How find the best step for the array.

2 visualizaciones (últimos 30 días)
Ivan Shorokhov
Ivan Shorokhov el 16 de Mzo. de 2016
Comentada: Ced el 17 de Mzo. de 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 comentarios
Ivan Shorokhov
Ivan Shorokhov el 17 de Mzo. de 2016
Editada: Ivan Shorokhov el 17 de Mzo. de 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 el 17 de Mzo. de 2016
Glad it helped. Done, thanks!

Iniciar sesión para comentar.

Respuesta aceptada

Ced
Ced el 17 de Mzo. de 2016
Editada: Ced el 17 de Mzo. de 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.

Más respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Etiquetas

Productos

Community Treasure Hunt

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

Start Hunting!

Translated by