options = optimset('Display','iter','MaxIter',3,'OutputFcn',#outfun);
[x,fval,~,output] = fminsearch(#(param) esm6(param,identi),result(k,1:end-1),options);
This code will find the local Minimum of my esm6 function and due to the 'Display' Option it will Output strings like this
Iteration Func-count min f(x) Procedure
0 1 36.9193
1 5 35.9815 initial simplex
2 7 35.4924 contract inside
3 9 35.4924 contract inside
4 11 33.0085 expand
So in the command window, i get the function Count for each Iteration step. The structure output, which is created by fminsearch has only the total amount of func-count in it. Is there a way to receive all the Information, that is outputed in the command window also in the Output-structure?
EDIT:
I think i'm pretty Close to the solution. I wrote this outputfunction:
function stop = outfun(x,optimvalues,state);
stop = false;
if state == 'iter'
history = evalin('base','history');
history = [history; optimvalues.iteration optimvalues.funcCount];
assignin('base','history',history);
end
end
due to http://de.mathworks.com/help/matlab/math/output-functions.html this should work, but in fact, matlab tells me,
??? Reference to non-existent field 'funcCount'.
any idea, why this happens?
Related
I have a program with 2 timer functions like this (p_maj_input and p_maj_output_motion are equal so the period of both timer is the same):
maj_input = timer('ExecutionMode','fixedRate','Period',p_maj_input,'TasksToExecute', ...
floor(Tmeasurement/p_maj_input)-1,'TimerFcn',{#time_append,p_maj_input,HR_all,freq_HR, ...
BVP_all,freq_BVP,TEMP_all,freq_TEMP,ACC_x_all,ACC_y_all,ACC_z_all,freq_ACC,EDA_all,freq_EDA, ...
folder_all,all_dir,num_dir}); start(maj_input);
maj_output_motion=timer('ExecutionMode','fixedRate','Period',p_maj_output_motion,'TasksToExecute', ...
floor(Tmeasurement/p_maj_output_motion)-1,'TimerFcn',{#output_motion_append, ...
freq_MOTION,freq_mvt_score,freq_walk,p_maj_output_motion,p_maj_output_walk,folder_all,all_dir,num_dir});%,'StartDelay',min(5,p_maj_output_motion)); %startDelay must be min 5 for walk detection start(maj_output_motion);
In each timer callback function there is a loop over subfolders contained in a folder that is selected at the beginning of the program.
output_motion_append(obj,event,freq_MOTION,freq_mvt_score,freq_walk,p_maj_output_motion,p_maj_output_walk,folder_all,all_dir,num_dir)
fprintf('motion %d\n',obj.TasksExecuted)
toc
for folder_index=1:num_dir
[folder_original,folder_fictive] = subfolderFunction(all_dir, folder_all, folder_index);
fileName=strcat(folder_fictive,'\ACC.csv');
[ACC_time, ACC_x, ACC_y, ACC_z] = loadValuesE4_acc(fileName);
% Motion Amount
[agitation,agitation_ts] = identifyAgitation(ACC_x,ACC_y,ACC_z,ACC_time);
agitation=agitation';
len1=length(agitation);
if len1<=freq_MOTION*p_maj_output_motion
i_init1=1;
elseif len1>freq_MOTION*p_maj_output_motion
i_init1=len1-freq_MOTION*p_maj_output_motion+1;
end
writematrix([agitation(i_init1:len1)],strcat(folder_fictive,'\MOTION_output.csv'),'WriteMode','Append');
writematrix([mvt_score(i_init2:len2)],strcat(folder_fictive,'\neurologicScore_output.csv'),'WriteMode','Append');
end
end
Everything works fine if the number of subfolders is lower than 4 : the good values appear in the files on which is carried out the writematrix function. And the timer callback function are are called one after the other, so both timer work simultaneously.
However if there are 5 subfolders or more, it is not the good values and using the debugging I noticed that the first callback function is triggered the number of 'TasksToExecute', and then only the second callback function seems to be called. That is to say the 2 timers don't work simultaneously.
I have tried to increase p_maj_input and p_maj_output_motion to see if the problem is that matlab can't finish to run before another timer callback function is called but still for 5 subfolders I get the same problem.
Does anyone know where my problem is coming from?
