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Error Using a loop: Conversion to double from cell is not possible.

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Hello everyone,
On executing following code:
for w = 1:10
PN = PName{w};
CS = CSize{w};
Project = [PN CS];
Project = sortrows(Project,1);
Project = Project(11:end,1:end);
Proj(w) = Project;
end
I am getting error: Conversion to double from cell is not possible..
Code is running well till Project = Project(11:end,1:end);
Size of Project is: 4227x2 cell.
Error is while assigning a values to a new variable using a loop at Proj(w) = Project;
Any help would be appreciated :-)
Regard,s
Waqar Ali
  3 Comments
Waqar Ali Memon
Waqar Ali Memon on 10 Jul 2019
Hello Guillaume,
Since, loop range is from 1 to 10 and once the required data is executed (till Project = Project(11:end,1:end); ). I want to create a new variable and save this output in that. It means if w = 1 is executed: I should get new variable with name Proj1 and should have the output of Project; Once it is saved. It should be done for 2,3,...10. I hope, you understand :-)
Thank you for your time.
Stephen23
Stephen23 on 10 Jul 2019
"I want to create a new variable ... I should get new variable with name Proj1..."
Dynamically accessing variable names is one way that beginners force themselves into writing slow, complex, obfuscated, buggy code that is hard to debug. Read this to know why:
In contrast indexing is simple, neat, easy to debug, and very efficient. You should use indexing, just like madhan ravi showed in their answer.

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Accepted Answer

Peter Jarosi
Peter Jarosi on 10 Jul 2019
Edited: Peter Jarosi on 10 Jul 2019
Perhaps what you need is struct. Try the following:
s = struct();
for w = 1:10
PN = PName{w};
CS = CSize{w};
Project = [PN CS];
Project = sortrows(Project,1);
Project = Project(11:end,1:end);
FieldName = strcat('Proj', num2str(w));
s.(FieldName) = cell2mat(Project);
end
You will get your 10 "variables" within structure s. You can refer them as s.Proj1, s.Proj2, etc.
Please let me know if it works for you.
  7 Comments
Peter Jarosi
Peter Jarosi on 10 Jul 2019
Edited: Peter Jarosi on 10 Jul 2019
madhan ravi : I read Guillaume 's comments carefully (again). I didn't find what I missed but I found the best part of it: "Sometimes, the best answer is: you don't actually want to do that." I really like it. :)

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More Answers (2)

madhan ravi
madhan ravi on 10 Jul 2019
Proj = cell(10,1); % outside loop
Proj{w} = ... % inside loop , leave the rest unchanged
  3 Comments

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Waqar Ali Memon
Waqar Ali Memon on 10 Jul 2019
Thank you so much madhan ravi, Guillaume and Stephen Cobeldick. I wanted a loop and the code from Peter Jarosi did same. Though, it won't be proper solution but it helped me :-)
Thank you everyone :-)
Regards,
Waqar Ali Memon
  1 Comment
Peter Jarosi
Peter Jarosi on 10 Jul 2019
Waqar Ali Memon : You're very welcome!
I totally agree with programmer gurus madhan ravi, Guillaume , and Stephen Cobeldick. It is not recommended using dinamically changed variable names, and string operations instead of numbers, especially when we develope big applications. I appreciate their comments.
But sometimes we have to break rules, we need ugly tricks in order to solve our small problems quickly. :-)
Good luck!

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