoretically, are a "for" cycle every time). This force me to conclude that MATLAB has some magic that permit (for example) to:
C = A .* B;
D = C .* C;
to be run faster than a single "for" cycle with some operation inside it.
So:
what is the magic that avoid the 1st part to be executed so fast?
when you write "D= A .* B" does MATLAB actually do a component wise computation with a "for" cycle, or simply keeps track that D contains some multiplication of "bla" and "bla"?
EDIT
suppose I want to implement the same computation using C++ (using maybe some library). Will be the first method of MATLAB be faster even than the third one implemented in C++? (I'll answer to this question myself, just give me some time.)
EDIT 2
As requested, here there are the experiment runtimes:
Part 1: 0.237143
Part 2: 4.440132
of which 0.195154 for allocation
Part 3: 2.280640
of which 0.057500 for allocation
and without JIT:
Part 1: 0.337259
Part 2: 149.602017
of which 0.033886 for allocation
Part 3: 82.167713
of which 0.010852 for allocation
The first one is the fastest because vectorized code can be easily interpreted to a small number of optimized C++ library calls. Matlab could also optimize it at more high level, for example, replace G*H+I with an optimized mul_add(G,H,I) instead of add(mul(G,H),I) in its core.
The second one can't be converted to C++ calls easily. It has to be interpreted or compiled. The most modern approach for scripting languages is JIT-compilation. The Matlab JIT compiler is not very good but it doesn't mean it has to be so. I don't know why MathWorks don't improve it. Thus #2 performs so slow that #1 is faster even it makes more "mathematical" operations.
Julia language was invented to be a compromise between Matlab expression and C++ speed. The same non-vectorized code (julia vs matlab) works very fast because JIT-compilation is very good.
Regarding performance optimization I follow #memyself suggestion using the profiler for both approaches as mentioned in 'for' loop vs vectorization in MATLAB.
For educational purposes it does make sense to experiment with numerical algorithms, for anything else I would go with well proven libraries.