How to fill matrix in for loop?

Question

I G 2019-1-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/439304-how-to-fill-matrix-in-for-loop

编辑： Stephen23 2019-1-11

I need to fill matrix r, where r=r(z), and ri is constant. But with this code I onlu get the first row where r=-0.7:0.7.

z=-1:0.001:0;
ri=0.7;
R=ri-z*(ri-1);
for z=-1:0.001:0;
r=(linspace(-(ri-z*(ri-1)),ri-z*(ri-1),1001))
end

My full matrix need to be in this shape, or transpose of this:

-0.7......  0.7
0.8  ...    0.8
.            .
.            .
.            .
0 .9 ...     0.9
.            .
.            .
.            .
1     ...    1

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Answer 1

Stephen23 2019-1-11

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/439304-how-to-fill-matrix-in-for-loop#answer_355987

编辑：Stephen23 2019-1-11

>> (0.7:0.05:1).'*(0:0.2:1)
ans =
00000   0.14000   0.28000   0.42000   0.56000   0.70000
00000   0.15000   0.30000   0.45000   0.60000   0.75000
00000   0.16000   0.32000   0.48000   0.64000   0.80000
00000   0.17000   0.34000   0.51000   0.68000   0.85000
00000   0.18000   0.36000   0.54000   0.72000   0.90000
00000   0.19000   0.38000   0.57000   0.76000   0.95000
00000   0.20000   0.40000   0.60000   0.80000   1.00000

Adjust the step sizes to suit your requirements.

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Answer 2

KSSV 2019-1-11

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/439304-how-to-fill-matrix-in-for-loop#answer_355962

编辑：KSSV 2019-1-11

在 MATLAB Online 中打开

z=-1:0.001:0;
ri=0.7;
R=ri-z*(ri-1);
[R,Z] = meshgrid(z,R) ;
r = R-Z.*(R-1) ;

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I G 2019-1-11

编辑：I G 2019-1-11

在 MATLAB Online 中打开

This does not work for me because I got matrix with values from 0.7 to 1 in this shape:

1    1     ....   1
.
.
1    0.997  .... 0.7

and it need to be with these values:

-0.7......  0.7
-0.8  ...   0.8
.            .
.            .
.            .
-0.9 ...    0.9
.            .
.            .
.            .
-1     ...   1

or with values:

0   ......  0.7
0    ...    0.8
.            .
.            .
.            .
0  ...      0.9
.            .
.            .
.            .
0     ...    1