/edit: The loop doesn't become slower. I didn't take the time correctly. See Rasman's answer.
I'm looping over 3 parameters for a somewhat long and complicated function and I noticed two things that I don't understand:
The execution gets slower with each successive iteration, although the function only returns one struct (of which I only need one field) that I overwrite with each iteration.
The profiler shows that the end statement for the innermost for takes a quite long time.
Consider the following example (I'm aware that this can easily be vectorized, but as far as I understand the function I call can't):
function stuff = doSomething( x, y, z )
stuff.one = x+y+z;
stuff.two = x-y-z;
end
and how I execute the function
n = 50;
i = 0;
currenttoc = 0;
output = zeros(n^3,4);
tic
for x = 1:n
for y = 1:n
for z = 1:n
i = i + 1;
output(i,1) = x;
output(i,2) = y;
output(i,3) = z;
stuff = doSomething(x,y,z);
output(i,4) = stuff.one;
if mod(i,1e4) == 0 % only for demonstration, not in final script
currenttoc = toc - currenttoc;
fprintf(1,'time for last 10000 iterations: %f \n',currenttoc)
end
end
end
end
How can I speed this up? Why does every iteration take longer than the one before? I'm pretty sure this is horrible programming, sorry for that.
When I replace the call to doSomething with output(i,4)=toc;, and I plot diff(output(:,4)), I see that it's the call to fprintf that takes longer and longer every time, apparently.
Removing the if-clause returns to every iteration taking about the same amount of time.
So, the problem gets largely eliminated when I replace the if statement with:
if mod(i,1e4) == 0 % only for demonstration, not in final script
fprintf(1,'time for last 10000 iterations: %f \n',toc); tic;
end
I think the operation on toc may be causing the problem
It's MUCH faster if doSomething returns multiple output variables rather than a struct
function [out1,out2] = doSomething( x, y, z )
out1 = x+y+z;
out2 = x-y-z;
end
The fact that it gets slower on each subsequent iteration is strange and i have no explanation for it but hopefully that gives you some speed up at least.
Related
Recently I wrote this statement
v = arrayfun(#(x) sum(randn(1, 4).^2), zeros(1, 1000000));
to create a new vector and now I'm asking if there exists a function in Matlab to avoid the creation of the unnecessary second vector zeros(1, 1000000). I'm looking for something like
v = FUN(#someInitFunction, [rows, cols]);
without loops, recursion and unnecessary allocation where someInitFunction is given and can't be changed. Does Matlab provide such a function FUN? A simple "No, it doesn't exist" would be a valid answer for me.
To summarize the function FUN: I want to create a new array by calling a function someInitFunction for each element of this new array. The array should be equivalent to
[
someInitFunction() someInitFunction() ...;
someInitFunction() someInitFunction() ...;
.
.
.
someInitFunction() someInitFunction() ...
]
As far as I know there is no builtin function for that. It's relatively easy to create your own however.
You asked a solution without loop but the current solution you are using (arrayfun) uses loop under the hood, and generally coding the same in a properly organised loop is actually faster than arrayfun.
For your case, the function GenArrayFun.m :
function out = GenArrayFun(initFunction , arraySize)
out = zeros(arraySize) ;
for k=1:numel(out)
out(k) = initFunction() ;
end
It has a loop, but no more than arrayfun, and seem to perform twice as fast (at least on my installation, R2016a, win10):
initFunction = #() sum(randn(1, 4).^2) ;
tic
v = arrayfun(#(x) sum(randn(1, 4).^2), zeros(1, 1000000));
toc
tic
out = GenArrayFun( initFunction , [1,1000000] );
toc
Sorry I did not take the time to build a proper timeit benchmark for such a small example, I think the results are significant enough to notice a difference:
Elapsed time is 6.815043 seconds. % arrayfun
Elapsed time is 3.060161 seconds. % GenArrayFun
And just to make sure it evaluate the initFunction for every element:
>> out = GenArrayFun( initFunction , [2,3] )
out =
6.25676106665387 6.52758807745462 2.99236122767462
0.386750258201569 0.566092999842791 2.21158011908878
I want to assign one value to a list of objects in Matlab without using a for-loop (In order to increase efficiency)
Basically this works:
for i=1:Nr_of_Objects
Objectlist(i,1).weight=0.2
end
But I would like something like this:
Objectlist(:,1).weight=0.2
Which is not working. I get this error:
Expected one output from a curly brace or dot indexing expression, but there were 5 results.