This behavior occurs because one timer hasn't finished executing by the time it triggers again, so the second timer never has a chance to execute until the first timer is finished. If you change the ExecutionMode from 'fixedRate' to 'fixedSpacing', then you'll guarantee that there's time for the second timer to execute.
function duelingTimers()
disp('With ExecutionMode = fixedRate')
t1 = timer('ExecutionMode','fixedRate','Period',0.1,'TasksToExecute',5,'TimerFcn',#timerFcn1);
t2 = timer('ExecutionMode','fixedRate','Period',0.1,'TasksToExecute',5,'TimerFcn',#timerFcn2);
cleanup = onCleanup(#()delete([t1,t2]));
function timerFcn1(~,~)
pause(0.2)
disp('Timer 1');
end
function timerFcn2(~,~)
pause(0.2)
disp('Timer 2');
end
start(t1),start(t2)
wait(t1)
wait(t2)
disp(newline)
disp('With ExecutionMode = fixedSpacing')
t1.ExecutionMode = 'fixedSpacing';
t2.ExecutionMode = 'fixedSpacing';
start(t1),start(t2)
wait(t1)
wait(t2)
end
I am new to Julia and trying to use the Julia package DifferentialEquations to simultaneously solve for several conditions of the same set of coupled ODEs. My system is a model of an experiment and in one of the conditions, I increase the amount of one of the dependent variables at mid-way through the process.
I would like to be able to adjust the condition of this single trajectory, however so far I am only able to adjust all the trajectories at once. Is it possible to access a single one using callbacks? If not, is there a better way to do this?
Here is a simplified example using the lorentz equations for what I want to be doing:
#Differential Equations setup
function lorentz!(du,u,p,t)
a,r,b=p
du[1]= a*(u[2]-u[1])
du[2]=u[1]*(r-u[3])-u[2]
du[3]=u[1]*u[2]-b*u[3];
end
#function to cycle through inital conditions
function prob_func(prob,i,repeat)
remake(prob; u0 = u0_arr[i]);
end
#inputs
t_span=[(0.0,100.0),(0.0,100.0)];
u01=[0.0;1.0;0.0];
u02=[0.0;1.0;0.0];
u0_arr = [u01,u02];
p=[10.,28.,8/3];
#initialising the Ensemble Problem
prob = ODEProblem(lorentz!,u0_arr[1],t_span[1],p);
CombinedProblem = EnsembleProblem(prob,
prob_func = prob_func, #-> (prob),#repeat is a count for how many times the trajectories had been repeated
safetycopy = true # determines whether a safetly deepcopy is called on the prob before the prob_func (sounds best to leave as true for user-given prob_func)
);
#introducing callback
function condition(u,t,repeat)
return 50 .-t
end
function affect!(repeat)
repeat.u[1]=repeat.u[1] +50
end
callback = DifferentialEquations.ContinuousCallback(condition, affect!)
#solving
sim=solve(CombinedProblem,Rosenbrock23(),EnsembleSerial(),trajectories=2,callback=callback);
# Plotting for ease of understanding example
plot(sim[1].t,sim[1][1,:])
plot!(sim[2].t,sim[2][1,:])
I want to produce something like this:
Example_desired_outcome
But this code produces:
Example_current_outcome
Thank you for your help!
You can make that callback dependent on a parameter and make the parameter different between problems. For example:
function f(du,u,p,t)
if p == 0
du[1] = 2u[1]
else
du[1] = -2u[1]
end
du[2] = -u[2]
end
condition(t,u,integrator) = u[2] - 0.5
affect!(integrator) = integrator.prob.p = 1
For more information, check out the FAQ on this topic: https://diffeq.sciml.ai/stable/basics/faq/#Switching-ODE-functions-in-the-middle-of-integration
I want to find some numbers in Matlab (denominated below p11,..., p119) satisfying a bunch of inequalities (specifically, 16 inequalities). I want Matlab to keep searching until it finds such numbers. I thought about using while as below but it does no work. What is wrong? How can I proceed?
clear
rng default
%% SOME INITIAL VALUES
p11=0.3;
p12=0.4;
p13=0.1;
p14=0.2;
p15=0.2;
p16=0.2;
p17=0.06;
p18=0.03;
p19=0.02;
p110=0.04;
p111=0.07;
p112=50;
p113=0.02;
p114=0.03;
p115=0.01;
p116=0.08;
p117=0.01;
p118=0.1;
p119=0.07;
while ... %CONDITION THAT SHOULD BE SATISFIED (16 CONDITIONS)
((p11<=(p15+p19+p110+p111+p115+p116+p117+p119))+...
(p12<=(p16+p19+p112+p113+p115+p117+p118+p119))+...
