I am trying to create a parallel-for loop in MATLAB to fill an mxm matrix as follows:
parfor i = 1 : m^2
A(i) = sum( x .^ %Sum of indices% );
end
If I was doing it without parfor I would simply do:
for i = 1 : m
for j = 1 : m
A(i,j) = sum( x.^(i+j) );
end
end
However unfortunately parfor does not allow for other indices within it's body and therefore I cannot do what I'd like to do.
I have tried extracting the sum of the indices by using ceil(i/3) + mod(i,3) however this doesn't work, because at i=3 it gives me 1 instead of 4 as I want.
Does anyone have a better way of extracting the sum of the indices?
r=sum( bsxfun(#power, x(:), m+1:2*m ) ,1);
c=sum( bsxfun(#power, x(:), m+1:-1:2 ) ,1);
A=fliplr(toeplitz(c,r));
[I,J]=ndgrid(1:m);
K=I+J;
A=zeros(m);
parfor i=1:m^2
A(i)=sum(x.^(K(i)));
end
Here is another vectorized variation:
[I,J] = ndgrid(1:m, 1:m);
K = I + J;
A = reshape(sum(bsxfun(#power, x, K(:).')), [m m]);
Related
I try to create a 3d-array with 3 indices k=0,...,K-1, m'=0,...,M-1, m=0,...,M-1. Can I generate such a 3d-array which the element at position (k,m',m) is k/K+m'-m in some quick way? My current approach is to use a 3-layer for-loop to calculate each element. It is a little clumsy and I believe there must be better way. So can anyone give me some inspiration? Grateful.
By the way another question. Can I generate a 3d-array from a matrix A and a vector b? I mean each element in b do some calculation with A so that we can derive a 3d-array? And how?
You need to create 3 vectors:
k of size [1 x K x 1] ,
mp of size [M x 1 x 1] and
m of size [1 x 1 x M].
Using implicit expansion you can create the desired array:
K = 5;
M = 7;
k = 0:K-1;
mp = (0:M-1).'
m = reshape(0:M-1,1,1,[]);
result = k/K+mp-m;
Note that it is more efficient than ndgrid solution.
I have a matrix A of dimension m-by-n composed of zeros and ones, and a matrix J of dimension m-by-1 reporting some integers from [1,...,n].
I want to construct a matrix B of dimension m-by-n such that for i = 1,...,m
B(i,j) = A(i,j) for j=1,...,n-1
B(i,n) = abs(A(i,n)-1)
If sum(B(i,:)) is odd then B(i,J(i)) = abs(B(i,J(i))-1)
This code does what I want:
m = 4;
n = 5;
A = [1 1 1 1 1; ...
0 0 1 0 0; ...
1 0 1 0 1; ...
0 1 0 0 1];
J = [1;2;1;4];
B = zeros(m,n);
for i = 1:m
B(i,n) = abs(A(i,n)-1);
for j = 1:n-1
B(i,j) = A(i,j);
end
if mod(sum(B(i,:)),2)~=0
B(i,J(i)) = abs(B(i,J(i))-1);
end
end
Can you suggest more efficient algorithms, that do not use the nested loop?
No for loops are required for your question. It just needs an effective use of the colon operator and logical-indexing as follows:
% First initialize B to all zeros
B = zeros(size(A));
% Assign all but last columns of A to B
B(:, 1:end-1) = A(:, 1:end-1);
% Assign the last column of B based on the last column of A
B(:, end) = abs(A(:, end) - 1);
% Set all cells to required value
% Original code which does not work: B(oddRow, J(oddRow)) = abs(B(oddRow, J(oddRow)) - 1);
% Correct code:
% Find all rows in B with an odd sum
oddRow = find(mod(sum(B, 2), 2) ~= 0);
for ii = 1:numel(oddRow)
B(oddRow(ii), J(oddRow(ii))) = abs(B(oddRow(ii), J(oddRow(ii))) - 1);
end
I guess for the last part it is best to use a for loop.
Edit: See the neat trick by EBH to do the last part without a for loop
Just to add to #ammportal good answer, also the last part can be done without a loop with the use of linear indices. For that, sub2ind is useful. So adopting the last part of the previous answer, this can be done:
% Find all rows in B with an odd sum
oddRow = find(mod(sum(B, 2), 2) ~= 0);
% convert the locations to linear indices
ind = sub2ind(size(B),oddRow,J(oddRow));
B(ind) = abs(B(ind)- 1);
I have two nested loops which I want to parallelize.
n=100;
x=rand(1,n);
m=5;
xx=rand(1,m);
r = zeros(1,m);
for i=1:n
q = ones(1,m);
for j=1:n
q = q .* (xx-x(j))/(x(i)-x(j));
end
r = r + q;
end
In order to prepare this function for palatalization, I changed local variables to global ones.
