I'm going to write a program in MATLAB that takes a function, sets the value D from 10 to 100 (the for loop), integrates the function with Simpson's rule (the while loop) and then displays it. Now, this works fine for the first 7-8 values, but then it takes longer time and eventually I run out of memory, and I don't understand the reason for this. This is the code so far:
global D;
s=200;
tolerance = 9*10^(-5);
for D=10:1:100
r = Simpson(#f,0,D,s);
error = 1;
while(error>tolerance)
s = 2*s;
error = (1/15)*(Simpson(#f,0,D,s)-r);
r = Simpson(#f,0,D,s);
end
clear error;
disp(r)
end
mtrw's comment probably already answers the question in part: s should be reinitialized inside the for loop. The posted code results in s increasing irreversibly every time the error was too large, so for larger values of D the largest s so far will be used.
Additionally, since the code re-evaluates the entire integration instead of reusing the previous integration from [0, D-1] you waste lots of resources unless you want to explicitly show the error tolerance of your Simpson function - s will have to increase a lot for large D to maintain the same low error (since you integrate over a larger range you have to sum up more points).
Finally, your implementation of Simpson could of course do funny stuff as well, which no one can tell without seeing it...
Related
So, I've recently started using Matlab's built-in profiler on a regular basis, and I've noticed that while its usually great at showing which lines are taking up the most time, sometimes it'll tell me a large chunk of time is being used on the end statement of a for loop.
Now, seeing as such a line is just used for denoting the end of the loop, I can't imagine how it could use anything other than a trivial amount of processing.
I've seen a specific version of this question asked on matlab central, but a consensus didn't seem to be reached.
EDIT: Here's a minimal example of this problem:
for i =1:1000
x = 1;
x = [x 1];
% clear x;
end
Even if you uncomment the clear, the end line still takes up a lot of computation (about 20%), and the clear actually increases the absolute amount of computation performed by the end line.
When I've seen this in my code, it's been the deallocation of large temporaries created in the loop. Each new variable created in the loop is deallocated at the end.
I wrote a MATLAB code for finding seismic signal (ex. P wave) from SAC(seismic) file (which is read via another code). This algorithm is called STA/LTA trigger algorithm (actually not that important for my question)
Important thing is that actually this code works well, but since my seismic file is too big (1GB, which is for two months), it takes almost 40 minutes for executing to see the result. Thus, I feel the need to optimize the code.
I heard that replacing loops with advanced functions would help, but since I am a novice in MATLAB, I cannot get an idea about how to do it, since the purpose of code is scan through the every time series.
Also, I heard that preallocation might help, but I have mere idea about how to actually do this.
Since this code is about seismology, it might be hard to understand, but my notes at the top might help. I hope I can get useful advice here.
Following is my code.
function[pstime]=classic_LR(tseries,ltw,stw,thresh,dt)
% This is the code for "Classic LR" algorithm
% 'ns' is the number of measurement in STW-used to calculate STA
% 'nl' is the number of measurement in LTW-used to calculate LTA
% 'dt' is the time gap between measurements i.e. 0.008s for HHZ and 0.02s for BHZ
% 'ltw' and 'stw' are long and short time windows respectively
% 'lta' and 'sta' are long and short time windows average respectively
% 'sra' is the ratio between 'sta' and 'lta' which will be each component
% for a vector containing the ratio for each measurement point 'i'
% Index 'i' denotes each measurement point and this will be converted to actual time
nl=fix(ltw/dt);
ns=fix(stw/dt);
nt=length(tseries);
aseries=abs(detrend(tseries));
sra=zeros(1,nt);
for i=1:nt-ns
if i>nl
lta=mean(aseries(i-nl:i));
sta=mean(aseries(i:i+ns));
sra(i)=sta/lta;
else
sra(i)=0;
end
end
[k]=find(sra>thresh);
if ~isempty(k)
pstime=k*dt;
else
pstime=0;
end
return;
If you have MATLAB 2016a or later, you can use movmean instead of your loop (this means you also don't need to preallocate anything):
lta = movmean(aseries(1:nt-ns),nl+1,'Endpoints','discard');
sta = movmean(aseries(nl+1:end),ns+1,'Endpoints','discard');
sra = sta./lta;
The only difference here is that you will get sra with no leading and trailing zeros. This is most likely to be the fastest way. If for instance, aseries is 'only' 8 MB than this method takes less than 0.02 second while the original method takes almost 6 seconds!
However, even if you don't have Matlab 2016a, considering your loop, you can still do the following:
Remove the else statement - sta(i) is already zero from the preallocating.
Start the loop from nl+1, instead of checking when i is greater than nl.
