I have a code that actually works but the output is supposed to be multiplied together. Also, the time of execution was not in milliseconds. This is the Matlab code
function product = prod(A)
tic;
A=input('matrix A =');
[rows, columns] = size(A);
for i=1:rows
prod=A(i,i)*A(i,end)
end
seconds=toc
For instance if given A=[1,2,3;4,5,6;7,8,9] when i=1 we have 1*3=3 when i=2 we have 2*6=12 when i=3 we have 9*9=81.
The output I want should be 3*12*81=2916 (the product of the values above) and the time of execution should be milliseconds.
When I extend the solution you proffer for the code below, it does not multiply them together and the time elapsed seems to be big
When the above code is used for the matrix A = [2,3,6,4;2,1,7,-2;6,8,1,-3;5,3,4,1].
My output is
out = -22, -23, -7, -3
Elapsed time is 10.200446 seconds.
I want all the output to multiple each other so that it will be
out = -22*-23*-7*-3 = 10626
Well, it took me some time to decipher what you have been trying to do, but finally, I was able to reconstruct your output. So first, there is no need for the prod function, it only makes things messy.
Here is a simple function that does the trick:
function out = testValue
A=input('matrix A =');
out=1;
if mod(size(A),2)==0
for i = 1:length(A)
out = out*(A(i,i)*A(i,end)-A(i,end-1)*A(end,i));
end
else
for i = 1:(length(A)-1)/2
out = out*(A(i,i)*A(i,end)-A(i,end-1)*A(end,i));
end
end
end
Some remarks on the code above:
No need for the columns variable
I omitted the tic toc - you can add them back if needed
No use for the prod function, instead the input for prod is simply what you consider as its output.
You have to initialize out=1 for a cumulative product
I'm sure there are cleverer ways to do this, but as #beaker wrote it is not clear what exactly is the purpose of this code.
Hope this helps :)
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.
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.
I run a lot of long simulations in Matlab, typically taking from a couple of minutes to a couple of hours, so to speed things up I decided to run the simulations simultaneously using a parfor loop.
arglist = [arg1, arg2, arg3, arg4];
parfor ii = 1:size(arglist, 2)
myfun(arglist(ii));
end
Everything worked just fine, except for one thing: the progress printing. Since each simulation takes a lot of time, I usually print progress using something like
prevlength = 0;
for ii = 1:tot_iter
% Some calculations here
msg = sprintf('Working on %d of %d, %.2f percent done', ii, tot_iter, ii/tot_iter);
fprintf(repmat('\b', 1, prevlength))
fprintf(msg);
prevlength = numel(msg);
end
but, as could be expected, when doing this inside a parfor loop, you get chaos.
I have googled a lot in search of a solution and have found a bunch of "parfor progress printers" like this one. However, all of them print the progress of the entire parfor loop instead of showing how far each of the individual iterations have come. Since I only have about 4-8 iterations in the parfor loop, but each iteration takes about an hour, this approach isn't very helpful to me.
The ideal solution for me would be something that looks like this
Working on 127 of 10000, 1.27 percent done
Working on 259 of 10000, 2.59 percent done
Working on 3895 of 10000, 38.95 percent done
Working on 1347 of 10000, 13.47 percent done
that is, each simulation gets one line showing how far it has run. I'm not sure though if this is possible at all, I can at least not imagine any way to do this.
Another way would be to do something like this
Sim 1: 1.27% Sim 2: 2.59% Sim 3: 38.95% Sim 4: 13.47%
that is, show all the progresses on the same line. To do this, you would need to keep track of what position on the line each simulation is to write on and write there, without erasing the other progresses. I can't figure out how this would be done, is this possible to do?
If there is some other solution to my problem (showing the progress of each individual iteration) that I haven't thought of, I'd be happy to hear about it.
Since this is the first time I ask a question here on SO it is quite possible that there is something that I have missed; if so, please feel free to comment below.
Edit
After receiving this answer, I thought that I should share how I used it to solve my problem since I didn't use it exactly as in the answer, in case someone else encounters the same problem.
