I want to calculate count of outside points which are not in the circle. But I got this problem. My circle is unit circle.My Error is this : The temporary variable outside will be cleared at the beginning of each iteration
of the parfor loop.
function [ ] = girkoson( N,n )
%UNTİTLED Summary of this function goes here
% Detailed explanation goes here
hold on
outside = 0;
parfor i=0:N
E=ones(N,n);
karekok = sqrt(n);
E = [E, eig(randn(n))/karekok];
a=real(E);
b= imag(E);
plot(a,b,'.r');
if (a>= -1) | (a<=1) | (b>=-1) | (b<=1)
outside = outside +1;
fprintf('%f',outside);
end
end
derece=0:0.01:2*pi;
xp=1*cos(derece);
yp=1*sin(derece);
x=0;y=0;
plot(x+xp,y+yp,'-b');
hold off
end
It looks like you're trying to treat outside as a parfor reduction variable. You cannot access the intermediate values of reduction variables during the loop - you can only perform the reductions. In other words, the line fprintf('%f', outside) is causing the problem, you must remove this for the parfor loop to work.
Also note that workers operating on the body of your parfor loop cannot display graphics to your desktop, so your plot calls will not show anything on-screen. (You can use print to emit graphics to a file if you wish).
Related
I have a main Matlab code which I use in order to calculate ode.
The main code contains many lines but I will explain only the ones needed.
The code looks as following (short version):
global array
t0=0
x0=[5,5]
dt=0.01
tfinal=10
[tout,xout]=rk4('equation',t0,x0,dt,tfinal) % the functions use the function 'equation' and calculates its values at every time from 0 to 20
Where 'equation' is another function from the following type:
global array
Dydx = equation(t,x)
Dydx=[x(1)
x(2)+array(j)]
j=j+1
However, as can be seen, inside the function "equation" there is a variable j which I want to increase every t iteration of rk4.
Which means that while I calculate the ode for time t, the rk4 should use function 'equation' and increase the j by one and so on.
The problem I get is that if I try to make 'equation' gives output of j like [Dydx,j] =equation(t,x) I recieve an error because of rk4.
How can I increase j in the inner function while the outside functions also get the value?(tried global but it kept showing j as [] in the work space).
I have tried to make it global as following but it didnt work:
global array jcount
t0=0
x0=[5,5]
dt=0.01
tfinal=10
jcount = 1;
[tout,xout]=rk4('equation',t0,x0,dt,tfinal)
global array jcount
Dydx = equation(t,x)
Dydx=[x(1)
x(2)+array(jcount)]
jcount=jcount+1
Thank you.
.
total newbie here. I'm having problems looping a function that I've created. I'm having some problems copying the code over but I'll give a general idea of it:
function[X]=Test(A,B,C,D)
other parts of the code
.
.
.
X = linsolve(K,L)
end
where K,L are other matrices I derived from the 4 variables A,B,C,D
The problem is whenever I execute the function Test(1,2,3,4), I can only get one answer out. I'm trying to loop this process for one variable, keep the other 3 variables constant.
For example, I want to get answers for A = 1:10, while B = 2, C = 3, D = 4
I've tried the following method and they did not work:
Function[X] = Test(A,B,C,D)
for A = 1:10
other parts of the code...
X=linsolve(K,L)
end
Whenever I keyed in the command Test(1,2,3,4), it only gave me the output of Test(10,2,3,4)
Then I read somewhere that you have to call the function from somewhere else, so I edited the Test function to be Function[X] = Test(B,C,D) and left A out where it can be assigned in another script eg:
global A
for A = 1:10
Test(A,2,3,4)
end
But this gives an error as well, as Test function requires A to be defined. As such I'm a little lost and can't seem to find any information on how can this be done. Would appreciate all the help I can get.