Writing an array to the right hand side is also not working.
I`m not very familiar with object oriented programming in Matlab, so I would be happy if someone could help me.
Your looking for the deal function:
S(1,1).a = 1
S(2,1).a = 2
S(1,2).a = 3
[S(:,1).a] = deal(4)
Now S(1,1).a and S(2,1).a equal to 4.
In matlab you can concatenate several output in one array using []. And deal(X) copies the single input to all the requested outputs.
So in your case:
[Objectlist(:,1).weight] = deal(0.2)
Should work.
Noticed that I'm not sure that it will be faster than the for loop since I don't know how the deal function is implemented.
EDIT: Benchmark
n = 1000000;
[S(1:n,1).a] = deal(1);
tic
for ii=1:n
S(ii,1).a = 2;
end
toc
% Elapsed time is 3.481088 seconds
tic
[S(1:n,1).a] = deal(2);
toc
% Elapsed time is 0.472028 seconds
Or with timeit
n = 1000000;
[S(1:n,1).a] = deal(1);
g = #() func1(S,n);
h = #() func2(S,n);
timeit(g)
% ans = 3.67
timeit(h)
% ans = 0.41
function func1(S,n)
for ii=1:n
S(ii,1).a = 2;
end
end
function func2(S,n)
[S(1:n,1).a] = deal(2);
end
So it seems that using the deal function reduce the computational time.
Suppose we are running an infinite for loop in MATLAB, and we want to store the iterative values in a vector. How can we declare the vector without knowing the size of it?
z=??
for i=1:inf
z(i,1)=i;
if(condition)%%condition is met then break out of the loop
break;
end;
end;
Please note first that this is bad practise, and you should preallocate where possible.
That being said, using the end keyword is the best option for extending arrays by a single element:
z = [];
for ii = 1:x
z(end+1, 1) = ii; % Index to the (end+1)th position, extending the array
end
You can also concatenate results from previous iterations, this tends to be slower since you have the assignment variable on both sides of the equals operator
z = [];
for ii = 1:x
z = [z; ii];
end
Sadar commented that directly indexing out of bounds (as other answers are suggesting) is depreciated by MathWorks, I'm not sure on a source for this.
If your condition computation is separate from the output computation, you could get the required size first
k = 0;
while ~condition
condition = true; % evaluate the condition here
k = k + 1;
end
z = zeros( k, 1 ); % now we can pre-allocate
for ii = 1:k
z(ii) = ii; % assign values
end
Depending on your use case you might not know the actual number of iterations and therefore vector elements, but you might know the maximum possible number of iterations. As said before, resizing a vector in each loop iteration could be a real performance bottleneck, you might consider something like this:
maxNumIterations = 12345;
myVector = zeros(maxNumIterations, 1);
for n = 1:maxNumIterations
myVector(n) = someFunctionReturningTheDesiredValue(n);
if(condition)
vecLength = n;
break;
end
end
% Resize the vector to the length that has actually been filled
myVector = myVector(1:vecLength);
By the way, I'd give you the advice to NOT getting used to use i as an index in Matlab programs as this will mask the imaginary unit i. I ran into some nasty bugs in complex calculations inside loops by doing so, so I would advise to just take n or any other letter of your choice as your go-to loop index variable name even if you are not dealing with complex values in your functions ;)
You can just declare an empty matrix with
z = []
This will create a 0x0 matrix which will resize when you write data to it.
In your case it will grow to a vector ix1.
Keep in mind that this is much slower than initializing your vector beforehand with the zeros(dim,dim) function.
