[Scilab-users] vector operation replacing for loop

[Heinz Nabielek]

Sorry, Scilab friends, I am still not fluid with vector operations.
Can someone rewrite the for loop for me into something much more efficient?

n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;

The transformation generates an autocorrelated (here rho=0.9) normal distribution V from an uncorrelated normal distribution Z and eventually I will need it for very much larger n values....

Heinz

[Collewet Guylaine]

Hello,

Could you try this? I think this gives the same result as with the loop
I think that the matrix mat could be probably built in a smarter way

V=zeros(n,1);
V(1)=Z(1);
expo=1:n-1;
vect=[zeros(1,n-1) r.^(expo-1)];
A=makematrix_toeplitz(vect);
mat=A(n:$,1:n-1);
V(2:$)=(Z(1)*r.^expo)'+sqrt(1-r^2)*mat*Z(2:$)';


Guylaine



-----Message d'origine-----
De : users [mailto:users-***@lists.scilab.org] De la part de Heinz
Nabielek
Envoyé : jeudi 1 novembre 2018 09:27
À : Users mailing list for Scilab
Objet : [Scilab-users] vector operation replacing for loop

Sorry, Scilab friends, I am still not fluid with vector operations.
Can someone rewrite the for loop for me into something much more
efficient?

n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;

The transformation generates an autocorrelated (here rho=0.9) normal
distribution V from an uncorrelated normal distribution Z and eventually I
will need it for very much larger n values....

Heinz

[Samuel Gougeon]

Hello,

Post by Heinz Nabielek
Sorry, Scilab friends, I am still not fluid with vector operations.
Can someone rewrite the for loop for me into something much more efficient?
n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;
The transformation generates an autocorrelated (here rho=0.9) normal distribution V from an uncorrelated normal distribution Z and eventually I will need it for very much larger n values....

You may use filter(), with a feedback component (since V(i) depends on
the previous state V(i-1) computed at the previous step).
However, as shown below, a quick trial shows an initial discrepancy
between filter() result and yours with the explicit loop.
I don't know why. May be the setting for the initial condition should be
carefully considered/tuned...

n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;

y = filter(sqrt(1-r^2),[1 -r], Z);

clf
subplot(3,1,1), plot(Z), ylabel('input Z')
subplot(3,1,2), plot(V), ylabel('V')
d = abs(y-V); d(d==0) = %nan;
subplot(3,1,3), plot2d("nl",d), title('filter() - V')

[Samuel Gougeon]

Post by Collewet Guylaine
Hello,

Post by Heinz Nabielek
Sorry, Scilab friends, I am still not fluid with vector operations.
Can someone rewrite the for loop for me into something much more efficient?
n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;
The transformation generates an autocorrelated (here rho=0.9) normal distribution V from an uncorrelated normal distribution Z and eventually I will need it for very much larger n values....

You may use filter(), with a feedback component (since V(i) depends on
the previous state V(i-1) computed at the previous step).
However, as shown below, a quick trial shows an initial discrepancy
between filter() result and yours with the explicit loop.
I don't know why. May be the setting for the initial condition should
be carefully considered/tuned...

This is certainly the reason. For i=1, V(i-1) is unknown and must be
somewhat provided. The last filter() option zi is not really documented
(i have no time to go to the reference to get clear about how "the
initial condition relative to a "direct form II transposed" state space
representation." works. Trying to provide a value V(0) such that
V(1)==Z(1) decreases the initial discrepancy by a factor 10. But it is
still non zero.

y = filter(sqrt(1-r^2), [1 -r], Z, Z(1)*(1-sqrt(1-r^2))/r);

[Samuel Gougeon]

Hello,

Is there anyone here that uses to use the last filter() optional argument?

In the example of the thread reminded below:

V = Z;
for i=2:n
V(i) = a*V(i-1) + b*Z(i);
end

the input is Z, the output is V, a and b are fixed parameters.

