Oftentimes I need to dynamically fill a vector in Matlab. However this is sligtly annoying since you first have to define an empty variable first, e.g.:
[a,b,c]=deal([]);
for ind=1:10
if rand>.5 %some random condition to emphasize the dynamical fill of vector
a=[a, randi(5)];
end
end
a %display result
Is there a better way to implement this 'push' function, so that you do not have to define an empty vector beforehand? People tell me this is nonsensical in Matlab- if you think this is the case please explain why.
related: Push a variable in a vector in Matlab, is-there-an-elegant-way-to-create-dynamic-array-in-matlab
In MATLAB, pre-allocation is the way to go. From the docs:
for and while loops that incrementally increase the size of a data structure each time through the loop can adversely affect performance and memory use.
As pointed out in the comments by m7913d, there is a question on MathWorks' answers section which addresses this same point, read it here.
I would suggest "over-allocating" memory, then reducing the size of the array after your loop.
numloops = 10;
a = nan(numloops, 1);
for ind = 1:numloops
if rand > 0.5
a(ind) = 1; % assign some value to the current loop index
end
end
a = a(~isnan(a)); % Get rid of values which weren't used (and remain NaN)
No, this doesn't decrease the amount you have to write before your loop, it's even worse than having to write a = []! However, you're better off spending a few extra keystrokes and minutes writing well structured code than making that saving and having worse code.
It is (as for as I known) not possible in MATLAB to omit the initialisation of your variable before using it in the right hand side of an expression. Moreover it is not desirable to omit it as preallocating an array is almost always the right way to go.
As mentioned in this post, it is even desirable to preallocate a matrix even if the exact number of elements is not known. To demonstrate it, a small benchmark is desirable:
Ns = [1 10 100 1000 10000 100000];
timeEmpty = zeros(size(Ns));
timePreallocate = zeros(size(Ns));
for i=1:length(Ns)
N = Ns(i);
timeEmpty(i) = timeit(#() testEmpty(N));
timePreallocate(i) = timeit(#() testPreallocate(N));
end
figure
semilogx(Ns, timeEmpty ./ timePreallocate);
xlabel('N')
ylabel('time_{empty}/time_{preallocate}');
% do not preallocate memory
function a = testEmpty (N)
a = [];
for ind=1:N
if rand>.5 %some random condition to emphasize the dynamical fill of vector
a=[a, randi(5)];
end
end
end
% preallocate memory with the largest possible return size
function a = testPreallocate (N)
last = 0;
a = zeros(N, 1);
for ind=1:N
if rand>.5 %some random condition to emphasize the dynamical fill of vector
last = last + 1;
a(last) = randi(5);
end
end
a = a(1:last);
end
This figure shows how much time the method without preallocating is slower than preallocating a matrix based on the largest possible return size. Note that preallocating is especially important for large matrices due the the exponential behaviour.
Related
I have large sets of 3D data consisting of 1D signals acquired in 2D space.
The first step in processing this data is thresholding all signals to find the arrival of a high-amplitude pulse. This pulse is present in all signals and arrives at different times.
After thresholding, the 3D data set should be reordered so that every signal starts at the arrival of the pulse and what came before is thrown away (the end of the signals is of no importance, as of now i concatenate zeros to the end of all signals so the data remains the same size).
Now, I have implemented this in the following manner:
First, i start by calculating the sample number of the first sample exceeding the threshold in all signals
M = randn(1000,500,500); % example matrix of realistic size
threshold = 0.25*max(M(:,1,1)); % 25% of the maximum in the first signal as threshold
[~,index] = max(M>threshold); % indices of first sample exceeding threshold in all signals
Next, I want all signals to be shifted so that they all start with the pulse. For now, I have implemented it this way:
outM = zeros(size(M)); % preallocation for speed
for i = 1:size(M,2)
for j = 1:size(M,3)
outM(1:size(M,1)+1-index(1,i,j),i,j) = M(index(1,i,j):end,i,j);
end
end
This works fine, and i know for-loops are not that slow anymore, but this easily takes a few seconds for the datasets on my machine. A single iteration of the for-loop takes about 0.05-0.1 sec, which seems slow to me for just copying a vector containing 500-2000 double values.
