Is it possible to do a for loop of this code?

Hi,

 sigma1cm0 = std(cm0_1); sigma1cmA = std(cmA_1);
 sigma1cmq = std(cmq_1); sigma1cmE = std(cmE_1);

The size of cm0/A/q/E its all the same

 >> size(cm0_1)

 ans =

     4     1

all 1 to 3 (three sets, from three different datas: cm0_1,cm0_2,cm0_3 and cmq_1,cmq_2 and so on ) and maybe each letter, as they have 'cm' in common? (0,A,q,E) Thanks

[Roger Stafford]

Your computations seem entirely normal, but could you please state your question in more detail. I don't understand what you are asking. In particular what is it you mean by "1 to 3 and maybe each letter? (0,A,q,E)" or "Is it possible to loop this" ?

[Andil Aboubakari]

That code is only a third of what it should be. I have another two set which is exactly the same but all the 1 changes to 2 (second set) and then 3.

So I was asking if its possible to avoid the repetition and loop one line through all the four letters three times, roughly like this:

 letter={'0';'A';'q';'E'}
 for ii=1:3
  for jj=1:numel(letter)
  sigma(ii)cm(letter(jj))=std(cm(letter(jj)_(ii))
  cm(letter(jj)(ii))=max(cm(letter(jj)_(ii))
  cm(letter(jj)(ii))=min(cm(letter(jj)_(ii))
  end
 end

Obviously that not the right way for sure.

Answer 1

[dpb]

...if its possible to avoid the repetition and loop one line through all the four letters...

"Possible", yes, but "there be dragons" that way...

matlab.wikia.com/wiki/FAQ#How_can_I_create_variables_A1.2C_A2.2C....2CA10_in_a_loop.3F

for why and alternatives. I think newer doc has a discussion under eval of the topic as well. For this case probably the alternative of dynamically-named structure is good choice.