Syntax
------
[y,zf] = filter(B, A, x [,zi])
Parameters
----------
B : real vector : the coefficients of the filter numerator in
decreasing power order, or
a polynomial.
A : real vector : the coefficients of the filter denominator in
decreasing power order,
or a polynomial.
x : real row vector : the input signal
zi : real row vector of length max(length(a),length(b))-1: the
initial condition
relative to a "direct form II transposed" state space
representation. The default
value is a vector filled with zeros.

So here i assume that zi is V(0) (so actually Z(1)). Then, how to get
[Z(1) V(2:n)]
as from the above loop? I have tried the following, without success:
y = filter(b, [1 -a], Z);
or
y = [Z(1) filter(b, [1 -a], Z(2:$), Z(1))];
or
y = filter(b, [1 -a], Z, Z(1)*(1-b)/a);

The zi option has no example, and unless filter() is bugged, the way zi
works/is taken into account is unclear.

Any hint (before analyzing its hard code..)?

Regards
Samuel

Post by Collewet Guylaine
Hello,

Post by Heinz Nabielek
Sorry, Scilab friends, I am still not fluid with vector operations.
Can someone rewrite the for loop for me into something much more efficient?
n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;
The transformation generates an autocorrelated (here rho=0.9) normal distribution V from an uncorrelated normal distribution Z and eventually I will need it for very much larger n values....

You may use filter(), with a feedback component (since V(i) depends
on the previous state V(i-1) computed at the previous step).
However, as shown below, a quick trial shows an initial discrepancy
between filter() result and yours with the explicit loop.
I don't know why. May be the setting for the initial condition should
be carefully considered/tuned...

This is certainly the reason. For i=1, V(i-1) is unknown and must be
somewhat provided. The last filter() option zi is not really
documented (i have no time to go to the reference to get clear about
how "the initial condition relative to a "direct form II transposed"
state space representation." works. Trying to provide a value V(0)
such that V(1)==Z(1) decreases the initial discrepancy by a factor 10.
But it is still non zero.
y = filter(sqrt(1-r^2), [1 -r], Z, Z(1)*(1-sqrt(1-r^2))/r);
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[Heinz Nabielek]

Post by Collewet Guylaine
Hello,
Is there anyone here that uses to use the last filter() optional argument?
V = Z;
for i=2:n
V(i) = a*V(i-1) + b*Z(i);
end

In my case, a^2 + b^2 =1, if that should make any difference?

Has the problem been sorted out, how to use "filter"?
Heinz

Post by Collewet Guylaine
the input is Z, the output is V, a and b are fixed parameters.

Syntax
------
[y,zf] = filter(B, A, x [,zi])
Parameters
----------
B : real vector : the coefficients of the filter numerator in decreasing power order, or
a polynomial.
A : real vector : the coefficients of the filter denominator in decreasing power order,
or a polynomial.
x : real row vector : the input signal
zi : real row vector of length max(length(a),length(b))-1: the initial condition
relative to a "direct form II transposed" state space representation. The default
value is a vector filled with zeros.

So here i assume that zi is V(0) (so actually Z(1)). Then, how to get
[Z(1) V(2:n)]
y = filter(b, [1 -a], Z);
or
y = [Z(1) filter(b, [1 -a], Z(2:$), Z(1))];
or
y = filter(b, [1 -a], Z, Z(1)*(1-b)/a);
The zi option has no example, and unless filter() is bugged, the way zi works/is taken into account is unclear.
Any hint (before analyzing its hard code..)?
Regards
Samuel

Post by Collewet Guylaine
Hello,

Post by Heinz Nabielek
Sorry, Scilab friends, I am still not fluid with vector operations.
Can someone rewrite the for loop for me into something much more efficient?
n=1000;
Z=grand(1,n,'nor',0,1);
r=0.9;
V=Z;
for i=2:n;
V(i)=r*V(i-1)+sqrt(1-r^2)*Z(i);
end;
The transformation generates an autocorrelated (here rho=0.9) normal distribution V from an uncorrelated normal distribution Z and eventually I will need it for very much larger n values....