Therefore, I have looked into the best way to tackle this, but for now I haven't found anything better.
I have tried several things: 3D masks, linear indexing, and parallel loops (parfor).
for 3D masks, I checked to see if any improvements are possible. Therefore i first contruct a logical mask, and then compare the speed of the logical mask indexing/copying to the double nested for loop.
%% set up for logical mask copying
AA = logical(ones(500,1)); % only copy the first 500 values after the threshold value
Mask = logical(zeros(size(M)));
Jepla = zeros(500,size(M,2),size(M,3));
for i = 1:size(M,2)
for j = 1:size(M,3)
Mask(index(1,i,j):index(1,i,j)+499,i,j) = AA;
end
end
%% speed comparison
tic
Jepla = M(Mask);
toc
tic
for i = 1:size(M,2)
for j = 1:size(M,3)
outM(1:size(M,1)+1-index(1,i,j),i,j) = M(index(1,i,j):end,i,j);
end
end
toc
The for-loop is faster every time, even though there is more that's copied.
Next, linear indexing.
%% setup for linear index copying
%put all indices in 1 long column
LongIndex = reshape(index,numel(index),1);
% convert to linear indices and store in new variable
linearIndices = sub2ind(size(M),LongIndex,repmat(1:size(M,2),1,size(M,3))',repelem(1:size(M,3),size(M,2))');
% extend linear indices with those of all values to copy
k = zeros(numel(M),1);
count = 1;
for i = 1:numel(LongIndex)
values = linearIndices(i):size(M,1)*i;
k(count:count+length(values)-1) = values;
count = count + length(values);
end
k = k(1:count-1);
% get linear indices of locations in new matrix
l = zeros(length(k),1);
count = 1;
for i = 1:numel(LongIndex)
values = repelem(LongIndex(i)-1,size(M,1)-LongIndex(i)+1);
l(count:count+length(values)-1) = values;
count = count + length(values);
end
l = k-l;
% create new matrix
outM = zeros(size(M));
%% speed comparison
tic
outM(l) = M(k);
toc
tic
for i = 1:size(M,2)
for j = 1:size(M,3)
outM(1:size(M,1)+1-index(1,i,j),i,j) = M(index(1,i,j):end,i,j);
end
end
toc
Again, the alternative approach, linear indexing, is (a lot) slower.
After this failed, I learned about parallelisation, and though this would for sure speed up my code.
By reading some of the documentation around parfor and trying it out a bit, I changed my code to the following:
gcp;
outM = zeros(size(M));
inM = mat2cell(M,size(M,1),ones(size(M,2),1),size(M,3));
tic
parfor i = 1:500
for j = 1:500
outM(:,i,j) = [inM{i}(index(1,i,j):end,1,j);zeros(index(1,i,j)-1,1)];
end
end
end
toc
I changed it so that "outM" and "inM" would both be sliced variables, as I read this is best. Still this is very slow, a lot slower than the original for loop.
So now the question, should I give up on trying to improve the speed of this operation? Or is there another way in which to do this? I have searched a lot, and for now do not see how to speed this up.
Sorry for the long question, but I wanted to show what I tried.
Thank you in advance!
Not sure if an option in your situation, but looks like cell arrays are actually faster here:
outM2 = cell(size(M,2),size(M,3));
tic;
for i = 1:size(M,2)
for j = 1:size(M,3)
outM2{i,j} = M(index(1,i,j):end,i,j);
end
end
toc
And a second idea which also came out faster, batch all data which have to be shifted by the same value:
tic;
for i = 1:unique(index).'
outM(1:size(M,1)+1-i,index==i) = M(i:end,index==i);
end
toc
It totally depends on your data if this approach is actually faster.