(p13<=(p17+p110+p112+p114+p116+p117+p118+p119))+...
(p14<=(p18+p111+p113+p114+p115+p116+p118+p119))+...
(p11+p12<=(p15+p19+p110+p111+p115+p116+p117+p119+...
p16+p112+p113+p118))+...
(p11+p13<=(p15+p19+p110+p111+p115+p116+p117+p119+...
p17+p112+p114+p118))+...
(p11+p14<=(p15+p19+p110+p111+p115+p116+p117+p119+...
p18+p113+p114+p118))+...
(p12+p13<=(p16+p19+p112+p113+p115+p117+p118+p119+...
p17+p110+p114+p116))+...
(p12+p14<=(p16+p19+p112+p113+p115+p117+p118+p119+...
p18+p111+p114+p116))+...
(p13+p14<=(p17+p110+p112+p114+p116+p117+p118+p119+...
p18+p111+p113+p115))+...
(p11+p12+p13<=(p15+p19+p110+p111+p115+p116+p117+p119+...
p16+p112+p113+p118+...
p17+p114))+...
(p11+p12+p14<=(p15+p19+p110+p111+p115+p116+p117+p119+...
p16+p112+p113+p118+...
p18+p114))+...
(p11+p13+p14<=(p15+p19+p110+p111+p115+p116+p117+p119+...
p17+p112+p114+p118+...
p18+p113))+...
(p12+p13+p14<=(p16+p19+p112+p113+p115+p117+p118+p119+...
p17+p110+p114+p116+...
p18+p111))+...
(p11+p12+p13+p14==1)+...
(p15+p16+p17+p18+p19+p110+p111+p112+p113+p114+p115+p116+p117+p118+p119==1))~=15
% IF THE CONDITION IS NOT SATISFIED KEEP SEARCHING BY GUESSING
% OTHER NUMBERS
p11=unifrnd(0,1);
p12=unifrnd(0,1);
p13=unifrnd(0,1);
p14=unifrnd(0,1);
p15=unifrnd(0,1);
p16=unifrnd(0,1);
p17=unifrnd(0,1);
p18=unifrnd(0,1);
p19=unifrnd(0,1);
p110=unifrnd(0,1);
p111=unifrnd(0,1);
p112=unifrnd(0,1);
p113=unifrnd(0,1);
p114=unifrnd(0,1);
p115=unifrnd(0,1);
p116=unifrnd(0,1);
p117=unifrnd(0,1);
p118=unifrnd(0,1);
p119=unifrnd(0,1);
end
The while loop will run while the condition is true. If false it terminates. Your test conditions is while .... ~= 15. This is false as the initial guesses result in 15 out of 16 trues. Since 15 ~= 15 is false, the while loop doesn't run.
One way to fix the issue is to change from ~= to ==. This will run through and find a solution to that condition.
You could have seen this by creating a variable called tests and populated it like this:
tests = [(p11<=(p15+p19+p110+p111+p115+p116+p117+p119));...
... skipped a bunch of stuff ...
(p15+p16+p17+p18+p19+p110+p111+p112+p113+p114+p115+p116+p117+p118+p119==1)];
sum(tests)
ans = 15
Or any other way of tracking that value.
I have the following codes which I wish to have an output matrix Rpp of (10201,3). I run this code (which takes a bit long) then I check the matrix size of Rpp and I see (1,3), I tried so many things I couldn't find any proper way. The logic of the codes is to take the 6 values (contain 4 constant values and 2 variable values (chosen from 101 values)) and make the calculation for 3 different i1 and store every output vector of 3 in a matrix with (101*101 (pairs of those 2 variable values)) rows and 3 (for each i1) columns.
I appreciate your help
Vp1=linspace(3000,3500,101);
Vp2=3850;
rho1=2390;
rho2=2510;
Vs1=linspace(1250,1750,101);
Vs2=2000;
i1=[10 25 40];
Rpp = zeros(length(Vp1)*length(Vs1),length (i1));
for n=1:length(Vp1)*length(Vs1)
for m=1:length (i1)
for l=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(l);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(l)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(l)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp=((b.*(cos(i1)/Vp1(l))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(l))).*(cos(j2)/Vs2)).*H.*p.^2)./D
end
end
end
end
Try this. You 2 outer loops didn't do anything. You never used m or n so I killed those 2 loops. Also you just kept overwriting Rpp on every loop so your initialization of Rpp didn't do anything. I added an index var to assign the results to the equation to what I think is the correct part of Rpp.
