c# - ILNumerics如何实现二维卷积？

Question

是否有可能像 matlab“CONV2”函数一样获得“二维卷积”ilnumerics？以及如何去做？ ilnumerics 可以像 matlab 一样方便地操作矩阵。所以如果我有 matlab 代码，我可以用 ilnumeris 做同样的事情，但是 mathlab 中的函数“CONV2”被称为“内置函数”。顺便说一下，我的矩阵可能是 1660×521 大小。 这里是matlab中“CONV2”的帮助文档。

% CONV2 Two dimensional convolution.

%   C = CONV2(A, B) performs the 2-D convolution of matrices A and B.

%   If [ma,na] = size(A), [mb,nb] = size(B), and [mc,nc] = size(C), then
   mc = max([ma+mb-1,ma,mb]) and nc = max([na+nb-1,na,nb]).

%   C = CONV2(H1, H2, A) first convolves each column of A with the vector
%   H1 and then convolves each row of the result with the vector H2.  If
%   n1 = length(H1), n2 = length(H2), and [mc,nc] = size(C) then
%   mc = max([ma+n1-1,ma,n1]) and nc = max([na+n2-1,na,n2]).
%   CONV2(H1, H2, A) is equivalent to CONV2(H1(:)*H2(:).', A) up to
%   round-off.
%
%   C = CONV2(..., SHAPE) returns a subsection of the 2-D
%   convolution with size specified by SHAPE:
%     'full'  - (default) returns the full 2-D convolution,
%     'same'  - returns the central part of the convolution
%               that is the same size as A.
%     'valid' - returns only those parts of the convolution
%               that are computed without the zero-padded edges.
%               size(C) = max([ma-max(0,mb-1),na-max(0,nb-1)],0).
%
%   See also CONV, CONVN, FILTER2 and, in the Signal Processing
%   Toolbox, XCORR2.

Answer 1

是的，当然可以!如果您查看 Matlab 的文档，您将看到“conv2”使用的是哪种算法。它被称为“空间形式的二维卷积方程”。您可以在 ILNumerics 中轻松实现这一点。以下代码实现了简单的形式方程:

ILArray<double> A = new double[,] { { 1, 4, 7 }, { 2, 5, 8 }, { 3, 6, 9 } };
ILArray<double> B = new double[,] { { 1, 5, 9, 13 }, { 2, 6, 10, 14 }, { 3, 7, 11, 15 }, { 4, 8, 12, 16 } };

// calc the size
int ma = A.S[0]; int na = A.S[1];
int mb = B.S[0]; int nb = B.S[1];

int mc = Math.Max(ma + mb - 1, Math.Max(ma, mb));
int nc = Math.Max(na + nb - 1, Math.Max(na, nb));
ILArray<double> C = ILMath.zeros(mc, nc);

for (int n1 = 0; n1 < mc; n1++)
{
    for (int n2 = 0; n2 < nc; n2++)
    {
        for (int k1 = 0; k1 < ma; k1++)
        {
            int bm = n1 - k1;

            // Make sure the outside of the boundaries 
            // are checked! 
            if (bm < 0 || bm >= mb)
            {
                continue;
            }

            for (int k2 = 0; k2 < na; k2++)
            {
                int bn = n2 - k2;

                // Here as well
                if (bn < 0 || bn >= nb)
                {
                    continue;
                }

                // If it is a fit - calculate and add
                C[n1, n2] = C[n1, n2] + A[k1, k2] * B[bm, bn];
            }
        }
    }
}

请注意，这不是最有效的实现方式。结果与其他实现完全相同。

我们正在开发信号处理工具箱，它将更有效地实现此类功能。

使用 ILNumerics Array Visualizer，您可以轻松地以文本和视觉方式检查结果，并且它会响应数组的即时更改: