用MATLAB进行稀疏矩阵插值 [英] Sparse Matrix Interpolation With MATLAB

问题描述

如果我有这样的矩阵

A = [1 2; 3 4];

我可以像这样用interp2对其进行插值

I can use interp2 to interpolate it like this

newA = interp2(A,2);

然后我得到一个5x5的插值矩阵.

and I get a 5x5 interpolated matrix.

但是如果我有一个像这样的矩阵怎么办?

But what if I have a matrix like this:

B = zeros(20);
B(3,2) = 5;
B(17,4) = 3;
B(16, 19) = 2.3;
B(5, 18) = 4.5;

我将如何对这个矩阵进行插值(或填补空白).我已经研究过interp2以及TriScatteredInterp，但这些都似乎无法完全满足我的需求.

How would I interpolate (or fill-in the blanks) this matrix. I've looked into interp2 as well as TriScatteredInterp but neither of these seem to fit my needs exactly.

推荐答案

一个好的解决方案是使用我的 inpaint_nans .只需提供不存在任何信息的NaN元素，然后使用inpaint_nans.它将对NaN元素进行插值，并将其填充以与数据点平滑一致.

A good solution is to use my inpaint_nans. Simply supply NaN elements where no information exists, then use inpaint_nans. It will interpolate for the NaN elements, filling them in to be smoothly consistent with the data points.

B = nan(20);
B(3,2) = 5;
B(17,4) = 3;
B(16, 19) = 2.3;
B(5, 18) = 4.5;
Bhat = inpaint_nans(B);

surf(B,'marker','o')
hold on
surf(Bhat)

对于那些对inpaint_nans是否可以处理更复杂的表面感兴趣的人，我曾经拍摄过一张数字化的莫奈画(在左侧看到，然后通过随机删除50％的像素来破坏它.最后，我应用inpaint_nans进行了观察如果我能很好地恢复图像，则右边的图像是已修复的图像.分辨率较低时，恢复的图像是不错的恢复.

For those interested in whether inpaint_nans can handle more complex surfaces, I once took a digitized Monet painting (seen on the left hand side, then corrupted it by deleting a random 50% of the pixels. Finally, I applied inpaint_nans to see if I could recover the image reasonably well. The right hand image is the inpainted one. While the resolution is low, the recovered image is a decent recovery.

作为另一个示例，请尝试以下操作:

As another example, try this:

[x,y] = meshgrid(0:.01:2);
z = sin(3*(x+y.^2)).*cos(2*x - 5*y);
surf(x,y,z)
view(-23,40)

现在，删除此数组中大约7/8个元素，将其替换为NaN.

Now, delete about 7/8 of the elements of this array, replacing them with NaNs.

k = randperm(numel(z));
zcorrupted = z;
zcorrupted(k(1:35000)) = NaN;

使用修复进行恢复. z轴具有不同的缩放比例，因为在边缘的+/- 1上下都存在微小的变化，但否则，后一个表面是一个很好的近似值.

Recover using inpainting. The z-axis has a different scaling because there are minor variations above and below +/-1 around the edges, but otherwise, the latter surface is a good approximation.

zhat = inpaint_nans(zcorrupted);
surf(x,y,zhat)
view(-23,40)

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