How to divide a constant value (input) to several parts and find average of the output

Question

Ope 2019-4-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/456426-how-to-divide-a-constant-value-input-to-several-parts-and-find-average-of-the-output

回答： Walter Roberson 2019-5-27

Hello, I have a problem and need help.

Equation 1 and 2 works fine when z (e.g. 2000) is applied as a constant value. But since there is an exponential decay in p; I will like to divide z into several parts (e.g. 0, 100, 200 .... 2000), for each part calculate D, then find average D. Sum of z (z1, z2, ... zn) is equal to the constant value.

How can I divide z into equal parts, such that I can determine average D? I have tried different models but it doesn't work.

for i1=1:n
p=(phi0)*(exp(-c*z));	------- 1
D=((p)*(rW))+((1-p)*(rSG)); ------- 2
end 

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Image Analyst 2019-5-25

Attach some data

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2019-5-27

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/456426-how-to-divide-a-constant-value-input-to-several-parts-and-find-average-of-the-output#answer_376713

在 MATLAB Online 中打开

Zn = floor(z/n) * ones(1,n);
left_over = z - sum(Zn);
enlarge_at = randperm(n, left_over);
Zn(enlarge_at) = Zn(enlarge_at) + 1;

Now sum(Zn) = z and Zn is the sizes of each partition. You might want to use

offsets = cumsum([0, Zn]);

Then D(offsets(k):offsets(k)-1) is the k'th section.

The partitions will not generally be exactly the same sizes through-out, because z will not generally be exactly divisible by n . This code randomly distributes an additional length of 1 to make sure that the total comes out correctly.

It would also be possible to distribute the extras semi-regularly. You could consider linspace(1,z,n+1) and then round() or floor() or ceil() to create the offsets.