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Compacting a For Loop

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Benedict Comerford
Benedict Comerford on 7 Oct 2020
Commented: Sindar on 7 Oct 2020
Hi all
I would love some help with compacting a for loop, i'm using it to find the smallest error in a matrix is there a way of finding the smallest value of the matrix as well as the name of the variable associated with the smallest value other then printing out a masive for loop.
Thanks Ben.

  2 Comments

Sindar
Sindar on 7 Oct 2020
  • post code in blocks, don't attach unless it's really long:
norm_er = 1.301;
exp_er = 9.126;
log_er = 1.301;
ray_er = 2.606;
Errors = [norm_er,exp_er,log_er,ray_er]
if norm_er<exp_er && norm_er<log_er && norm_er<ray_er
disp('Normal distribution is best')
elseif exp_er<norm_er && exp_er<log_er && exp_er<ray_er
disp('Exponential distribution is best')
elseif log_er<norm_er && log_er<exp_er && log_er<ray_er
disp('Log distribution is best')
elseif ray_er<norm_er && ray_er<exp_er && ray_er<log_er
disp('Rayleigh distribution is best')
elseif norm_er==exp_er
disp('Normal and Exponential distributions are best')
elseif norm_er==log_er
disp('Normal and Log distributions are best')
elseif norm_er==ray_er
disp('Normal and Rayleigh distributions are best')
elseif exp_er==log_er
disp('Exponential and Log distributions are best')
elseif exp_er==ray_er
disp('Exponential and Rayleigh distributions are best')
elseif log_er==ray_er
disp('Log and Rayleigh distributions are best')
end
  • that's not a for loop
  • is there an actual appreciable chance that two methods will give the same error?
Benedict Comerford
Benedict Comerford on 7 Oct 2020
Hi Sindar
Sorry yeah should have said an if statement
Yes there is a strong possibility that 2 will have the same error

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Accepted Answer

Sindar
Sindar on 7 Oct 2020
Edited: Sindar on 7 Oct 2020
norm_er = 1.301;
exp_er = 9.126;
log_er = 1.301;
ray_er = 2.606;
Errors = [norm_er,exp_er,log_er,ray_er];
% set a tolerance to define equal error
tol = 1e-3;
% find minimum error
min_error = min(Errors);
% find all methods with this error value
idxs = find((Errors-min_error) < tol);
error_str = ["Normal";"Exponential";"Log";"Rayleigh"];
% print based on how many errors are equal (extends to any number of distributions)
% none? that's not good
if length(idxs)==0
error('something went wrong')
% e.g., Rayleigh distribution is best
elseif length(idxs)=1
fprintf('%s Distribution is best\n',error_str(idxs))
elseif length(idxs)=length(Errors)
fprintf('All Distributions are equally good\n')
% e.g., Log and Rayleigh distributions are best
elseif length(idxs)=2
fprintf('%s and %s Distributions are best\n',error_str(idxs))
% e.g., Log, Exponential, and Rayleigh distributions are best
else
tmp = strjoin(error_str(idxs(1:end-1)),", ") + ", and " + error_str(idxs(end);
fprintf('%s Distributions are best',tmp)
end

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More Answers (1)

Benedict Comerford
Benedict Comerford on 7 Oct 2020
Hi Sindar
Thanks so much for that i think that will work just getting an error atm because (val) is undefined

  2 Comments

Benedict Comerford
Benedict Comerford on 7 Oct 2020
All good it was just the min values
Sindar
Sindar on 7 Oct 2020
yup, had a typo / holdover from a first draft

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