So if there is any way to figure out the max value of i you should initialize it withz = zeros(i,1)
cheers,
Simon
You can initialize z to be an empty array, it'll expand automatically during looping ...something like:
z = [];
for i = 1:Inf
z(i) = i;
if (condition)
break;
end
end
However this looks nasty (and throws a warning: Warning: FOR loop index is too large. Truncating to 9223372036854775807), I would do here a while (true) or the condition itself and increment manually.
z = [];
i = 0;
while !condition
i=i+1;
z[i]=i;
end
And/or if your example is really what you need at the end, replace the re-creation of the array with something like:
while !condition
i=i+1;
end
z = 1:i;
As mentioned in various times in this thread the resizing of an array is very processing intensive, and could take a lot of time.
If processing time is not an issue:
Then something like #Wolfie mentioned would be good enough. In each iteration the array length will be increased and that is that:
z = [];
for ii = 1:x
%z = [z; ii];
z(end+1) = ii % Best way
end
If processing time is an issue:
If the processing time is a large factor, and you want it to run as smooth as possible, then you need to preallocating.If you have a rough idea of the maximum number of iterations that will run then you can use #PluginPenguin's suggestion. But there could still be a change of hitting that preset limit, which will break (or severely slow down) the program.
My suggestion:
If your loop is running infinitely until you stop it, you could do occasional resizing. Essentially extending the size as you go, but only doing it once in a while. For example every 100 loops:
z = zeros(100,1);
for i=1:inf
z(i,1)=i;
fprintf("%d,\t%d\n",i,length(z)); % See it working
if i+1 >= length(z) %The array as run out of space
%z = [z; zeros(100,1)]; % Extend this array (note the semi-colon)
z((length(z)+100),1) = 0; % Seems twice as fast as the commented method
end
if(condition)%%condition is met then break out of the loop
break;
end;
end
This means that the loop can run forever, the array will increase with it, but only every once in a while. This means that the processing time hit will be minimal.
Edit:
As #Cris kindly mentioned MATLAB already does what I proposed internally. This makes two of my comments completely wrong. So the best will be to follow what #Wolfie and #Cris said with:
z(end+1) = i
Hope this helps!
I'm looping in parallel and changing a variable if a condition is met. Super idiomatic code that I'm sure everyone has written a hundred times:
trials = 100;
greatest_so_far = 0;
best_result = 0;
for trial_i = 1:trials
[amount, result] = do_work();
if amount > greatest_so_far
greatest_so_far = amount;
best_result = result;
end
end
If I wanted to replace for by parfor, how can I ensure that there aren't race conditions when checking whether we should replace greatest_so_far? Is there a way to lock this variable outside of the check? Perhaps like:
trials = 100;
greatest_so_far = 0;
best_result = 0;
parfor trial_i = 1:trials
[amount, result] = do_work();
somehow_lock(greatest_so_far);
if amount > greatest_so_far
greatest_so_far = amount;
best_result = result;
end
somehow_unlock(greatest_so_far);
end
Skewed answer. It does not exactly solve your problem, but it might help you avoiding it.
If you can afford the memory to store the outputs of your do_work() in some vectors, then you could simply run your parfor on this function only, store the result, then do your scoring at the end (outside of the loop):
amount = zeros( trials , 1 ) ;
result = zeros( trials , 1 ) ;
parfor trial_i = 1:trials
[amount(i), result(i)] = do_work();
end
[ greatest_of_all , greatest_index ] = max(amount) ;
best_result = result(greatest_index) ;
Edit/comment : (wanted to put that in comment of your question but it was too long, sorry).
I am familiar with .net and understand completely your lock/unlock request. I myself tried many attempts to implement a kind of progress indicator for very long parfor loop ... to no avail.
If I understand Matlab classification of variable correctly, the mere fact that you assign greatest_so_far (in greatest_so_far=amount) make Matlab treat it as a temporary variable, which will be cleared and reinitialized at the beginning of every loop iteration (hence unusable for your purpose).
So an easy locked variable may not be a concept we can implement simply at the moment. Some convoluted class event or file writing/checking may do the trick but I am afraid the timing would suffer greatly. If each iteration takes a long time to execute, the overhead might be worth it, but if you use parfoor to accelerate a high number of short execution iterations, then the convoluted solutions would slow you down more than help ...