You may use filter(), with a feedback component (since V(i) depends on the previous state V(i-1) computed at the previous step).
However, as shown below, a quick trial shows an initial discrepancy between filter() result and yours with the explicit loop.
I don't know why. May be the setting for the initial condition should be carefully considered/tuned...

This is certainly the reason. For i=1, V(i-1) is unknown and must be somewhat provided. The last filter() option zi is not really documented (i have no time to go to the reference to get clear about how "the initial condition relative to a "direct form II transposed" state space representation." works. Trying to provide a value V(0) such that V(1)==Z(1) decreases the initial discrepancy by a factor 10. But it is still non zero.
y = filter(sqrt(1-r^2), [1 -r], Z, Z(1)*(1-sqrt(1-r^2))/r);
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http://lists.scilab.org/mailman/listinfo/users

_______________________________________________
users mailing list
http://lists.scilab.org/mailman/listinfo/users

[Samuel Gougeon]

Post by Heinz Nabielek

Post by Collewet Guylaine
Hello,
Is there anyone here that uses to use the last filter() optional argument?
V = Z;
for i=2:n
V(i) = a*V(i-1) + b*Z(i);
end

In my case, a^2 + b^2 =1, if that should make any difference?

It does not.

Post by Heinz Nabielek
Has the problem been sorted out, how to use "filter"?

It hasn't been clarified. We likely must go in the code to see what it
does... and document the option.

[kjubo]

Hello,

precalculate the know values, I gain 40% faster execution time.
Hope helps a bit...

BR

clc, clear, mode(0)

n=1e6
Z=grand(1,n,'nor',0,1);
Z=Z(:);
r=0.9;

tic()
V1=Z;
for ii=2:n;
V1(ii)=r*V1(ii-1)+sqrt(1-r^2)*Z(ii);
end;
toc()

tic()
V2 = Z;
k1 = sqrt(1-r^2).*Z;
for ii=2:n;
V2(ii) = r*V2(ii-1) + k1(ii);
end;
toc()

isequal(V1,V2)




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[Heinz Nabielek]

Great. Thanks. Should have thought of it myself...

Still: is there a vector way?
Heinz

Post by Collewet Guylaine
Hello,
precalculate the know values, I gain 40% faster execution time.
Hope helps a bit...
BR
clc, clear, mode(0)
n=1e6
Z=grand(1,n,'nor',0,1);
Z=Z(:);
r=0.9;
tic()
V1=Z;
for ii=2:n;
V1(ii)=r*V1(ii-1)+sqrt(1-r^2)*Z(ii);
end;
toc()
tic()
V2 = Z;
k1 = sqrt(1-r^2).*Z;
for ii=2:n;
V2(ii) = r*V2(ii-1) + k1(ii);
end;
toc()
isequal(V1,V2)

[Samuel Gougeon]

Post by Heinz Nabielek
Great. Thanks. Should have thought of it myself...
Still: is there a vector way?

Didn't you read Guylaine's answer?
She proposes a tricky algorithm. Unfortunately, it cannot be used for
very long Z input, due to memory considerations.
I am afraid that there is no other better "vectorized" solution.

This is why a builtin function like filter() -- that aims to do what you
want -- is required.

BR

[Heinz Nabielek]

Thanks

Post by Samuel Gougeon

Post by Heinz Nabielek
Great. Thanks. Should have thought of it myself...
Still: is there a vector way?

Didn't you read Guylaine's answer?
She proposes a tricky algorithm. Unfortunately, it cannot be used for very long Z input, due to memory considerations.
I am afraid that there is no other better "vectorized" solution.
This is why a builtin function like filter() -- that aims to do what you want -- is required.
BR
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