And yes integer valued and logical indexing can be mixed
This question already has answers here:
'for' loop vs vectorization in MATLAB
(5 answers)
Closed 3 years ago.
In Matlab, I am trying to vectorise my code to improve the simulation time. However, the result I got was that I deteriorated the overall efficiency.
To understand the phenomenon I created 3 distinct functions that does the same thing but with different approach :
The main file :
clc,
clear,
n = 10000;
Value = cumsum(ones(1,n));
NbLoop = 10000;
time01 = zeros(1,NbLoop);
time02 = zeros(1,NbLoop);
time03 = zeros(1,NbLoop);
for test = 1 : NbLoop
tic
vector1 = function01(n,Value);
time01(test) = toc ;
tic
vector2 = function02(n,Value);
time02(test) = toc ;
tic
vector3 = function03(n,Value);
time03(test) = toc ;
end
figure(1)
hold on
plot( time01, 'b')
plot( time02, 'g')
plot( time03, 'r')
The function 01:
function vector = function01(n,Value)
vector = zeros( 2*n,1);
for k = 1:n
vector(2*k -1) = Value(k);
vector(2*k) = Value(k);
end
end
The function 02:
function vector = function02(n,Value)
vector = zeros( 2*n,1);
vector(1:2:2*n) = Value;
vector(2:2:2*n) = Value;
end
The function 03:
function vector = function03(n,Value)
MatrixTmp = transpose([Value(:), Value(:)]);
vector = MatrixTmp (:);
end
The blue plot correspond to the for - loop.
n = 100:
n = 10000:
When I run the code with n = 100, the more efficient solution is the first function with the for loop.
When n = 10000 The first function become the less efficient.
Do you have a way to know how and when to properly replace a for-loop by a vectorised counterpart?
What is the impact of index searching with array of tremendous dimensions ?
Does Matlab compute in a different manner an array of dimensions 3 or higher than a array of dimension 1 or 2?
Is there a clever way to replace a while loop that use the result of an iteration for the next iteration?
Using MATLAB Online I see something different:
n 10000 100
function01 5.6248e-05 2.2246e-06
function02 1.7748e-05 1.9491e-06
function03 2.7748e-05 1.2278e-06
function04 1.1056e-05 7.3390e-07 (my version, see below)
Thus, the loop version is always slowest. Method #2 is faster for very large matrices, Method #3 is faster for very small matrices.
The reason is that method #3 makes 2 copies of the data (transpose or a matrix incurs a copy), and that is bad if there's a lot of data. Method #2 uses indexing, which is expensive, but not as expensive as copying lots of data twice.
I would suggest this function instead (Method #4), which transposes only vectors (which is essentially free). It is a simple modification of your Method #3:
function vector = function04(n,Value)
vector = [Value(:).'; Value(:).'];
vector = vector(:);
end
Do you have a way to know how and when to properly replace a for-loop by a vectorised counterpart?
In general, vectorized code is always faster if there are no large intermediate matrices. For small data you can vectorize more aggressively, for large data sometimes loops are more efficient because of the reduced memory pressure. It depends on what is needed for vectorization.
What is the impact of index searching with array of tremendous dimensions?
This refers to operations such as d = data(data==0). Much like everything else, this is efficient for small data and less so for large data, because data==0 is an intermediate array of the same size as data.
Does Matlab compute in a different manner an array of dimensions 3 or higher than a array of dimension 1 or 2?
No, not in general. Functions such as sum are implemented in a dimensionality-independent waycitation needed.
Is there a clever way to replace a while loop that use the result of an iteration for the next iteration?
It depends very much on what the operations are. Functions such as cumsum can often be used to vectorize this type of code, but not always.
