Invalid use of operator

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Ikhsan Romli
Ikhsan Romli am 9 Feb. 2023
Kommentiert: Ikhsan Romli am 15 Feb. 2023
Please help me!. I have a problem with this code. When I execute this program. The result is invalid use of operator.
WorkName='Battery'; % work results are saved by WorkName
TimeUnit='weeks'; % time unit name
dt=5; % time interval (five weeks)
measuData=[1.0000 0.9351 0.8512 0.9028 0.7754 0.7114 0.6830 0.6147 ...
0.5628 0.7090]'; % measured data at every time intervals (k1 x 1)
thres=0.3; % threshold - critical value
ParamName=['x'; 'b'; 's']; % model parameters' name to be estimated
initDisPar=[0.9 1.1; 0 0.05; 0.01 0.1];
% probability parameters of initial distribution, pxq
% (p: num. of unknown param, q: num. of probability param)
n=5e3; % number of particles
signiLevel=5; % significance level for C.I and P.I
%==========================================================================
% % % PROGNOSTICS using PARTICLE FILTER
p=size (ParamName,1);
for j=1:p; % Initial Distribution
param(j,:)=unifrnd(initDisPar(j,1),initDisPar(j,2),1,n);
ParamResul(j,:)=[ParamName(j,:) 'Resul'];
eval([ParamResul(j,:) '=param(j,:);']);
end;
kl=length(measuData); k=1; % Update Process or Prognosis
if measuData (end)-measuData(1)<0; cofec=-1; else cofec =1; end
while min(eval([ParamResul(1,:) '(k,:)'])*cofec)<thres*cofec; k=k+1;
% stepl. predlotion (prior)
paramPredi=param;
for j=1:p; eval([ParamName(j,:) '=paramPredi(j.:);']); end
paramPredi(1,:)=...%====== PROBLEM DEFINITION: MODEL DEFINITION=============
exp(-b.*dt).*x;
%==========================================================================
if k<=k1 % (Update Process)
% step2. update (likelihood)
like1=normpdf(measuData(k),paramPredi(1,:),paramPredi(end,:));
% step3. resampling
cdf=cumsum(like1)./max(sum(likel));
for i=1:n;
u=rand;
loca=find(cdf >= u); param(:,i)=paramPredi(:,loca(1));
end;
else % (Prognosis)
param=paramPredi;
end
for j=1:p; eval([ParamResul(j,:) '(k,:)=param(j,:);']); end;
if k>k1;
eval([ParamResul(1,:) '(k,:)=normrnd(param(1,:),param(end,:));']);
end
end
% % % POST-PROCESSING
time=[0:dt:dt*(k-1)]'; % RUL Calculation
perceValue=[50 signiLevel 100-signiLevel];
for i=1:n;
loca=find(eval([ParamResul(1,:) '(:,i)'])*cofec>=thres*cofec);
RUL(i)=time(loca(1))-time(k1);
end;
RULPrece=prctile(RUL',perceValue);
figure; set(gca,'fontsize',14); hist(RUL,30); % RUL Results Display
xlim([min(RUL) max(RUL)]);
xlabel(['RUL' ' (' TimeUnit ')']);
titleName=['at ' num2str(time(k1)) ' ' TimeUnit]; title(titleName)
fprintf('\n # Percentiles of RUL atg weeks \n', time(k1))
fprintf('\n %g, median: %g, %gprct: %g \n', perceValue(2), ...
RULPerce(2), RULPerce(1), perceValue(3), RULPerce(3))
Name=[WorkName ' at ' num2str(time(k1)) '.mat']; save(Name); % Work Save
The following is the error that appears in the command window
>> Prognosis
Error using Prognosis
Invalid use of operator.
  1 Kommentar
Sarvesh Kale
Sarvesh Kale am 9 Feb. 2023
The error is in the following line
p=size (ParamName,1);
for j=1:p;
param(j,:)=unifrnd(initDisPar(j,1),initDisPar(j,2),1,n);
ParamResul(j,:)=[ParamName(j,:) 'Resul'];
eval([ParamResul(j,:) '=param(j,:);']); % this line gives error
end;
what are you exactly trying to do with the eval there ?

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Antworten (1)

Jan
Jan am 9 Feb. 2023
Bearbeitet: Jan am 9 Feb. 2023
The old problem: Strange methods to define names of variables dynamically using eval() demands for even stranger methods to access these variables later on. This is the typical shot in the own knee. See: FAQ: Why and how to avoid EVAL .
Use dynamic field names instead:
% Replace:
% ParamResul(j,:)=[ParamName(j,:) 'Resul'];
% eval([ParamResul(j,:) '=param(j,:);']);
% by:
Resul.(ParamName(j)) = param(j,:);
Afterwards all eval calls can be replaced by using the fields directly. This makes the code cleaner, easier to debug and to maintain and much faster.
  3 Kommentare
Jan
Jan am 14 Feb. 2023
I cannot fix the errors in your code without seeing the code. So please post zu current version of your code.
Changing
eval([ParamResul(j,:) '=param(j,:);']);
to
Resul.(ParamName(j)) = param(j,:);
seems to be more or less trivial, so I cannot guess, which problems you have with it.
Ikhsan Romli
Ikhsan Romli am 15 Feb. 2023
Here I attach my version of the code

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