简化计算，因此可以使用矩阵运算来完成

Question

The basic operation I have is an operation on two probability vectors of the same length. let's call them A,B. in R the formula is:

t = 1-prod(1-A*B)

that is, the result is a scalar, the (1-AB) is a point-wise operation, whose result is a vector whose i'th element is 1-a_i*b_i . The prod operator gives the product of the elements of the vector.
The meaning of this (as you could guess) is this: suppose A is the probability for each of N sources of a disease (or other signal) to have a certain disease. B is the vector of probabilities for each of sources to transmit the disease, if they have it, to the target. The outcome is the probability of the target to acquire the disease from (at least one of) the sources.

Ok, so now I have many types of signals, so I have many "A" vectors. and for each type of signal I have many targets, each with different probability of transmission (or many "B" vectors), and I want to compute the "t" outcome for each pair.
Ideally, a matrix multiplication can do the trick if the operation was an "inner product" of the vectors. but my operation is not such (I think).

What I look for is some kind of a transformation on the vectors A and B, so I could use matrix multiplication. Any other suggestion to simplify my computation is welcome.

Here is an example (code in R)

A = rbind(c(0.9,0.1,0.3),c(0.7,0.2,0.1))
A 
# that is, the probability of source 2 to have disease/signal 1 is 0.1 (A[1,2]
# neither rows nor columns need to sum to 1.
B = cbind(c(0,0.3,0.9),c(0.9,0.6,0.3),c(0.3,0.8,0.3),c(0.4,0.5,1))
B
# that is, the probability of target 4 to acquire a disease from source 2 is 0.5 B[2,4]
# again, nothing needs to sum to 1 here

# the outcome should be:
C = t(apply(A,1,function(x) apply(B,2,function(y) 1-prod(1-x*y))))
# which basically loops on every row in A and every column in B and 
# computes the required formula
C
# while this is quite elegant, it is not very efficient, and I look for transformations
# on my A,B matrices so I could write, in principle
# C = f(A)%*%g(B), where f(A) is my transformed A, g(B) is my transformed(B),
# and %*% is matrix multiplication

# note that if replace (1-prod(1-xy)) in the formula above with sum(x*y), the result
# is exactly matrix multiplication, which is why I think, I'm not too far from that
# and want to enjoy the benefits of already implemented optimizations of matrix
# multiplications.

Answer 1

This a job where Rcpp excels. Nested loops are straight forward to implement and you don't need much C++ experience. (I like RcppEigen, but you don't really need it for this. You could use "pure" Rcpp.)

library(RcppEigen)
library(inline)

incl <- '
using  Eigen::Map;
using  Eigen::MatrixXd;
typedef  Map<MatrixXd>  MapMatd;
'

body <- '
const MapMatd        A(as<MapMatd>(AA)), B(as<MapMatd>(BB));
const int            nA(A.rows()), mA(A.cols()), mB(B.cols());
MatrixXd             R = MatrixXd::Ones(nA,mB);
for (int i = 0; i < nA; ++i) 
{
  for (int j = 0; j < mB; ++j) 
  {
    for (int k = 0; k < mA; ++k) 
    {
      R(i,j) *= (1 - A(i,k) * B(k,j));
    }
    R(i,j) = 1 - R(i,j);
  }
}
return                wrap(R);
'

funRcpp <- cxxfunction(signature(AA = "matrix", BB ="matrix"), 
                         body, "RcppEigen", incl)

Now, lets put your code in an R function:

doupleApply <- function(A, B) t(apply(A,1,
                               function(x) apply(B,2,function(y) 1-prod(1-x*y))))

Compare the results:

all.equal(doupleApply(A,B), funRcpp(A,B))
#[1] TRUE

Benchmarks:

library(microbenchmark)
microbenchmark(doupleApply(A,B), funRcpp(A,B))

# Unit: microseconds
#             expr     min       lq   median       uq     max neval
#doupleApply(A, B) 169.699 179.2165 184.4785 194.9290 280.011   100
#    funRcpp(A, B)   1.738   2.3560   4.6885   4.9055  11.293   100

set.seed(42)
A <- matrix(rnorm(3*1e3), ncol=3)
B <- matrix(rnorm(3*1e3), nrow=3)

all.equal(doupleApply(A,B), funRcpp(A,B))
#[1] TRUE
microbenchmark(doupleApply(A,B), funRcpp(A,B), times=5)