Cheers guys
I think this is what you're looking for:
A=1:10; B=2; C=3; D=4;
%Do pre-allocation for X according to the dimensions of your output
for iter = 1:length(A)
X(:,:,iter)= Test(A(iter),B,C,D);
end
X
where
function [X]=Test(A,B,C,D)
%other parts of the code
X = linsolve(K,L)
end
Try this:
function X = Test(A,B,C,D)
% allocate output (it is faster than changing the size in every loop)
X = {};
% loop for each position in A
for i = 1:numel(A);
%in the other parts you have to use A(i) instead of just A
... other parts of code
%overwrite the value in X at position i
X{i} = linsolve(K,L);
end
end
and run it with Test(1:10,2,3,4)
To answer what went wrong before:
When you loop with 'for A=1:10' you overwrite the A that was passed to the function (so the function will ignore the A that you passed it) and in each loop you overwrite the X calculated in the previous loop (that is why you can only see the answer for A=10).
The second try should work if you have created a file named Test.m with the function X = (A,B,C,D) as the first code in the file. Although the global assignment is unnecessary. In fact I would strongly recommend you not to use global variables as it gets very messy very fast.
When you are using function in MATLAB you have just the output of the function in the work space and all the other variables that maybe created or used in the body of that function are not shown. I am wondering how does function work? Does it clear all other variables from memory and just save the output?
function acts like a small, isolated programming environment. At the front end you insert your input (e.g. variables, strings, name-value pairs etc). After the function has finished, only the output is available, discarding all temporarily created variables.
function [SUM] = MySum(A)
for ii = 1:length(A)-1
SUM(ii) = A(ii)+A(ii+1);
kk(ii) = ii;
end
end
>> A=1:10
>> MySum(A)
This code just adds two consecutive values for the input array A. Note that the iteration number, stored in kk, is not output and is thus discarded after the function has completed. In MATLAB kk(ii) = ii; will be underlined orange, since it 'might be unused'.
Say you want to also retain kk, just add it to the function outputs:
function [SUM,kk] = MySum(A)
and keep the rest the same.
If you have large variables that you only use up to a certain point and wish them not clogging up your memory whilst the function is running, use clear for that:
function [SUM] = MySum(A)
for ii = 1:length(A)-1
SUM(ii) = A(ii)+A(ii+1);
kk(ii) = ii;
end
clear kk
end
I am having trouble using struct arrays in Matlab's parfor loop. The following code has 2 problems I do not understand:
s=struct('a',{},'b',{});
if matlabpool('size')==0
matlabpool open local 2
end
for j = 1:2
parfor k=1:4
fprintf('[%d,%d]\n',k,j)
s(j,k).a = k;
s(j,k).b = j;
end
end
matlabpool close
It fails with an error Error using parallel_function (line 589)
Insufficient number of outputs from right hand side of equal sign to satisfy assignment.
On output, variable s is a vector, not an array (as it should be, even if the code breaks before finishing).
EDIT the problem is solved if I initialize the struct arrays to the correct size, by:
s=struct('a',cell(2,4),'b',cell(2,4));
However, I would still be happy to get insights about the problem (e.g is it rally a bug, as suggested by Oleg Komarov)
It was originally working fine for me but then I don't know what happens. In general you need to be careful with parfor loops and there are ample documentation on how to align everything. Two different words of advice.
First and more importantly, the parfor loop is on the outside loop:
function s = foo
s=struct('a',{},'b',{});
parfor j = 1:2
for k=1:4
fprintf('[%d,%d]\n',k,j)
s(j,k).a = k;
s(j,k).b = j;
end
end
Two, Matlab gets very picky about writing the main exit variable (i.e. the variable contained in the parfor loop which is indexed to the loop, in your case, s). You first want to create a dummy variable that holds all the innerloop information, and then writes to it once at the end of the loops. Example:
function s = khal
s=struct('a',{},'b',{});
parfor j = 1:2
dummy=struct('a',{},'b',{});
for k=1:4
fprintf('[%d,%d]\n',k,j)
dummy(k).a = k;
dummy(k).b = j;
end
s(j,:) = dummy;
end
You don't have a problem here, but it can get complicated in other instances
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