Azzera filtri
Azzera filtri

How find the best step for the array.

2 visualizzazioni (ultimi 30 giorni)
Ivan Shorokhov
Ivan Shorokhov il 16 Mar 2016
Commentato: Ced il 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 Commenti
Ivan Shorokhov
Ivan Shorokhov il 17 Mar 2016
Modificato: Ivan Shorokhov il 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 il 17 Mar 2016
Glad it helped. Done, thanks!

Accedi per commentare.

Risposta accettata

Ced
Ced il 17 Mar 2016
Modificato: Ced il 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.

Più risposte (0)

Categorie

Scopri di più su Graphics Performance in Help Center e File Exchange

Prodotti

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by