Vp1=linspace(3000,3500,101);
Vp2=3850;
rho1=2390;
rho2=2510;
Vs1=linspace(1250,1750,101);
Vs2=2000;
i1=[10 25 40];
Rpp = zeros(length(Vp1)*length(Vs1),length (i1));
index = 1;
for l=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(l);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(l)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(l)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp(index,:)=((b.*(cos(i1)/Vp1(l))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(l))).*(cos(j2)/Vs2)).*H.*p.^2)./D;
index = index+1;
end
end
Results:
>> size(Rpp)
ans =
10201 3
The way you use the for loop is wrong. You're running the calculation for length(Vp1)*length(Vs1) * length (i1) * length(Vp1) * length(Vs1) times. Here's the correct way. I changed l into lll just so I won't confuse it with the number 1. In each iteration of the first for loop, you're running length(Vs1) times, and you need to assign the result (a 1X3 array) to the Rpp by using a row number specified by k+(lll-1)*length(Vp1).
for lll=1:length(Vp1)
for k=1:length(Vs1)
p=sin(i1)/Vp1(lll);
i2=asin(p*Vp2);
j1=asin(p*Vs1(k));
j2=asin(p*Vs2);
a=rho2*(1-2*Vs2^2*p.^2)-rho1*(1-2*Vs1(k).^2*p.^2);
b=rho2*(1-2*Vs2^2*p.^2)+2*rho1*Vs1(k)^2*p.^2;
c=rho1*(1-2*Vs1(k)^2*p.^2)+2*rho2*Vs2^2*p.^2;
d=2*(rho2*Vs2^2-rho1*Vs1(k)^2);
E=b.*cos(i1)./Vp1(lll)+c.*cos(i2)/Vp2;
F=b.*cos(j1)./Vs1(k)+c.*cos(j2)/Vs2;
G=a-d*(cos(i1)/Vp1(lll)).*(cos(j2)/Vs2);
H=a-d*(cos(i2)/Vp2).*(cos(j1)/Vs1(k));
D=E.*F+G.*H.*p.^2;
Rpp(k+(lll-1)*length(Vp1),:)=((b.*(cos(i1)/Vp1(lll))-c.*cos((i2)/Vp2)).*F-(a+d*((cos(i1)/Vp1(lll))).*(cos(j2)/Vs2)).*H.*p.^2)./D;
end
end
Sorry for pretty basic question.
Just starting out. Using flowgorithm to write code that gives out a calculation of exponential numbers.
The code to do it is:
function exponential(base, power) {
var answer;
answer = 1;
var i;
for (i = 1 ; i <= power ; i+= 1) {
answer = answer * base;
}
return answer;
f
then it loops up to the number of power. And i just understand that in the flowgorithm chart but i dont understand the code for it.
what does each section of the for statement mean?
i = 1 to power, i just need help understanding how it is written? What is the 1+= 1 bit?
Thanks.
The exponential function will take in 2 parameters, base and power.
You can create this function and call (fire) it when ever it is needed like so exponential(2,4).
The for (i = 1; 1 <= power; i+=1) is somewhat of an ugly for loop.
for loops traditionaly take three parameters. The first parameter in this case i =1 is the assignment parameter, the next one 1 <= power is the valadation parameter. So if we call the function like so...exponential(2,4) is i less than 4? The next parameter is an increment/decrement parameter. but this doesnt get executed until the code inside the for loop gets executed. Once the code inside the for loop is executed then this variable i adds 1 to itself so it is now 2. This is usefull because once i is no longer less than or equal to power it will exit the for loop. So in the case of exponential(2,4) once the code inside this for loop is ran 5 times it will exit the for loop because 6 > 5.
So if we look at a variable answer, we can see that before this for loop was called answer was equal to 1. After the first iteration of this for loop answer = answer times base. In the case of exponential(2,4) then answer equals 1 times 2, now answer =2. But we have only looped through the foor loop once , and like i said a for loop goes like (assignment, validator, "code inside the foor loop". then back up to increment/decrement). So since we to loop through this for loop 5 times in the case of exponential(2,4) it will look like so.
exponential(2,4)
answer = 1 * 2
now answer = 2
answer = 2 * 2
now answer = 4
answer = 4 * 2
now answer = 8
answer = 8 * 2
now answer = 16
answer = 16 * 2
now answer = 32
So if we could say... var int ans = exponential(2,4)
Then ans would equal 32 hence, the return answer; at the last line of your code.