This is my timing code, I hope it shows how to properly use timeit:
%n = 10000;
n = 100;
Value = cumsum(ones(1,n));
vector1 = function01(n,Value);
vector2 = function02(n,Value);
vector3 = function03(n,Value);
vector4 = function04(n,Value);
assert(isequal(vector1,vector2))
assert(isequal(vector1,vector3))
assert(isequal(vector1,vector4))
timeit(#()function01(n,Value))
timeit(#()function02(n,Value))
timeit(#()function03(n,Value))
timeit(#()function04(n,Value))
function vector = function01(n,Value)
vector = zeros(2*n,1);
for k = 1:n
vector(2*k-1) = Value(k);
vector(2*k) = Value(k);
end
end
function vector = function02(n,Value)
vector = zeros(2*n,1);
vector(1:2:2*n) = Value;
vector(2:2:2*n) = Value;
end
function vector = function03(n,Value)
MatrixTmp = transpose([Value(:), Value(:)]);
vector = MatrixTmp(:);
end
function vector = function04(n,Value)
vector = [Value(:).'; Value(:).'];
vector = vector(:);
end
I have the following calculations in two steps:
Initially, I create a set of 4 grid vectors, each spanning from -2 to 2:
u11grid=[-2:0.1:2];
u12grid=[-2:0.1:2];
u22grid=[-2:0.1:2];
u21grid=[-2:0.1:2];
[ca, cb, cc, cd] = ndgrid(u11grid, u12grid, u22grid, u21grid);
u11grid=ca(:);
u12grid=cb(:);
u22grid=cc(:);
u21grid=cd(:);
%grid=[u11grid u12grid u22grid u21grid]
sg=size(u11grid,1);
Next, I have an algorithm assigning the same index (equalorder) to the rows of grid sharing a specific structure:
U1grid=[-u11grid -u21grid -u12grid -u22grid Inf*ones(sg,1) -Inf*ones(sg,1)];
U2grid=[u21grid-u11grid -u21grid u22grid-u12grid -u22grid Inf*ones(sg,1) -Inf*ones(sg,1)];
s1=size(U1grid,2);
s2=size(U2grid,2);
%-------------------------------------------------------
%sortedU1grid gives U1grid with each row sorted from smallest to largest
%for each row i of sortedU1grid and for j=1,2,...,s1 index1(i,j) gives
%the column position 1,2,...,s1 in U1grid(i,:) of sortedU1grid(i,j)
[sortedU1grid,index1] = sort(U1grid,2);
%for each row i of sortedU1grid, d1(i,:) is a 1x(s1-1) row of ones and zeros
% d1(i,j)=1 if sortedU1grid(i,j)-sortedU1grid(i,j-1)=0 and d1(i,j)=0 otherwise
d1 = diff(sortedU1grid,[],2) == 0;
%-------------------------------------------------------
%Repeat for U2grid
[sortedU2grid,index2] = sort(U2grid,2);
d2 = diff(sortedU2grid,[],2) == 0;
%-------------------------------------------------------
%Assign the same index to the rows of grid sharing the same "ordering"
[~,~,equalorder] = unique([index1 index2 d1 d2],'rows', 'stable'); %sgx1
My question: is there a way to compute the algorithm in step 2 without the initial construction of the grid vectors in step 1? I am asking this because step 1 takes a lot of memory given that it basically generates the Cartesian product of 4 sets.
A solution should not rely on the specific content of U1grid and U2grid as that part changes in my actual code. To be more clear: U1grid and U2grid are ALWAYS derived from u11grid, ..., u21grid; however, the way in which they are derived from u11grid, ..., u21grid is slightly more complicated in my actual code from what I have reported here.
As Cris Luengo mentions in a comment, you're always going to be dealing with a trade-off between speed and memory. That said, one option you have is to only compute each of your 4 grid variables (u11grid u12grid u22grid u21grid) when needed instead of computing them once and storing them. You will save on memory but will lose speed if you are recomputing each one multiple times.