So your new loop will be:
for i=nl+1:nt-ns
lta = mean(aseries(i-nl:i));
sta = mean(aseries(i:i+ns));
sra(i)=sta/lta;
end
But it won't be so faster.
I run a Matlab code with an iteration loop, and during each iteration it sample random numbers and uses the function intlinprog. My issue is that, due to the large amount of data I provide to the intlinprog function and to the stochastic values I assign to part of its variables, some of the iterations take a really long time.
My code is more or less like this:
rounds = 1E3;
Total_PF = zeros(rounds,4893);
for i=1:rounds
i
cT = zeros (4894,1);
cT(4894,1) = 1;
xint = linspace(1,4893,4893);
xint = xint';
AT = rand(4,4894);
beT = ones(4,1);
lb = zeros(4894,1);
ub = ones (4894,1);
ub(4894,1) = Inf;
[x] = intlinprog(cT,xint,AT,beT,[],[],lb,ub);
Total_PF(i,:)= (x(1:length(x)-1)');
end
Now in the minimal working example I provided, all the iterations are quite fast, but in my real code, sometimes intlinprog takes really long time ( I mean hours) to do a single iteration.
Therefore, I was wondering: is there a way to break the intlinprog while the intlinprog line is being executed? I was thinking that it may be done by modifying the matlab function but first of all I do not know if I am allowed to do it, secondly I am afraid that may be very dangerous.
This is very difficult to do effectively.
You could try using a timer object to watch the value. However, since you are using inherent Matlab functions and not executing external functions, you could set a time limit value and utilize tic and toc in a while loop while you execute the intlinprog and checking the value of toc against your time limit and break your code if toc exceeds the limit.
I am running a very large meta-simulation where I go through two hyperparameters (lets say x and y) and for each set of hyperparameters (x_i & y_j) I run a modest sized subsimulation. Thus:
for x=1:I
for y=1:j
subsimulation(x,y)
end
end
For each subsimulation however, about 50% of the data is common to every other subsimulation, or subsimulation(x_1,y_1).commondata=subsimulation(x_2,y_2).commondata.
This is very relevant since so far the total simulation results file size is ~10Gb! Obviously, I want to save the common subsimulation data 1 time to save space. However, the obvious solution, being to save it in one place would screw up my plotting function, since it directly calls subsimulation(x,y).commondata.
I was wondering whether I could do something like
subsimulation(x,y).commondata=% pointer to 1 location in memory %
If that cant work, what about this less elegant solution:
subsimulation(x,y).commondata='variable name' %string
and then adding
if(~isstruct(subsimulation(x,y).commondata)),
subsimulation(x,y).commondata=eval(subsimulation(x,y).commondata)
end
What solution do you guys think is best?
Thanks
DankMasterDan
You could do this fairly easily by defining a handle class. See also the documentation.
An example:
classdef SimulationCommonData < handle
properties
someData
end
methods
function this = SimulationCommonData(someData)
% Constructor
this.someData = someData;
end
end
end
Then use like this,
commonData = SimulationCommonData(something);
subsimulation(x, y).commondata = commonData;
subsimulation(x, y+1).commondata = commonData;
% These now point to the same reference (handle)
As per my comment, as long as you do not modify the common data, you can pass it as third input and still not copy the array in memory on each iteration (a very good read is Internal Matlab memory optimizations). This image will clarify:
As you can see, the first jump in memory is due to the creation of common and the second one to the allocation of the output c. If the data were copied on each iteration, you would have seen many more memory fluctuations. For instance, a third jump, then a decrease, then back up again and so on...
Follows the code (I added a pause in between each iteration to make it clearer that no big jumps occur during the loop):
function out = foo(a,b,common)
out = a+b+common;
end
for ii = 1:10; c = foo(ii,ii+1,common); pause(2); end
I have a program which I copied from a textbook, and which times the difference in program execution runtime when calculating the same thing with uninitialized, initialized array and vectors.
However, although the program runs somewhat as expected, if running several times every once in a while it will give out a crazy result. See below for program and an example of crazy result.
