How to fill matrix in for loop?

21 views (last 30 days)
I G
I G on 11 Jan 2019
Edited: Stephen23 on 11 Jan 2019
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

Accepted Answer

Stephen23
Stephen23 on 11 Jan 2019
Edited: Stephen23 on 11 Jan 2019
>> (0.7:0.05:1).'*(0:0.2:1)
ans =
0.00000 0.14000 0.28000 0.42000 0.56000 0.70000
0.00000 0.15000 0.30000 0.45000 0.60000 0.75000
0.00000 0.16000 0.32000 0.48000 0.64000 0.80000
0.00000 0.17000 0.34000 0.51000 0.68000 0.85000
0.00000 0.18000 0.36000 0.54000 0.72000 0.90000
0.00000 0.19000 0.38000 0.57000 0.76000 0.95000
0.00000 0.20000 0.40000 0.60000 0.80000 1.00000
Adjust the step sizes to suit your requirements.

More Answers (1)

KSSV
KSSV on 11 Jan 2019
Edited: KSSV on 11 Jan 2019
z=-1:0.001:0;
ri=0.7;
R=ri-z*(ri-1);
[R,Z] = meshgrid(z,R) ;
r = R-Z.*(R-1) ;
  1 Comment
I G
I G on 11 Jan 2019
Edited: I G on 11 Jan 2019
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

Sign in to comment.

Categories

Find more on Loops and Conditional Statements in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by