Function ranges in Matalb

Question

Ahmed Elsherif 2020-8-17

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/580650-function-ranges-in-matalb

评论： Ahmed Elsherif 2020-8-21

Hello everybody,

I am trying to use Gui matlab to determine the output of a function in two different ranges which are compared with user-entered values. I set the axis Z as

Z=0.000001:500;

The I used the entered values by user in calculations:

Z03= (4.*E.^1.6)/(r3.*0.8862269);% E and r3 are given by the user 
Z04= (4.*E.^1.6)/(r4.*0.8862269);% E and r4 are given by the user 
delat3=0;
delat4=0;
p3=-exp((-(Z-delat3)./Z03).^2);
p4=-exp((-(Z-delat4)./Z04).^2);

After that I defined the function's ranges as follows:

for i=1:numel(Z)
    if Z(i)<=d3 % d3 is given by the user 
        P2L=(p3.*(2*(Z-delat3)./Z03))./((Z03).^2);
    elseif (Z(i) > d3) && (Z(i) < d4) % d4 is given by the user 
        P2L=(p4.*(2*(Z-delat4)./Z04))./((Z04).^2);
    end
end

and plot the results by

plot(Z, P2L,'linewidth',2)

When I press 'Plot' I got this error

Operands to the || and && operators must be convertible to logical scalar values.

Would you please help me to slove this error?

Thanks in advance,

Ahmed

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dpb 2020-8-17

在 MATLAB Online 中打开

for i=1:numel(Z)
    if Z(i)<=d3 % d3 is given by the user 
        P2L=(p3.*(2*(Z-delat3)./Z03))./((Z03).^2);
    elseif (Z(i) > d3) && (Z(i) < d4) % d4 is given by the user 
        P2L=(p4.*(2*(Z-delat4)./Z04))./((Z04).^2);
    end
end

in

elseif (Z(i) > d3) && (Z(i) < d4) % d4 is given by the user

can't get there unless Z(i)>d3 because all cases of Z(i)<=d3 are in the other branch already...so, j r is pointing out the code is same as

 for i=1:numel(Z)
   if Z(i)<=d3 % d3 is given by the user 
     P2L=(p3.*(2*(Z-delat3)./Z03))./((Z03).^2);
   else 
     if Z(i)<d4
       P2L=(p4.*(2*(Z-delat4)./Z04))./((Z04).^2);
     end
   end
 end   

to separate out the two ... or, of course, you can leave the elseif in and just remove the superfluous test.

The error as you wrote is owing to the use of '&&' and that d3 which is not defined in the supplied code snippet must have been a vector, not a single element...observe at the command line if write:

 >> v0  % a variable happened to be in my current workspace for convenience...
v0 =
         10.00         11.00         12.00         13.00         14.00         15.00         16.00         17.00         18.00         19.00         20.00
>> v0(1) && 3  % short-circuit && and a constant
ans =
  logical
   1
>> v0(1) && [3 4]  % short-circuit && and a vector
Operands to the || and && operators must be convertible to logical scalar values. 
>> 

NB: the latter goes "boom"...

dpb 2020-8-17

编辑：dpb 2020-8-17

Well, whenever it was that the code posted before was run, the error message is clear that wasn't a scalar.

Use the debugger to see in context.

But, as written there's no need for the & operator anyways...and it is "&" you would want here, not "&&" if did need the compound test.

Ahmed Elsherif 2020-8-21

Thanks dpd, I already tried with Piecewise and it works nicely. Thanks for your help.

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Answer 1

Walter Roberson 2020-8-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/580650-function-ranges-in-matalb#answer_483026

在 MATLAB Online 中打开

for i=1:numel(Z)
  if Z(i)<=d3 % d3 is given by the user 
      P2L(i)=(p3.*(2*(Z(i)-delat3)./Z03))./((Z03).^2);
  elseif (Z(i) < d4) % d4 is given by the user 
      P2L(i)=(p4.*(2*(Z(i)-delat4)./Z04))./((Z04).^2);
  else
      P2L(i) = nan;  %undefined outside the range given
  end
end