clear all; clc;
% Purpose:
% This program calculates the time required to calculate the squares of
% all integers from 1 to 10000 in three different ways:
% 1. using a for loop with an uninitialized output array
% 2. Using a for loop with a pre-allocated output array
% 3. Using vectors
% PERFORM CALCULATION WITH AN UNINITIALIZED ARRAY
% (done only once because it is so slow)
maxcount = 1;
tic;
for jj = 1:maxcount
clear square
for ii = 1:10000
square(ii) = ii^2;
end
end
average1 = (toc)/maxcount;
% PERFORM CALCULATION WITH A PRE-ALLOCATED ARRAY
% (averaged over 10 loops)
maxcount = 10;
tic;
for jj = 1:maxcount
clear square
square = zeros(1,10000);
for ii = 1:10000
square(ii) = ii^2;
end
end
average2 = (toc)/maxcount;
% PERFORM CALCULATION WITH VECTORS
% (averaged over 100 executions)
maxcount = 100;
tic;
for jj = 1:maxcount
clear square
ii = 1:10000;
square = ii.^2;
end
average3 = (toc)/maxcount;
% Display results
fprintf('Loop / uninitialized array = %8.6f\n', average1)
fprintf('Loop / initialized array = %8.6f\n', average2)
fprintf('Vectorized = %8.6f\n', average3)
Result - normal:
Loop / uninitialized array = 0.195286
Loop / initialized array = 0.000339
Vectorized = 0.000079
Result - crazy:
Loop / uninitialized array = 0.203350
Loop / initialized array = 973258065.680879
Vectorized = 0.000102
Why is this happening ?
(sometimes the crazy number is on vectorized, sometimes on loop initialized)
Where did MATLAB "find" that number?
That is indeed crazy. Don't know what could cause it, and was unable to reproduce on my own Matlab R2010a copy over several runs, invoked by name or via F5.
Here's an idea for debugging it.
When using tic/toc inside a script or function, use the "tstart = tic" form that captures the output. This makes it safe to use nested tic/toc calls (e.g. inside called functions), and lets you hold on to multiple start and elapsed times and examine them programmatically.
t0 = tic;
% ... do some work ...
te = toc(t0); % "te" for "time elapsed"
You can use different "t0_label" suffixes for each of the tic and toc returns, or store them in a vector, so you preserve them until the end of your script.
t0_uninit = tic;
% ... do the uninitialized-array test ...
te_uninit = toc(t0_uninit);
t0_prealloc = tic;
% ... test the preallocated array ...
te_prealloc = toc(t0_prealloc);
Have the script break in to the debugger when it finds one of the large values.
if any([te_uninit te_prealloc te_vector] > 5)
keyboard
end
Then you can examine the workspace and the return values from tic, which might provide some clues.
EDIT: You could also try testing tic() on its own to see if there's something odd with your system clock, or whatever tic/toc is calling. tic()'s return value looks like a native timestamp of some sort. Try calling it many times in a row and comparing the subsequent values. If it ever goes backwards, that would be surprising.
function test_tic
t0 = tic;
for i = 1:1000000
t1 = tic;
if t1 <= t0
fprintf('tic went backwards: %s to %s\n', num2str(t0), num2str(t1));
end
t0 = t1;
end
On Matlab R2010b (prerelease), which has int64 math, you can reproduce a similar ridiculous toc result by jiggering the reference tic value to be "in the future". Looks like an int rollover effect, as suggested by gary comtois.
>> t0 = tic; toc(t0+999999)
Elapsed time is 6148914691.236258 seconds.
This suggests that if there were some jitter in the timer that toc were using, you might get rollover if it occurs while you're timing very short operations. (I assume toc() internally does something like tic() to get a value to compare the input to.) Increasing the number of iterations could make the effect go away because a small amount of clock jitter would be less significant as part of longer tic/toc periods. Would also explain why you don't see this in your non-preallocated test, which takes longer.
UPDATE: I was able to reproduce this behavior. I was working on some unrelated code and found that on one particular desktop with a CPU model we haven't used before, a Core 2 Q8400 2.66GHz quad core, tic was giving inaccurate results. Looks like a system-dependent bug in tic/toc.
On this particular machine, tic/toc will regularly report bizarrely high values like yours.
>> for i = 1:50000; t0 = tic; te = toc(t0); if te > 1; fprintf('elapsed: %.9f\n', te); end; end
elapsed: 6934787980.471930500
elapsed: 6934787980.471931500
elapsed: 6934787980.471899000
>> for i = 1:50000; t0 = tic; te = toc(t0); if te > 1; fprintf('elapsed: %.9f\n', te); end; end
>> for i = 1:50000; t0 = tic; te = toc(t0); if te > 1; fprintf('elapsed: %.9f\n', te); end; end
elapsed: 6934787980.471928600
elapsed: 6934787980.471913300
>>
It goes past that. On this machine, tic/toc will regularly under-report elapsed time for operations, especially for low CPU usage tasks.
>> t0 = tic; c0 = clock; pause(4); toc(t0); fprintf('Wall time is %.6f seconds.\n', etime(clock, c0));
Elapsed time is 0.183467 seconds.
Wall time is 4.000000 seconds.