You can have a look at this stack exchange question, you may find something of interest for your case: Semaphores and locks in MATLAB
The solution from Hoki is the right way to solve the problem as stated. However, as you asked about race conditions and preventing them when loop iterations depend on each other you might want to investigate spmd and the various lab* functions.
You need to use SPMD to do this - SPMD allows communication between the workers. Something like this:
bestResult = -Inf;
bestIndex = NaN;
N = 97;
spmd
% we need to round up the loop range to ensure that each
% worker executes the same number of iterations
loopRange = numlabs * ceil(N / numlabs);
for idx = 1:numlabs:loopRange
if idx <= N
local_result = rand(); % obviously replace this with your actual function
else
local_result = -Inf;
end
% Work out which index has the best result - use a really simple approach
% by concatenating all the results this time from each worker
% allResultsThisTime will be 2-by-numlabs where the first row is all the
% the results this time, and the second row is all the values of idx from this time
allResultsThisTime = gcat([local_result; idx]);
% The best result this time - consider the first row
[bestResultThisTime, labOfBestResult] = max(allResultsThisTime(1, :));
if bestResultThisTime > bestResult
bestResult = bestResultThisTime;
bestIndex = allResultsThisTime(2, labOfBestResult);
end
end
end
disp(bestResult{1})
disp(bestIndex{1})
I am running a parfor loop in Matlab that takes a lot of time and I would like to know how many iterations are left. How can I get that info?
I don't believe you can get that information directly from MATLAB, short of printing something with each iteration and counting these lines by hand.
To see why, recall that each parfor iteration executes in its own workspace: while incrementing a counter within the loop is legal, accessing its "current" value is not (because this value does not really exist until completion of the loop). Furthermore, the parfor construct does not guarantee any particular execution order, so printing the iterator value isn't helpful.
cnt = 0;
parfor i=1:n
cnt = cnt + 1; % legal
disp(cnt); % illegal
disp(i); % legal ofc. but out of order
end
Maybe someone does have a clever workaround, but I think that the independent nature of the parfor iterations belies taking a reliable count. The restrictions mentioned above, plus those on using evalin, etc. support this conclusion.
As #Jonas suggested, you could obtain the iteration count via side effects occurring outside of MATLAB, e.g. creating empty files in a certain directory and counting them. This you can do in MATLAB of course:
fid = fopen(['countingDir/f' num2str(i)],'w');
fclose(fid);
length(dir('countingDir'));
Try this FEX file: http://www.mathworks.com/matlabcentral/fileexchange/32101-progress-monitor--progress-bar--that-works-with-parfor
You can easily modify it to return the iteration number instead of displaying a progress bar.
Something like a progress bar could be done similar to this...
Before the parfor loop :
fprintf('Progress:\n');
fprintf(['\n' repmat('.',1,m) '\n\n']);
And during the loop:
fprintf('\b|\n');
Here we have m is the total number of iterations, the . shows the total number of iterations and | shows the number of iterations completed. The \n makes sure the characters are printed in the parfor loop.
With Matlab 2017a or later you can use a data queue or a pollable data queue to achieve this. Here's the MathWorks documentation example of how to do a progress bar from the first link :
function a = parforWaitbar
D = parallel.pool.DataQueue;
h = waitbar(0, 'Please wait ...');
afterEach(D, #nUpdateWaitbar);
N = 200;
p = 1;
parfor i = 1:N
a(i) = max(abs(eig(rand(400))));
send(D, i);
end
function nUpdateWaitbar(~)
waitbar(p/N, h);
p = p + 1;
end
end
End result :
If you just want to know how much time is left approximately, you can run the program once record the max time and then do this
tStart = tic;
parfor i=1:n
tElapsed = toc(tStart;)
disp(['Time left in min ~ ', num2str( ( tMax - tElapsed ) / 60 ) ]);
...
end
I created a utility to do this:
http://www.mathworks.com/matlabcentral/fileexchange/48705-drdan14-parforprogress