The solution I came up with involves creating an anonymous function equivalent for each of the 4 grid variables, using combinations of repmat and repelem to compute each individually instead of ndgrid to compute them all together:
u11gridFcn = #() repmat((-2:0.1:2).', 41.^3, 1);
u12gridFcn = #() repmat(repelem((-2:0.1:2).', 41), 41.^2, 1);
u22gridFcn = #() repmat(repelem((-2:0.1:2).', 41.^2), 41, 1);
u21gridFcn = #() repelem((-2:0.1:2).', 41.^3);
sg = 41.^4;
You would then use these by replacing every usage of your 4 grid variables in U1grid and U2grid with their corresponding function call. For your specific example above, this would be the new code for U1grid and U2grid (note also the use of inf(...) instead of Inf*ones(...), a small detail):
U1grid = [-u11gridFcn() ...
-u21gridFcn() ...
-u12gridFcn() ...
-u22gridFcn() ...
inf(sg, 1) ...
-inf(sg, 1)];
U2grid = [u21gridFcn()-u11gridFcn() ...
-u21gridFcn() ...
u22gridFcn()-u12gridFcn() ...
-u22gridFcn() ...
inf(sg, 1) ...
-inf(sg, 1)];
In this example, you avoid the memory needed to store the 4 grid variables, but the values for u11grid and u12grid will each be computed twice while the values for u21grid and u22grid will each be computed three times. Likely a small time trade-off for a potentially significant memory savings.
You may be able to remove the ndgrid, but it is not the memory bottleneck of this code, which is the call to unique on the large matrix A = [index1 index2 d1 d2]. The size of A is 2825761 by 22 (much larger than the grids), and it seems that unique may even internally copy A. I was able to avoid this call using
[sorted, ind] = sortrows([index1 index2 d1 d2]);
change = [1; any(diff(sorted), 2)];
uniqueInd = cumsum(change);
equalorder(ind) = uniqueInd;
[~, ~, equalorder] = unique(equalorder, 'stable');
where the last line is still the memory bottleneck and is only needed if you want the same numbering as your code produces. If any unique ordering is okay, you can skip it. You may be able to further reduce the memory footprint by carefully clearing variables are soon as they are no longer needed.
I have to construct the following function in MATLAB and am having trouble.
Consider the function s(t) defined for t in [0,4) by
{ sin(pi*t/2) , for t in [0,1)
s(t) = { -(t-2)^3 , for t in [1,3)*
{ sin(pi*t/2) , for t in [3,4)
(i) Generate a column vector s consisting of 512 uniform
samples of this function over the interval [0,4). (This
is best done by concatenating three vectors.)
I know it has to be something of the form.
N = 512;
s = sin(5*t/N).' ;
But I need s to be the piecewise function, can someone provide assistance with this?
If I understand correctly, you're trying to create 3 vectors which calculate the specific function outputs for all t, then take slices of each and concatenate them depending on the actual value of t. This is inefficient as you're initialising 3 times as many vectors as you actually want (memory), and also making 3 times as many calculations (CPU), most of which will just be thrown away. To top it off, it'll be a bit tricky to use concatenate if your t is ever not as you expect (i.e. monotonically increasing). It might be an unlikely situation, but better to be general.
Here are two alternatives, the first is imho the nice Matlab way, the second is the more conventional way (you might be more used to that if you're coming from C++ or something, I was for a long time).
function example()
t = linspace(0,4,513); % generate your time-trajectory
t = t(1:end-1); % exclude final value which is 4
tic
traj1 = myFunc(t);
toc
tic
traj2 = classicStyle(t);
toc
end
function trajectory = myFunc(t)
trajectory = zeros(size(t)); % since you know the size of your output, generate it at the beginning. More efficient than dynamically growing this.
