How find the best step for the array.
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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 个评论
Ivan Shorokhov
2016-3-17
编辑：Ivan Shorokhov
2016-3-17
Ced
2016-3-17
Glad it helped. Done, thanks!
采纳的回答
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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