How do I stop the calculation in a for loop?

2021 6 13
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更新时间：2021 6 14
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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.

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Walter Roberson
Walter Roberson 2021-6-13
Ran in:

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Ran in:
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

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gamer
gamer 2021-6-13
Tank you very much Walter! Thats exactly what I was asking for!
Could you help me maybe one more time
That was your code. At the begiining I have a starting position (of spheres)
p=[r + (a-2*r)*rand(n,1),r + (b-2*r)*rand(n,1)
Then I will change the position and slow the spheres down. ( Thats why I was asking for the code)
v = rand(n,2);
for f = 1:1000
temp = v - 0.001 ;
temp(temp < 0.001) = 0 ;
v = temp ;
if ~any(v(:)); break;
end
The positon is always updating but unfortunately its not slowing down. It always uses the first matrix and not the other 999. Is it maybe possiible to index the matrix?
for f = 1:1000
p = p +v % is it possible to index v(f)?
end
Walter Roberson
Walter Roberson 2021-6-13
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

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