I am trying to using a while loop inside a for loop in Matlab. The while loop will repeat the same action until it satifies some criteria. The outcome from the while loop is one iteration in the for loop. I am having a problem to get that correctly.
n=100;
for i=1:n
while b<0.5
x(i)=rand;
b=x(i);
end
end
I am not sure what i am doing wrongly.
Thanks
Approach the problem differently. There's no need to try again if rand doesn't give you the value you want. Just scale the result of rand to be in the range you want. This should do it:
x = 0.5 + 0.5*rand(1, 100);
With the example you showed, you have to initialize b or the while-statement cannot be evaluated when it is first called.
Do it inside the for-loop to avoid false positives after the first for-iteration:
n=100;
for ii=1:n
b = 0;
while b<0.5
x(ii)=rand;
b=x(ii);
end
end
Or, without b:
n=100;
x = zeros(1,100);
for ii=1:n
while x(ii)<0.5
x(ii)=rand;
end
end
Related
This is my attempt of a simple example (it seems pointless) but the idea is bigger than this simple code.
During a for loop, if something happens, I want to skip this step of the for loop and then add an extra step on the end.
I am trying to create a list of numbers, that do not include the number 8.
If the code creates an 8, this will mean that exitflag is equal to 1.
Can I adapt this program so that if the exitflag=1, the it will remove that result and add another loop.
The code:
for i = 1:1000
j = 1+round(rand*10)
if j == 8
exitflag = 1
else
exitflag = 0
end
storeexit(i)=exitflag;
storej(i)=j;
end
sum(storeexit)
I would ideally like a list of numbers, 1000 long which does not contain an 8.
If what you want to do is 1000 iterations of the loop, but repeat a loop iteration if you don't like its result, instead of tagging the repeat at the end, what you can do is loop inside the for loop until you do like the result of that iteration:
stores = zeros(1000,1); % Note that it is important to preallocate arrays, even in toy examples :)
for i = 1:1000
success = false; % MATLAB has no do..while loop, this is slightly more awkward....
while ~success
j = 1+round(rand*10);
success = j ~= 8;
end
storej(i) = j; % j guaranteed to not be 8
end
No.
With for loop, the number of loops is determined on the loop start and it is not dynamic.
For doing what you want, you need to use a while loop.
I am trying to force a for loop to restart if a condition is not stratified. I can do it with while since I want the loop to run for a certain number of iterations. I tried to set iter=iter-1 inside the if statement but it did not work. Any suggestions?
R=2*10^3;
lamda= 0.00001;
h=100;
a = 9.6117;
b = 0.1581;
for iter=1:10
M=poissrnd(lamda*R^2);
xx=R*rand(1,M);
yy=R*rand(1,M);
zz=ones(1,M)*h;
BS=[xx' yy' zz'];
user=[0,0, 0];
s=pdist2(BS(:,1:2),user(1,1:2));
anga=atand(h./s);
PL=1./(1+(a*exp(b*(a-anga))));
berRV=binornd(1,PL);
if berRV(1)==1
% do something
else
% repeat
end
end
You can accomplish this by using a while loop, with a comparison to whether the number of needed results have been identified. See the comment about saving your found values, as you hadn't specified what needs to be done when the condition you're searching for is satisfied.
R=2*10^3;
lamda= 0.00001;
h=100;
a = 9.6117;
b = 0.1581;
total_results_found = 0;
needed_results_found = 10;
while total_results_found < needed_results_found
M=poissrnd(lamda*R^2);
xx=R*rand(1,M);
yy=R*rand(1,M);
zz=ones(1,M)*h;
BS=[xx' yy' zz'];
user=[0,0, 0];
s=pdist2(BS(:,1:2),user(1,1:2));
anga=atand(h./s);
PL=1./(1+(a*exp(b*(a-anga))));
berRV=binornd(1,PL);
if berRV(1)==1
% save the result here
% iterate the counter
total_results_found = total_results_found + 1;
end
end
The simplest approach here would be with a while loop inside the for loop:
for iter=1:10
berRV(1) = 0
while berRV(1)~=1
% original loop code here
end
% do something
end
[Sadly, MATLAB does not have a do...while loop, it would make the above a little cleaner.]
While the other two solutions are perfectly valid, I wanted to give you another solution, which I think is the closest to the logic you provided in the question.
for iter=1:10
while 1
% loop code here
if berRV(1) == 1
break
end
end
end
The idea is similar to the one Cris presents, namely that you repeat the body of the for-loop until some condition is met. The difference lies solely in how you terminate the while loop,
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!
In MATLAB, I have a variable proba and I have a parfor loop as showing below:
parfor f = 1:N
proba = (1/M)*ones(1, M);
% rest of the code
end
pi_proba = proba;
MATLAB said that: "The temporary variable 'proba' is used after the PARFOR loop, but its value is nondeterministic"
I do not understand how to correct this error. I need to use a parallel loop and I need proba after the loop. How to do this?
When using parfor the classes are classified according to these categories. Make sure every variable matches one of these categories. For non-writing access to proba a Broadcast-Variable would be the best choice:
proba = (1/M)*ones(1, M);
parfor f = 1:N
% rest of the code
end
pi_proba = proba;
In case of writing access within the loop, a sliced variable is nessecary:
proba=cell(1,N)
parfor f = 1:N
%now use proba{f} inside the loop
proba{f}=(1/M)*ones(1, M);
% rest of the code
end
%get proba from whatever iteration you want
pi_proba = proba{N};
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