How to fill matrix in for loop?

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I G
I G am 11 Jan. 2019
Bearbeitet: Stephen23 am 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

Akzeptierte Antwort

Stephen23
Stephen23 am 11 Jan. 2019
Bearbeitet: Stephen23 am 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.

Weitere Antworten (1)

KSSV
KSSV am 11 Jan. 2019
Bearbeitet: KSSV am 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 Kommentar
I G
I G am 11 Jan. 2019
Bearbeitet: I G am 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

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