n=100;
x=rand(1,n);
m=5;
xx=rand(1,m);
r = ones(n,m);
for i=1:n
for j=1:n
r(i,:) = r(i,:) .* (xx-x(j))/x(i)-x(j))
end
end
r = sum(r,1);
Instead of transforming a whole vector at once, let's try it with only one scalar. Also use the simplest element of x which depends on i and j. I also removed the sum in the end. We can add it back later.
n=100;
x=rand(1,n);
r = ones(n,1);
for i=1:n
for j=1:n
y = x(i)+x(j);
r(i) = r(i) * y;
end
end
The code above is the example function, I want to parallelize.
The inner loop always needs to access the same vector r(i) for one iteration of the outer loop i. This access is a write operation (*=), but the order doesn't matter for this operation.
Since nested parfor loops are not allowed in Matlab, I tried to pack everything in one parfor loop.
n=100;
x=rand(1,n);
r = ones(n,1);
parfor k=1:(n*n)
%i = floor((k-1)/n)+1; % outer loop
%j = mod(k-1,n)+1; % inner loop
[j,i] = ind2sub([n,n],k);
y = x(i)+x(j);
r(i) = r(i) * y; % ERROR here
end
Since indies are calculated, Matlab still doesn't know hot to slice it.
So, I decided to move the multiplication operation outside and use linear indices.
n=100;
x=rand(1,n);
r = ones(n,n);
parfor k=1:(n*n)
[j,i] = ind2sub([n,n],k);
y = x(i)+x(j);
r(k) = y;
end
r = prod(r,1);
r = squeeze(r); % remove singleton dimensions
While this does work for scalar values in the inner loop, it doesn't work for vectors in the inner loop since indices must be again calculated.
n=100;
x=rand(1,n);
m=5;
r = ones(n,n,m);
parfor k=1:(n*n)
[j,i] = ind2sub([n,n],k);
y = x(i)+x(j);
r((k-1)*m+1:k*m) = y.*(1:m); % ERROR here
end
r = prod(r,1);
r = squeeze(r); % remove singleton dimensions
Although it does work, when I reshape the array.
n=100;
x=rand(1,n);
m=5;
r = ones(n*n,m);
parfor k=1:(n*n)
[j,i] = ind2sub([n,n],k);
y = x(i)+x(j);
r(k,:) = y.*(1:m); % ERROR here
end
r = reshape(r,n,n,m);
r = prod(r,2);
r = squeeze(r); % remove singleton dimensions
This way, I can transform a vector xx to another vector r.
n=100;
x=rand(1,n);
m=5;
xx=rand(1,m);
r = ones(n*n,m);
parfor k=1:(n*n)
[j,i] = ind2sub([n,n],k);
y = x(i)+x(j);
r(k,:) = y.*xx; % ERROR here
end
r = reshape(r,n,n,m);
r = prod(r,2);
r = sum(r,1);
r = reshape(r,size(xx)); % reshape output vector to input vector
For my parallel solution, I need an n*n*m array instead of a n*m array which seems quite inefficient.
Is there a better way of doing what I want?
What are the advantages of other ways (prettier code, less CPU, less RAM, ...)?
UPDATE
In the order of trying to simplify the task and reduce it to the minimum working example of the problem, I omitted the check of i~=j to make it easier, although resulting in an all NaN result. Further, the nature of the code results in an all 1 result when adding this check. In order for the code to make sense, the factors are just weights for another vector z.
The more elaborate problem looks as follows:
n=100;
x=rand(1,n);
z=rand(1,n);
m=5;
xx=rand(1,m);
r = zeros(1,m);
for i=1:n
q = ones(1,m);
for j=1:n
if i~=j
q = q .* (xx-x(j))/(x(i)-x(j));
end
end
r = r + z(i) .* q;
end
This problem does not need any parallel for loop to execute. One problem is that x(i)-x(j) is redundandly calculated a lot of times. This is inefficient. The approach suggested calculates every number exactly once and it vectorize the operations for each element in xx. Since xx is the shortest vector by far it is almost completely vectorized. In case you want to vectorize the last loop as well this will probably just be like a hidden for loop as well, it will much more memory and the code would be more complicated (like 3D matrices and so). I took the freedom to switch minus to plus in the denominator just for testing. Minus would generate NaN for all numbers. The last approach is slightly faster. About 10 times for n=10000. I suggest you try a bit more elaborate benchmark.