% you could put an assert for t>0 and t<3, otherwise you could end up with 0s wherever t is outside your expected range
% find the indices for each piecewise segment you care about
idx1 = find(t<1);
idx2 = find(t>=1 & t<3);
idx3 = find(t>=3 & t<4);
% now calculate each entry apprioriately
trajectory(idx1) = sin(pi.*t(idx1)./2);
trajectory(idx2) = -(t(idx2)-2).^3;
trajectory(idx3) = sin(pi.*t(idx3)./2);
end
function trajectory = classicStyle(t)
trajectory = zeros(size(t));
% conventional way: loop over each t, and differentiate with if-else
% works, but a lot more code and ugly
for i=1:numel(t)
if t(i)<1
trajectory(i) = sin(pi*t(i)/2);
elseif t(i)>=1 & t(i)<3
trajectory(i) = -(t(i)-2)^3;
elseif t(i)>=3 & t(i)<4
trajectory(i) = sin(pi*t(i)/2);
else
error('t is beyond bounds!')
end
end
end
Note that when I tried it, the 'conventional way' is sometimes faster for the sampling size you're working on, although the first way (myFunc) is definitely faster as you scale up really a lot. In anycase I recommend the first approach, as it is much easier to read.
I have a code that repeatedly calculates a sparse matrix in a loop (it performs this calculation 13472 times to be precise). Each of these sparse matrices is unique.
After each execution, it adds the newly calculated sparse matrix to what was originally a sparse zero matrix.
When all 13742 matrices have been added, the code exits the loop and the program terminates.
The code bottleneck occurs in adding the sparse matrices. I have made a dummy version of the code that exhibits the same behavior as my real code. It consists of a MATLAB function and a script given below.
(1) Function that generates the sparse matrix:
function out = test_evaluate_stiffness(n)
ind = randi([1 n*n],300,1);
val = rand(300,1);
[I,J] = ind2sub([n,n],ind);
out = sparse(I,J,val,n,n);
end
(2) Main script (program)
% Calculate the stiffness matrix
n=1000;
K=sparse([],[],[],n,n,n^2);
tic
for i=1:13472
temp=rand(1)*test_evaluate_stiffness(n);
K=K+temp;
end
fprintf('Stiffness Calculation Complete\nTime taken = %f s\n',toc)
I'm not very familiar with sparse matrix operations so I may be missing a critical point here that may allow my code to be sped up considerably.
Am I handling the updating of my stiffness matrix in a reasonable way in my code? Is there another way that I should be using sparse that will result in a faster solution?
A profiler report is also provided below:
If you only need the sum of those matrices, instead of building all of them individually and then summing them, simply concatenate the vectors I,J and vals and call sparse only once. If there are duplicate rows [i,j] in [I,J] the corresponding values S(i,j) will be summed automatically, so the code is absolutely equivalent. As calling sparse involves an internal call to a sorting algorithm, you save 13742-1 intermediate sorts and can get away with only one.
This involves changing the signature of test_evaluate_stiffness to output [I,J,val]:
function [I,J,val] = test_evaluate_stiffness(n)
and removing the line out = sparse(I,J,val,n,n);.
You will then change your other function to:
n = 1000;
[I,J,V] = deal([]);
tic;
for i = 1:13472
[I_i, J_i, V_i] = test_evaluate_stiffness(n);
nE = numel(I_i);
I(end+(1:nE)) = I_i;
J(end+(1:nE)) = J_i;
V(end+(1:nE)) = rand(1)*V_i;
end
K = sparse(I,J,V,n,n);
fprintf('Stiffness Calculation Complete\nTime taken = %f s\n',toc);
If you know the lengths of the output of test_evaluate_stiffness ahead of time, you can possibly save some time by preallocating the arrays I,J and V with appropriately-sized zeros matrices and set them using something like:
I((i-1)*nE + (1:nE)) = ...
J((i-1)*nE + (1:nE)) = ...
V((i-1)*nE + (1:nE)) = ...
The biggest remaining computation, taking 11s, is the sparse operation
on the final I,J,V vectors so I think we've taken it down to the bare
bones.
Nearly... but one final trick: if you can create the vectors so that J is sorted ascending then you will greatly improve the speed of the sparse call, about a factor 4 in my experience.
(If it's easier to have I sorted, then create the transpose matrix sparse(J,I,V) and un-transpose it afterwards.)