Here is a small test program with basically the same structure as my program, making use of the progress bar (parfor_progress) mentioned in the answer:
function parfor_progress_test()
cpus = feature('numCores');
matlabpool('open', cpus);
cleaner = onCleanup(#mycleaner);
args = [1000, 1000, 1000, 1000];
m = sum(args);
parfor_progress(m);
parfor ii = 1:size(args,2)
my_fun(args(ii));
end
parfor_progress(0);
end
function my_fun(N)
for ii = 1:N
pause(rand*0.01);
parfor_progress;
end
end
function mycleaner
matlabpool close;
fclose all;
end
Simple 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.
Progress Bar and Percentage Completion
Otherwise you could try this: http://www.mathworks.com/matlabcentral/fileexchange/32101-progress-monitor--progress-bar--that-works-with-parfor
It will display a progress bar and percentage completion but can be easily modified to just include the percentage completion or progress bar.
This function amends a character to a file on every iteration and then reads the number of characters written to that file which indicates the number of iterations completed. This file accessing method is allowed in parfor's.
Assuming you correctly add the above to your MATLAB path somehow, you can then use the following:
arglist = [arg1, arg2, arg3, arg4];
parfor_progress(size(arglist, 2)); % Set the total number of iterations
parfor ii = 1:size(arglist, 2)
myfun(arglist(ii));
parfor_progress; % Increment the progress counter
end
parfor_progress(0); % Reset the progress counter
Time to Completion and Percentage Completion
There is also a function called showTimeToCompletion() which is available from: https://www.soundzones.com/software/sound-zone-tools/
and works alongside parfor_progress. This function allows you to print a detailed summary of the progress of a parfor loop (or any loop for that matter) which contains the start time, length of time running, estimated finish time and percentage completion. It makes smart use of the \b (backspace) character so that the command window isn't flooded with text. Although not strictly a progress 'bar' it is perhaps more informative.
The third example in the header of the function file,
fprintf('\t Completion: ');
showTimeToCompletion; startTime=tic;
len=1e2;
p = parfor_progress( len );
parfor i = 1:len
pause(1);
p = parfor_progress;
showTimeToCompletion( p/100, [], [], startTime );
end
outputs the following to the command window:
Completion: 31.00%
Remaining: 00:00:23
Total: 00:00:33
Expected Finish: 3:30:07PM 14-Nov-2017
Useful for estimating the completion of a running simulation, especially one that may take hours or days.
Starting in R2013b, you can use PARFEVAL to evaluate your function asynchronously and have the client display progress updates. (Obviously this approach is not quite as simple as adding stuff to your PARFOR loop). There's an example here.
The Diary property of the Future returned by PARFEVAL is updated continuously while processing, so that might also be useful if you have a small number of large tasks.
Starting in R2017a, you can use parallel.pool.DataQueue and afterEach to implement a waitbar for parfor, like so:
if isempty(gcp('nocreate'))
parpool('local', 3);
end
dq = parallel.pool.DataQueue;
N = 10;
wb = waitbar(0, 'Please wait...');
% Use the waitbar's UserData to track progress
wb.UserData = [0 N];
afterEach(dq, #(varargin) iIncrementWaitbar(wb));
afterEach(dq, #(idx) fprintf('Completed iteration: %d\n', idx));
parfor idx = 1:N
pause(rand());
send(dq, idx);
end
close(wb);
function iIncrementWaitbar(wb)
ud = wb.UserData;
ud(1) = ud(1) + 1;
waitbar(ud(1) / ud(2), wb);
wb.UserData = ud;
end
after exploring #Edric answer I found that there is an example in the Matlab documentation that exactly implements a waitbar for a pareval loop. Check help FetchNext
N = 100;
for idx = N:-1:1
% Compute the rank of N magic squares
F(idx) = parfeval(#rank, 1, magic(idx));
end
% Build a waitbar to track progress
h = waitbar(0, 'Waiting for FevalFutures to complete...');
results = zeros(1, N);
for idx = 1:N
[completedIdx, thisResult] = fetchNext(F);
% store the result
results(completedIdx) = thisResult;
% update waitbar
waitbar(idx/N, h, sprintf('Latest result: %d', thisResult));
end
% Clean up
delete(h)
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