# Unit: milliseconds
#              expr        min         lq     median         uq        max neval
# doupleApply(A, B) 4483.46298 4585.18196 4587.71539 4672.01518 4712.92597     5
#     funRcpp(A, B)   24.05247   24.08028   24.48494   26.32971   28.38075     5

Answer 2

First I should note that the R code might be misleading to some Matlab users because A*B in R is equivalent to A.*B in Matlab (element-wise multiplication). I used symbolic variables in my calculations so that the operations that take place are clearer.

syms a11 a12 a21 a22 b11 b12 b21 b22
syms a13 a31 a23 a32 a33
syms b13 b31 b23 b32 b33

First consider the easiest case we have only 1 vector A and 1 vector B :

A1 = [a11;a21] ;
B1 = [b11;b21] ;

The result you want is

1 - prod(1-A1.*B1)
=
1 - (a11*b11 - 1)*(a12*b12 - 1)

Now assume we have 3 vectors A and 2 vectors B stacked one next to the other in columns:

A3 = [a11 a12 a13;a21 a22 a23; a31 a32 a33];
B2 = [b11 b12 ;b21 b22 ; b31 b32];

In order to get the indices of all the possible combinations of columns vectors of A3 paired with all the possible combinations of column vectors of B2 you can do the following:

[indA indB] = meshgrid(1:3,1:2);

Now since for pairwise product of two vectors a,b it holds that a.*b = b.*a we can just keep the unique pairs of indices. You can do that as follows:

indA = triu(indA); indB = triu(indB);
indA = reshape(indA(indA>0),[],1); indB = reshape(indB(indB>0),[],1);

Now the result that you want could be calculated:

result = 1 - prod(1-A3(:,indA).*B2(:,indB))

Just for better readability:

pretty(result.')

=

  +-                                               -+ 
  |  (a11 b11 - 1) (a21 b21 - 1) (a31 b31 - 1) + 1  | 
  |                                                 | 
  |  (a12 b11 - 1) (a22 b21 - 1) (a32 b31 - 1) + 1  | 
  |                                                 | 
  |  (a12 b12 - 1) (a22 b22 - 1) (a32 b32 - 1) + 1  | 
  |                                                 | 
  |  (a13 b11 - 1) (a23 b21 - 1) (a33 b31 - 1) + 1  | 
  |                                                 | 
  |  (a13 b12 - 1) (a23 b22 - 1) (a33 b32 - 1) + 1  | 
  +-                                               -+

Answer 3

If I understand amit's question, what you can do in Matlab is the following:

Data:

M = 4e3;    % M different cases
N = 5e2;    % N sources
K = 5e1;    % K targets
A = rand(M, N);    % M-by-N matrix of random numbers
A = A ./ repmat(sum(A, 2), 1, N);    % M-by-N matrix of probabilities (?)
B = rand(N, K);    % N-by-K matrix of random numbers
B = B ./ repmat(sum(B), N, 1);    % N-by-K matrix of probabilities (?)

First solution

% One-liner solution:
tic
C = squeeze(1 - prod(1 - repmat(A, [1 1 K]) .* permute(repmat(B, [1 1 M]), [3 1 2]), 2));
toc
% Elapsed time is 6.695364 seconds.

Second solution

% Partial vectorization 1
tic
D = zeros(M, K);
for hh = 1:M
  tmp = repmat(A(hh, :)', 1, K);
  D(hh, :) = 1 - prod((1 - tmp .* B), 1);
end
toc
% Elapsed time is 0.686487 seconds.

Third solution

% Partial vectorization 2
tic
E = zeros(M, K);
for hh = 1:M
  for ii = 1:K
    E(hh, ii) = 1 - prod(1 - A(hh, :)' .* B(:, ii));
  end
end
toc
% Elapsed time is 2.003891 seconds.

Fourth solution

% No vectorization at all
tic
F = ones(M, K);
for hh = 1:M
  for ii = 1:K
    for jj = 1:N
      F(hh, ii) = F(hh, ii) * prod(1 - A(hh, jj) .* B(jj, ii));
    end
    F(hh, ii) = 1 - F(hh, ii);
  end
end
toc
% Elapsed time is 19.201042 seconds.

The solutions are equivalent …

chck1 = C - D;
chck2 = C - E;
chck3 = C - F;
figure
plot(sort(chck1(:)))
figure
plot(sort(chck2(:)))
figure
plot(sort(chck3(:)))

… but apparently the approaches with partial vectorization, without repmat and permute, are more efficient in terms of memory and execution time.