How to repeat run
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Mustafa Vural
il 6 Dic 2020
Modificato: Mustafa Vural
il 7 Dic 2020
I am generating random numbers and after that I estimate them. In my code, there is a function for 2 and 3 parameterkombination (but thats not important for the question). But my problem is, to estimate for example 30 random numbers (n=30), I "run" the code then I get one number. But if I want to have for example 50 estimated numbers, I need to click 50 times "run"-button. Is there a code where it estimate automatically 50 times but everytime it estimate with other random numbers, like I would "run" the code manually. (sry for my bad english).
clear all;
n = 30;
t0 = 0.9;
b = 2;
T = 2;
for v_T = 1:length(T)
for v_b = 1:length(b)
T_A = T(v_T)
b_A = b(v_b)
pdf2p = @(x) (b_A/T_A).*(x/T_A).^(b_A-1).*(exp(-(x/T_A).^(b_A)));
integral_pdf2p = integral(pdf2p,0,Inf);
pdf2p_norm = @(x) (b_A/T_A).*(x/T_A).^(b_A-1).*(exp(-(x/T_A).^(b_A)))/integral_pdf2p;
pdf2p_norm_VZW = @(x) -(b_A/T_A).*(x/T_A).^(b_A-1).*(exp(-(x/T_A).^(b_A)))/integral_pdf2p;
min2p=fminbnd(pdf2p_norm_VZW,0,2*T_A); %returns a value x that is a local minimizer
max2p=pdf2p_norm(min2p) % X-wert min2p in die normierte funktion einsetzen = hochpunkt
%Grenze der Funktion 2P
Grenze_pdf2p_norm=2*T_A; %rechter grenzwert. 2*T um schneller die 99% der Fläche zu bekommen.
while (integral(pdf2p_norm,0,Grenze_pdf2p_norm)<0.9999) %solange das integral(die fläche) der normierten funktion(zwischen 0 bis 2*T_A) kleiner als 99% ist
Grenze_pdf2p_norm=Grenze_pdf2p_norm+0.1; %grenze wird um 0,1 nach rechts versetzt
end
Grenze_pdf2p_norm;
i=1;
while(i<=n) %ist diese bedingung wahr, dann wird die untere zeilen ausgeführt
ZufallszahlenX2p = random('unif',0,Grenze_pdf2p_norm); % xachse zufallszahlen generieren
ZufallszahlenY2p = rand*max2p; % yachse zufallszahlen generieren
if(ZufallszahlenY2p<pdf2p_norm(ZufallszahlenX2p)) %wenn zufallszahl in yachse < normierte funktion (mit zufallszahl der xachse)
data2p(i,v_b,v_T)=ZufallszahlenX2p; %nimm den wert an
% i um 1 erhöhen und wdh
i=i+1; % i um 1 erhöhen und wdh
end
end
end
end
for v_T = 1:length(T)
for v_b = 1:length(b)
T_A = T(v_T)
b_A = b(v_b)
% 2P schätzen
params2p(v_b,1:2,v_T) = wblfit(data2p(:,v_b,v_T));
params2p(v_b,4,v_T) = T_A;
params2p(v_b,5,v_T) = b_A;
params2p(v_b,7,v_T) = n;
end
Ergebnis2p((v_T-1) *length(b)+1:v_T*length(b), 1:size(params2p, 2)) = params2p(:,:,v_T);
Ergebnis3p((v_T-1) *length(b)+1:v_T*length(b), 1:size(params3p, 2)) = params3p(:,:,v_T);
end
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Risposta accettata
per isakson
il 6 Dic 2020
Put your script in a function and call it 50 times
for jj = 1 : 50
[ Ergebnis2p, Ergebnis3p ] = good_name( );
% store or display the result somewhere
end
function [ Ergebnis2p, Ergebnis3p ] = good_name( )
% your code, except "clear all"
end
5 Commenti
per isakson
il 7 Dic 2020
I am investigating more combinations of T and b In that case you should let T and b be input arguments
function [ Ergebnis2p, Ergebnis3p ] = good_name( T, b )
n = 30;
t0 = 0.9;
% b = 1:5;
% T = 2:6;
...
Test runs
>> [ Ergebnis2p, Ergebnis3p ] = good_name( 2, (1:5) );
>> whos
Name Size Bytes Class Attributes
Ergebnis2p 5x7 280 double
Ergebnis3p 5x7 280 double
>> [ Ergebnis2p, Ergebnis3p ] = good_name( (2:6), (1:5) );
Warning: Maximum likelihood estimation did not converge. Function evaluation
limit exceeded.
> In mlecustom (line 239)
In mle (line 245)
In good_name (line 102)
>> whos
Name Size Bytes Class Attributes
Ergebnis2p 25x7 1400 double
Ergebnis3p 25x7 1400 double
And maybe it's better to put together the desired row inside the function, good_name, and return it as output.
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