How do I stop the calculation in a for loop?

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gamer
gamer le 13 Juin 2021
Commenté : gamer le 14 Juin 2021
Hello community!
First of all thanks for your help:)
n = 4;
v = rand(n,2);
for f = 1:1000
v = v-0.001
end
Thats already what I ve got. It gives me 1000 of matrixes back and alyways calculated by - 0.001. Now I ve got problem. If one element of the matrix (for example matrix Number 94) is already smaller than 0.001, then this element should be for all other matrixes ( 95 - 1000) equal to zero. I hope you have some advice! Thanks.

Réponse acceptée

Walter Roberson
Walter Roberson le 13 Juin 2021
The division by 50 is just to make the output shorter for this demonstration.
format long g
n = 4;
v = rand(n,2) / 50;
for f = 2:1000
temp = v - 0.001;
temp(temp < 0.001) = 0;
v = temp
if ~any(v(:)); break; end %all 0
end
v = 4×2
0.0117169155857542 0.00385373115144157 0.00322072350669969 0.0123980988902088 0.00540972017890342 0.00679499574461224 0.0119082492555772 0
v = 4×2
0.0107169155857542 0.00285373115144157 0.00222072350669969 0.0113980988902088 0.00440972017890342 0.00579499574461224 0.0109082492555772 0
v = 4×2
0.00971691558575421 0.00185373115144157 0.00122072350669969 0.0103980988902088 0.00340972017890342 0.00479499574461224 0.00990824925557724 0
v = 4×2
0.00871691558575421 0 0 0.00939809889020883 0.00240972017890342 0.00379499574461224 0.00890824925557724 0
v = 4×2
0.00771691558575421 0 0 0.00839809889020883 0.00140972017890342 0.00279499574461224 0.00790824925557724 0
v = 4×2
0.00671691558575421 0 0 0.00739809889020883 0 0.00179499574461224 0.00690824925557724 0
v = 4×2
0.00571691558575421 0 0 0.00639809889020883 0 0 0.00590824925557724 0
v = 4×2
0.00471691558575421 0 0 0.00539809889020883 0 0 0.00490824925557724 0
v = 4×2
0.00371691558575421 0 0 0.00439809889020883 0 0 0.00390824925557724 0
v = 4×2
0.00271691558575421 0 0 0.00339809889020883 0 0 0.00290824925557724 0
v = 4×2
0.00171691558575421 0 0 0.00239809889020883 0 0 0.00190824925557724 0
v = 4×2
0 0 0 0.00139809889020883 0 0 0 0
v = 4×2
0 0 0 0 0 0 0 0
  3 commentaires
Walter Roberson
Walter Roberson le 13 Juin 2021
p = [r + (a-2*r)*rand(n,1),r + (b-2*r)*rand(n,1)];
v = rand(n,2);
for f = 1:1000
temp = v - 0.001 ;
temp(temp < 0.001) = 0 ;
v = temp ;
if ~any(v(:)); break; end
p = p + v
end
gamer
gamer le 14 Juin 2021
Thank you so much!

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