So it looks like this is a bug in tic/toc that is related to particular CPU models (or something else specific to the system configuration). I've reported the bug to MathWorks.
This means that tic/toc is probably giving you inaccurate results even when it doesn't produce those insanely large numbers. As a workaround, on this machine, use etime() instead, and time only longer chunks of work to compensate for etime's lower resolution. You could wrap it in your own tick/tock functions that use the for i=1:50000 test to detect when tic is broken on the current machine, use tic/toc normally, and have them warn and fall back to using etime() on broken-tic systems.
UPDATE 2012-03-28: I've seen this in the wild for a while now, and it's highly likely due to an interaction with the CPU's high resolution performance timer and speed scaling, and (on Windows) QueryPerformanceCounter, as described here: http://support.microsoft.com/kb/895980/. It is not a bug in tic/toc, the issue is in the OS features that tic/toc is calling. Setting a boot parameter can work around it.
Here's my theory about what might be happening, based on these two pieces of data I found:
There is a function maxNumCompThreads which controls the maximum number of computational threads used by MATLAB to perform tasks. Quoting the documentation:
By default, MATLAB makes use of the
multithreading capabilities of the
computer on which it is running.
Which leads me to think that perhaps multiple copies of your script are running at the same time.
This newsgroup thread discusses a bug in an older version of MATLAB (R14) "in the way that MATLAB accelerates M-code with global structure variables", which it appears the TIC/TOC functions may use. The solution there was to disable the accelerator using the undocumented FEATURE function:
feature accel off
Putting these two things together, I'm wondering if the multiple versions of your script that are running in the workspace may be simultaneously resetting global variables used by the TIC/TOC functions and screwing one another up. Maybe this isn't a problem when converting your script to a function as Amro did since this would separate the workspaces that the two programs are running in (i.e. they wouldn't both be running in the main workspace).
This could also explain the exceedingly large numbers you get. As gary and Andrew have pointed out, these numbers appear to be due to an integer roll-over effect (i.e. an integer overflow) whereby the starting time (from TIC) is larger than the ending time (from TOC). This would result in a huge number that is still positive because TIC/TOC are internally using unsigned 64-bit integers as time measures. Consider the following possible scenario with two scripts running at the same time on different threads:
The first thread calls TIC, initializing a global variable to a starting time measure (i.e. the current time).
The first thread then calls TOC, and the immediate action the TOC function is likely to make is to get the current time measure.
The second thread calls TIC, resetting the global starting time measure to the current time, which is later than the time just measured by the TOC function for the first thread.
The TOC function for the first thread accesses the global starting time measure to get the difference between it and the measure it previously took. This difference would result in a negative number, except that the time measures are unsigned integers. This results in integer overflow, giving a huge positive number for the time difference.
So, how might you avoid this problem? Changing your scripts to functions like Amro did is probably the best choice, as that seems to circumvent the problem and keeps the workspace from becoming cluttered. An alternative work-around you could try is to set the maximum number of computational threads to one:
maxNumCompThreads(1);
This should keep multiple copies of your script from running at the same time in the main workspace.
There are at least two possible error sources. Can you try to differentiate between 'tic/toc' and 'fprintf' by just looking at the computed values without formatting them.
I don't understand the braces around 'toc' but they shouldn't do any harm.
Here is a hypothesis which is testable. Matlab's tic()/toc() have to be using some high-resolution timer. On Windows, because their return value looks like clock cycles, I think they're using the Win32 QueryPerformanceCounter() call, or maybe something else hitting the CPU's RDTSC time stamp counter. These apparently have glitches on some multiprocessor systems, mentioned in the linked articles. Perhaps your machine is one of those, getting different results if the Matlab process is moved from core to core by the process scheduler.
http://msdn.microsoft.com/en-us/library/ms644904(VS.85).aspx
http://www.virtualdub.org/blog/pivot/entry.php?id=106
This would be hardware and system configuration dependent, which would explain why other posters haven't been able to reproduce it.
Try using Windows Task Manager to set the affinity on your Matlab.exe process to a single CPU. (On the Processes tab, right-click MATLAB.exe, "Set affinity...", un-check all but CPU 0.) If the crazy timing goes away while affinity is set, looks like you found the cause.
Regardless, the workaround looks like to just increase maxcount so you're timing longer pieces of work, and the noise you're apparently getting in tic()/toc() is small compared to the measured value. (You don't want to have to muck around with CPU affinity; Matlab is supposed to be easy to run.) If there's a problem in there that's causing int overflow, the other small positive numbers are a bit suspect too. Besides, hi-res timing in a high level language like Matlab is a bit problematic. Timing workloads down to a couple hundred microseconds subjects them to noise from other transient conditions in your machine's state.