How to fill matrix in for loop?

5 visualizzazioni (ultimi 30 giorni)
I G
I G il 11 Gen 2019
Modificato: Stephen23 il 11 Gen 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

Risposta accettata

Stephen23
Stephen23 il 11 Gen 2019
Modificato: Stephen23 il 11 Gen 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.

Più risposte (1)

KSSV
KSSV il 11 Gen 2019
Modificato: KSSV il 11 Gen 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 Commento
I G
I G il 11 Gen 2019
Modificato: I G il 11 Gen 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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