## Avoid for following loops

Asked by hung lequang

### hung lequang (view profile)

on 18 Oct 2012
Accepted Answer by Sean de Wolski

### Sean de Wolski (view profile)

Hello everyone,

Can anyone show me how I can avoid following for loops. Thanks!

```N1=10;
N2=10;
N3=10;
```
```l1 = 100;
l2 = 100;
t=1;
```
```n1 = [1:10];
n2 = [1:10];
```
```x3 = rand(N3,1);
```
```for ik1=1:N1
for ik2=1:N2
for ix3=1:N3
k1= 2*pi*n1(ik1)/l1;
k2= 2*pi*n2(ik2)/l2;
s = sqrt(k1^2+k2^2);
if s~=0
R11(ik1,ik2,ix3)=1/(exp(s*t)-exp(-s*t))*k1*exp(s*x3(ix3));
end
end
end
end
```

Matt Kindig

### Matt Kindig (view profile)

on 18 Oct 2012

Perhaps, but it could get a little complicated-- you will need to use meshgrid() to define k1 and k2 over the full n1xn2 (10x10) grid, and defining the third dimension in R11 could get a bit complicated using purely vector operations.

Why are you doing this? Are you doing this for speed reasons? Oftentimes, for loops are actually faster than vectorized loops. Your best bet to improve speed is to preallocate R11 prior to entering the loop. That is, add this line to right before your for loop:

```R11 = NaN(N1, N2, N3);   %this preallocates memory for R11
```

Now each iteration of the loop just fills in data in the already existing R11 variable.

## Products

No products are associated with this question.

### Sean de Wolski (view profile)

Answer by Sean de Wolski

### Sean de Wolski (view profile)

on 18 Oct 2012
Edited by Sean de Wolski

### Sean de Wolski (view profile)

on 18 Oct 2012

```k1 = (2*pi.*n1(1:N1).')./l1;
S=bsxfun(@hypot,k1,2*pi*n2(1:N2)./l2);
P=bsxfun(@(s,ix3)(1./(exp(s.*t)-exp(-s.*t))).*exp(s.*ix3),S,x3(reshape(1:N3,1,1,N3)));
R22 = bsxfun(@times,k1,P);
```

Check to make sure it's close:

```norm(R22(:)-R11(:))
```

The difference arises because I used hypot in places of sqrt(x^2+y^2), it's a more numerically stable implementation of this.

Also note, that just redoing your for loops with some simple preallocation and calculation rearranging would help a lot.

preallocate R11 before the loop:

```R11 = zeros(N1,N2,N3);
```

And move things that don't change into their respective loops

```for ix1 = etc
k1 = etc.
for ix2 = etc.
k2 = etc.
```

Since k1 and k2 are independent of the inner loops.

hung lequang

### hung lequang (view profile)

on 18 Oct 2012

Thanks again Sean de Wolski for your answer.

Answer by Matt J

### Matt J (view profile)

on 18 Oct 2012

```[ik1,ik2,ix3]=ndgrid(1:N1,1:N2,1:N3);
```
```              k1= 2*pi*n1(ik1)/l1;
k2= 2*pi*n2(ik2)/l2;
s = sqrt(k1.^2+k2.^2);```
```R11=1./(exp(s.*t)-exp(-s.*t)).*k1.*exp(s.*x3(ix3));
```
```R11(isnan(R11))=0;
```

Matt J

### Matt J (view profile)

on 18 Oct 2012

Matt's solution is certainly easier to understand. However, it will be slower, especially when N1,N2,N3 get large

As I said, that's probably true unless possibly, the ngrid output will be recycled later for further operations. Once you've paid the overhead of ndgrid execution time, further operations tend to be fast element-wise operations on the grid variables.

Also, you have not one but several calls to BSXFUN. Not sure how that adds up...

Sean de Wolski

### Sean de Wolski (view profile)

on 18 Oct 2012

Well, I stand corrected, the ndgrid solution is still faster up to 150. At this point a smart for-loop wins:

```    tic
tau = 2*pi;
R11 = zeros(N1,N2,N3);
for ik1=1:N1
k1= tau*n1(ik1)/l1;
for ik2=1:N2
k2 = tau*n2(ik2)/l2;
s = sqrt(k1^2+k2^2);
st = s*t;
R11(ik1,ik2,:)=1/(exp(st)-exp(-st))*k1*exp(s*x3(1:N3));
end
end```
hung lequang

### hung lequang (view profile)

on 19 Oct 2012

#### Join the 15-year community celebration.

Play games and win prizes!

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi