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How find the best step for the array.

조회 수: 2 (최근 30일)
Ivan Shorokhov
Ivan Shorokhov 2016년 3월 16일
댓글: Ced 2016년 3월 17일
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
  댓글 수: 3
Ivan Shorokhov
Ivan Shorokhov 2016년 3월 17일
편집: Ivan Shorokhov 2016년 3월 17일
@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 2016년 3월 17일
Glad it helped. Done, thanks!

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Ced
Ced 2016년 3월 17일
편집: Ced 2016년 3월 17일
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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