function test()
% Initiate variables
n=100;
x=rand(1,n);
m=5;
xx=rand(1,m);
tic;
% Alternative 1
r = zeros(1,m);
for i=1:n
q = ones(1,m);
for j=1:n
q = q .* (xx-x(j))/(x(i)+x(j));
end
r = r + q;
end
toc;
tic;
% Alternative 2
xden = bsxfun(#plus, x, x.'); % Calculate denominator
xnom = repmat(x,n,1); % Calculate nominator
xfull = (xnom./xden).'; % calculate right term on rhs.
for (k = 1:m)
tmp= prod(xx(k)./xden - xfull); % Split in 2 calculations
r2(k) = sum(tmp); % "r = r + xx(k)"
end
toc;
disp(r);
disp(r2);
Just a note in the end. Alternative 2 is faster but it is also memory expensive, so in case of memory issues a loop is to prefer. Further, there is no need for global variables in case of parallelization. In case you need this you probably have to look over your design (but in case the code is short there is not some critical, so then you should not need to bother so much).
pvec = 1:3;
for i = 1:3
p=pvec(i);
for m = 1:p
erfun=erfc(5/(2*sqrt(p-m)));
suma(m) = sum(erfun)
end
end
I want to save sum of all values of erfun for every p i.e I want to have 3 values in final array but every value in the array should be sum of all the values of erfun for one p.
Similar questions have been addressed but I could not apply them in my case.
Minimal fix method (to your own code)
The following modification to your code will yield your requested results
suma = zeros(3,1);
pvec = 1:3;
for i = 1:3
p=pvec(i);
for m = 1:p
erfun=erfc(5/(2*sqrt(p-m)));
suma(i) = suma(i) + erfun; %// <-- modified here
end
end
Where I've also included suma = zeros(3,1), which I assume that you also have in your code (however not shown in your question); pre-allocating suma with sufficient entries.
Alternative method (arrayfun)
Another solution, you can make use of the arrayfun command to get rid of the inner for loop:
suma = zeros(1,3);
pvec = 1:3;
for i = 1:3
p=pvec(i);
suma(i) = sum(arrayfun(#(x) erfc(5/(2*sqrt(p-x))), 1:p));
end
Alternative method #2 (arrayfun)
An even more condensed solution, including also the purpose of the outer for loop in your arrayfun call:
suma = arrayfun(#(x) ...
sum(erfc(5./(2*sqrt(kron(x, ones(1,x-1)) - 1:(x-1))))), pvec)
Here we've made use of the kron command, which will be implicitly used in the arrayfun command above as follows
kron(1, []) = [] %// empty array
kron(2, [1]) = 2
kron(3, [1 1]) = [3 3]
and used the fact that erfc addition from 1/sqrt(0) is always 0 (i.e., erfc(Inf) = 0, and hence we needn't evaluate the case m=p as it yields no addition to our sum).
Result
All of the above methods yield the result
suma =
0
0.0004
0.0128
I have recently attempted to concisely draw several graphs in a plot using gnuplot and the plot for ... syntax. In this case, I needed nested loops because I wanted to pass something like the following index combinations (simplified here) to the plot expression:
i = 0, j = 0
i = 1, j = 0
i = 1, j = 1
i = 2, j = 0
i = 2, j = 1
i = 2, j = 2
and so on.
So i loops from 0 to some upper limit N and for each iteration of i, j loops from 0 to i (so i <= j). I tried doing this with the following:
# f(i, j, x) = ...
N = 5
plot for [i=0:N] for [j=0:i] f(i, j, x) title sprintf('j = %d', j)
but this only gives five iterations with j = 0 every time (as shown by the title). So it seems that gnuplot only evaluates the for expressions once, fixing i = 0 at the beginning and not re-evaluating to keep up with changing i values. Something like this has already been hinted at in this answer (“in the plot for ... structure the second index cannot depend on the first one.”).
Is there a simple way to do what I want in gnuplot (i.e. use the combinations of indices given above with some kind of loop)? There is the do for { ... } structure since gnuplot 4.6, but that requires individual statements in its body, so it can’t be used to assemble a single plot statement. I suppose one could use multiplot to get around this, but I’d like to avoid multiplot if possible because it makes things more complicated than seems necessary.
I took your problem personally. For your specific problem you can use a mathematical trick. Remap your indices (i,j) to a single index k, such that
(0,0) -> (0)
(1,0) -> (1)
(1,1) -> (2)
(2,0) -> (3)
...
It can be shown that the relation between i and j and k is
k = i*(i+1)/2 + j
which can be inverted with a bit of algebra
i(k)=floor((sqrt(1+8.*k)-1.)/2.)
j(k)=k-i(k)*(i(k)+1)/2
Now, you can use a single index k in your loop
N = 5
kmax = N*(N+1)/2 + N
plot for [k=0:kmax] f(i(k), j(k), x) title sprintf('j = %d', j(k))