How find the best step for the array.

Question

Ivan Shorokhov 2016-3-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/273885-how-find-the-best-step-for-the-array

评论： Ced 2016-3-17

采纳的回答： Ced

在 MATLAB Online 中打开

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
显示 1更早的评论隐藏 1更早的评论

Ivan Shorokhov 2016-3-17

编辑：Ivan Shorokhov 2016-3-17

@Ced, Great, thank you, could you please add your comment in the answer section, so I can select it as the best one.

Ced 2016-3-17

Glad it helped. Done, thanks!

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Ced 2016-3-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/273885-how-find-the-best-step-for-the-array#answer_214029

编辑：Ced 2016-3-17

在 MATLAB Online 中打开

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.