how to extract subvectors of a vector or matrix?

Question

Jorge Rodriguez 2020-12-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/697810-how-to-extract-subvectors-of-a-vector-or-matrix

编辑： Cris LaPierre 2020-12-19

My case is like this

n=100;
x=[1 2 3 4 5 6 7 8 9 10];
y=[1 4 9 16 25 36 49 64 81 100];
A=[x' y'];

I want to extract the data from the vectors x and y, but have it accumulated and stored in subvectors. The result will be subvectors

xx= [1 2] [1 2 3] [1 2 3 4] [1 2 3 4 5] [1 2 3 4 5 6] [1 2 3 4 5 6 7] [1 2 3 4 5 6 7 8] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9 10]...;
yy= [1 4] [1 4 9] [1 4 9 16] [1 4 9 16 25] [1 4 9 16 25 36] [1 4 9 16 25 36 49] [1 4 9 16 25 36 49 64] [1 4 9 16 25 36 49 64 81] [1 4 9 16 25 36 49 64 81 100]...;    

I need to generate the subvectors xx and yy to plot the data for each of the subvectors. I need to graph the evolution when 1 data is increased.

Example

xx(1)=[1 2];
yy(1)=(1,4);
plot(xx(1),(yy(1))

Finally plot all pairs xx vs yy for loop

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Cris LaPierre 2020-12-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/697810-how-to-extract-subvectors-of-a-vector-or-matrix#answer_579635

编辑：Cris LaPierre 2020-12-19

This sounds like an assignment designed to teach you how to use a for loop. If so, it doesn't sound like you are thinking about how to take advantage of the loop - specifically that the loop counter increases by one each time.

Think about what code would be repetitive here, and put that inside your loop. This page on array indexing may help you. One hint - your for loop counter does not have to start at 1.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

KSSV 2020-12-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/697810-how-to-extract-subvectors-of-a-vector-or-matrix#answer_579640

在 MATLAB Online 中打开

n=100;
x=[1 2 3 4 5 6 7 8 9 10];
y=[1 4 9 16 25 36 49 64 81 100];
m = length(x) ; 
xx = cell(m-1,1) ; 
yy = cell(m-1,1) ; 
for i = 1:m-1
  xx{i} = x(1:i+1) ; 
  yy{i} = y(1:i+1) ; 
end