Vectorized array operation which depends on previous array value

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Dylan Marques
Dylan Marques 2018-1-23
编辑： Jan 2018-1-23
Hi,
There is any way of making this operation vectorized?
a = ones(1000,1);
b = rand(1000,1);
c = rand(1000,1);
for i=2:1000
a(i) = b(i) + a(i-1) .* c(i);
end
I really appreciate your help, Many thanks, Dylan

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采纳的回答

Jan
Jan 2018-1-23
编辑：Jan 2018-1-23
Do you have a C compiler installed? Then a C-mex would be the best option.
// File: YourFcn.c
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[]) {
size_t n, i;
double a1, *b, *c, *r;
n = mxGetNumberOfElements(prhs[1]);
a1 = mxGetScalar(prhs[0]);
b = mxGetPr(prhs[1]);
c = mxGetPr(prhs[2]);
plhs[0] = mxCreateDoubleMatrix(n, 1, mxREAL);
r = mxGetPr(plhs[0]);
r[0] = a1;
for (i = 1; i < n; i++) {
r[i] = b[i] + r[i-1] * c[i];
}
}
UNTESTED! Running this function might crash your computer. Call it like:
a = YourFcn(a1, b, c);
[EDITED] Now the first element of a is provided as input.
For productive usage, add some checks of the inputs arguments: if the numbers of elements of b and c are equal, is the inputs are doubles and do not have an imaginary part, if they are sparse arrays, ... These checks are very cheap in a Mex function, but forgetting them might cost you hours for debugging.
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Walter Roberson
Walter Roberson 2018-1-23
a(1) is needed for input. The value can make quite a bit of difference to the final result.
Jan
Jan 2018-1-23
@Walter: In the original code a is set to ones completely, but actually only a(1) matters. I have added the input of a(1) now instead of setting it to 1.0 .

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更多回答（1 个）

Torsten
Torsten 2018-1-23
a(2:end)=b(2:end)+a(1:end-1).*c(2:end);
Best wishes
Torsten.
  2 个评论
Walter Roberson
Walter Roberson 2018-1-23
This would copy out the original a values and use them in the computation. The original code uses the a values that were just set in the previous iteration.

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