Re: [Scilab-users] interp and memory
It has been fixed by Antoine some minutes ago:
https://codereview.scilab.org/#/c/21708/
S.
Le 08/03/2021 à 10:42, Antoine Monmayrant a écrit :
Hello Jean-Yves,
There is a memory leak, in interp, not splin.
Could you create a bug report?
Antoine
///
test=%t; // no leak if true, leak if false
niter=100;
mems=zeros(1:niter+1)*%nan;
is=0:niter;
h=scf();
mem=evstr(tokens(unix_g('free -b| grep ''Mem:'''))(3));
mems(1)=mem;
plot(is,mems,'k.');
xlabel('iteration of splin/interp')
ylabel('memory used')
for i=1:niter
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
if test then
d=splin(x,y);
z=rand(xp);
else
d=splin(x,y);
z=interp(xp,x,y,d);
end
mem=evstr(tokens(unix_g('free -b| grep ''Mem:'''))(3));
mems(i+1)=mem;
plot(is,mems,'k.');
if test then
xs2png(h,"memory_leak_test.png");
else
xs2png(h,"memory_leak.png");
end
end
Le 08/03/2021 à 09:56, Jean-Yves Baudais a écrit :
Hi,
But I guess splin() gets the derivatives [...]
Oh, maybe I was not clear. The memory issue is not here. There is not
problem to compute z. The loop is used to show the problem: it seems
that interp doesn't free the used memory after each call. So after
hundreds of call Scilab is killed!
for i=1:1000
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
d=splin(x,y);
z=interp(xp,x,y,d);
end
--Jean-Yves
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Re: [Scilab-users] interp and memory
Hello Jean-Yves,
There is a memory leak, in interp, not splin.
Could you create a bug report?
Antoine
///
test=%t; // no leak if true, leak if false
niter=100;
mems=zeros(1:niter+1)*%nan;
is=0:niter;
h=scf();
mem=evstr(tokens(unix_g('free -b| grep ''Mem:'''))(3));
mems(1)=mem;
plot(is,mems,'k.');
xlabel('iteration of splin/interp')
ylabel('memory used')
for i=1:niter
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
if test then
d=splin(x,y);
z=rand(xp);
else
d=splin(x,y);
z=interp(xp,x,y,d);
end
mem=evstr(tokens(unix_g('free -b| grep ''Mem:'''))(3));
mems(i+1)=mem;
plot(is,mems,'k.');
if test then
xs2png(h,"memory_leak_test.png");
else
xs2png(h,"memory_leak.png");
end
end
Le 08/03/2021 à 09:56, Jean-Yves Baudais a écrit :
Hi,
But I guess splin() gets the derivatives [...]
Oh, maybe I was not clear. The memory issue is not here. There is not
problem to compute z. The loop is used to show the problem: it seems
that interp doesn't free the used memory after each call. So after
hundreds of call Scilab is killed!
for i=1:1000
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
d=splin(x,y);
z=interp(xp,x,y,d);
end
--Jean-Yves
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Re: [Scilab-users] interp and memory
Hello Jean-Yves,
Yes, you are right, it does look like a memory leak: the memory
increases linearly with the number of iterations (see attached file and
modified script below).
But keep in mind that the plot might integrate some growing overhead due
to the plotting...
Antoine
/
niter=1000;
mems=zeros(1:niter+1)*%nan;
is=0:niter;
h=scf();
mem=evstr(tokens(unix_g('free -b| grep ''Mem:'''))(3));
mems(1)=mem;
plot(is,mems,'k.');
xlabel('iteration of splin/interp')
ylabel('memory used')
for i=1:niter
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
d=splin(x,y);
z=interp(xp,x,y,d);
mem=evstr(tokens(unix_g('free -b| grep ''Mem:'''))(3));
mems(i+1)=mem;
plot(is,mems,'k.');
xs2png(h,"memory_leak.png");
end
Le 08/03/2021 à 09:56, Jean-Yves Baudais a écrit :
for i=1:1000
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
d=splin(x,y);
z=interp(xp,x,y,d);
end
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Re: [Scilab-users] interp and memory
Hi,
But I guess splin() gets the derivatives [...]
Oh, maybe I was not clear. The memory issue is not here. There is not
problem to compute z. The loop is used to show the problem: it seems
that interp doesn't free the used memory after each call. So after
hundreds of call Scilab is killed!
for i=1:1000
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
d=splin(x,y);
z=interp(xp,x,y,d);
end
--Jean-Yves
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Re: [Scilab-users] interp and memory
Jean-Ives,
It seems that the variables are not so huge, xp has 100 components
as well as z, while x and y have 1 components.
But I guess splin() gets the derivatives through solving a linear
equation system of, in this case, 1 x 1, and even if the
system's matrix is quasi diagonal (it has only the diagonal,
sub-diagonal and supra-diagonal components different from 0) and this
surely reduces the computational load, it seems quite a large system and
probably it gets stuck here.
Probably a lighter (and better) way to do what you seem to be looking
for is to try to resample your x-y data by a factor of 100 using
intdec(). However, doing that 1000 times may take quite a long time (I
haven't tested it). If you really need that, it would probably be better
to do the oversampling algorithm from scratch in such a way to design
only once the smoothing filter instead of letting intdec() design the
same filter over and over again.
Regards,
Federico Miyara
On 08/03/2021 04:33, Jean-Yves Baudais wrote:
Hello,
Is my code wrong or is there a real memory problem with the function
interp in the following code? (Scilab fulfills the memory and of
course stops.)
for i=1:1000
mprintf("%d\n",i);
n=1e6;
xp=1:n;
x=1:100:n;
y=rand(x);
d=splin(x,y);
z=interp(xp,x,y,d);
end
Thanks,
Jean-Yves
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