Effacer les filtres
Effacer les filtres

How find the best step for the array.

2 vues (au cours des 30 derniers jours)
Ivan Shorokhov
Ivan Shorokhov le 16 Mar 2016
Commenté : Ced le 17 Mar 2016
Hello,
I have following array:
first_tt= 1;step = 7;last_tt = 27;
tt= first_tt:step:last_tt;
Answer:
tt= [1,8,15,22];
So what I want is to include the last number 27, by slightly changing the step, BUT still meet the second array value of original array, in this case 1+7= 8. tt= [1, 8,15,22];
Currently done:
first_tt= 1; step = 7;last_tt = 27; tt= first_tt:step:last_tt; new_step = (last_tt-first_tt)/length(tt); new_tt= first_tt:new_step:last_tt;
Answer:
new_tt = [1,7.5,14,20.5,27];
So what I need is to include second array value of original array, i.e 8, so I'm wondering, if there are any ways of doing it?
new_tt = [1,..., *8 (?)*,....,27];
Even +/-2% would be ok.
i.e
new_tt = [1,..., *7.84 (?)*,....,26.46];
Best Regards,
Ivan
  3 commentaires
Ivan Shorokhov
Ivan Shorokhov le 17 Mar 2016
Modifié(e) : Ivan Shorokhov le 17 Mar 2016
@Ced, Great, thank you, could you please add your comment in the answer section, so I can select it as the best one.
Ced
Ced le 17 Mar 2016
Glad it helped. Done, thanks!

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Réponse acceptée

Ced
Ced le 17 Mar 2016
Modifié(e) : Ced le 17 Mar 2016
You need to decide whether you want equidistant steps, or matching numbers.
I'm sure I'm missing something, but if you just want the second and last, then
tt = first_tt:step:last_tt;
if ( tt(end) < last_tt )
tt(end+1) = last_tt;
end
Or, if you only need to match the two numbers, then:
first_tt= 1;step = 7;last_tt = 27;
second_tt = first_tt + step;
n_elements = floor((last_tt-second_tt)/step)+1;
tt = linspace(second_tt,last_tt,n_elements);
The point is, there are a million ways of doing this, but none of them will manage to go from one number to another in equidistant steps without some other compromise.

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