Hello, please help me with and advice about conditional statement

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Cristian Paduraru 2016-9-8

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评论： Cristian Paduraru 2016-9-8

Hello, I encountered the following issue: I was defined a vector with a specific length. I want to insert a conditional statement: if value of z<than a specific value then flow to be calculate with a specific formula, else (z>than a specific value) then a flow to be calculate with another formula. Bellow you should view the code:

clc,clf,close all 
% Calcul GF ca si canal deschis
b=2, % b -latimea conductei rectangulare
zr=116.50; %zr- cota radier
zc=133.10,
zg=117.50; %zg- cota ax golire de fund
minval=117.50,
z=zr:0.01:zc;
A=z(z<minval);
t=length(A)
B=z(z>=minval)
n=0.014; % n- coeficient de rugozitate
i=(116.50-116.40)/54.76; % i - panta hidraulica
 if z(z<minval)
z=A;
A=b*(z-zr).*(z>=zr); %aria sectiunii transversale
C=(1/n)*((b*(z-zr).*(z>=zr))/(b+2*(z-zr).*(z>=zr)))^(1/6);
R=((b*(z-zr).*(z>=zr))/(b+2*(z-zr).*(z>=zr)));
disp('1')
Q=A*C*R^0.5*sqrt(i)
   elseif z(z>=minval)
   z=B;    
  % ===================
  zz=zg:0.01:zc;
  zt=zr:0.01:zc;
  lambda=0.025; % lambda- coef. de rezistenta liniara 
   L=54.76; % L=72.25;  L- lungimea conductei
   omega=2*2; % A- aria sectiunii rectangulare a conductei
   D=2*sqrt(omega/pi); % D- diametrul efectiv al conductei rectangulare
   beta=2.42;s=0.001;b=0.005;alfa=90;
   Zita_g=beta*(s/b)^4/3*sin(alfa) %Zita_g- pierdere de sarcina gratar
   Zita_v=0.10; % Zita_v- pierdere de sarcina vana
   Zita_in=0.5; % Zita_in- pierdere de sarcina orif. de intrare
   SumZ=Zita_g+Zita_v+Zita_in
   miu=1/sqrt(1+SumZ+lambda*L/D)
kg=miu*omega*sqrt(2*9.81); 
disp('2')
Q=kg.*(z-zg).^0.5
   end
h=figure;plot(Q,z,'linewidth',3);grid; ;xlabel('Q [mc/s)');ylabel('Cota [mdMN]');title('Cheie limnimetrica golire de fund acumulare Pungesti')
disp('test 1')
for k=1:length(z)
    disp(sprintf('%10.2f',z(k),Q(k)))
end

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Answer 1

Joost 2016-9-8

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There is indeed a problem with the if-elseif-end statement.

As soon as the if condition is evaluated true, the elseif is not considered anymore. The if condition is in your case not a boolean expression, but a vector with doubles: all values of z lower than minval. Matlab interprets this as true, as long as there is no zero present in the vector. This seems to be the case in your data.

A potential solution is to make a for loop over all values in z. Inside the for loop you can create an if-else-end statement where the if condition only looks at 1 element in z. Based on the value ( < minval or not) the appropriate action is to be taken, and the results vector can be constructed.

In general it is better to avoid for loops, and use Matlab in a vectorized way. However, the for loop can be a good starting point to get your algorithm to work, and to get the job done. At a later stage, vectorization can be considered.

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Cristian Paduraru 2016-9-8

Hello Joost, I agree with the fact that if statement is complicated to setting out. I try to insert a loop with for instruction followed bt if conditional statement and I've noticed that the first condition is considered if I compress length of z, other solution didn't work. Or the second eq. is considered despite the first. I don't know what could I try else...

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Answer 2

Thorsten 2016-9-8

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在 MATLAB Online 中打开

I would program it like this

% compute Q1 with formula 1
...
Q1 = ...
% compute Q2 with formula 2
...
Q2 = 
% compute Q based on condition
Q = Q1;
Q(z>=minval) = Q2(z>=minval);

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Cristian Paduraru 2016-9-8

Hello Thorsten, I cannot put the equal sign on those two eqs. because they aren't same length. If you could be more specific I'l appreciate. Thank You.

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