将Matlab interp2移植到SciPy interp2d时出现问题 [英] Problems porting Matlab interp2 to SciPy interp2d

问题描述

我正在用Python语言重写Matlab的代码. 在Matlab代码中，我具有以下功能:gt= interp2(I(:,:,i)',xi,yi,'cubic')';，其中I是RGB图像，xi和yi是具有相同形状的2D矩阵，定义了x和y坐标. 在将gt(isnan(gt))=0;设置为边界之外的值之后. 此功能可在Matlab上完美运行.

I'm rewriting a Matlab's code in Python Language. In Matlab code I have this function: gt= interp2(I(:,:,i)',xi,yi,'cubic')';, where I is a RGB image, xi and yi are 2D matrixes with same shape defining the x and y coordinates. After I set gt(isnan(gt))=0; for the values outside the boundaries. This function runs perfectly on Matlab.

在Python中，我编写了以下代码:gt=interpolate.interp2d(x,y,img.T,kind='cubic',fill_value=0)，其中x和y与Matlab中的xi和yi相同，而img是灰度图像. 无论如何，我得到以下异常:"Invalid length for input z for non rectangular grid") ValueError: Invalid length for input z for non rectangular grid". 怎么了? 非常感谢.

In Python I wrote the following code: gt=interpolate.interp2d(x,y,img.T,kind='cubic',fill_value=0), where x and y are the same as the xi and yi in Matlab, and img is a gray-scale image. Anyway i get the following exception: "Invalid length for input z for non rectangular grid") ValueError: Invalid length for input z for non rectangular grid". What is wrong? Thank you very much.

推荐答案

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Looking at the documentation you should note a few things:

如果x和y代表规则网格，请考虑使用 RectBivariateSpline .

If x and y represent a regular grid, consider using RectBivariateSpline.

x和y应该是一维矢量，与Matlab不同，它们应与Matlab的x和y和 NOT 与xi和yi匹配.那些会晚一点.

x and y should be 1D vectors unlike in Matlab and they match with Matlab's x and y and NOT with xi and yi. Those come later.

SciPy将返回一个函数，供以后插值使用. Matlab立即调整插值向量.

SciPy will return a function to use to interpolate later. Matlab retuns the interpolated vector immediately.

您已经在Matlab中使用了它，但隐含地假设了

You've used it in Matlab but implicitly assuming that

[X, Y] = meshgrid(1:size(I,1), 1:size(I,2))
gt= interp2(X, Y, I(:,:,i)',xi,yi,'cubic')';

因此实际上是您需要传递给X和Y的插值.interp2d而非NOT xi和yi，它们会稍后出现.

So it's actually this X and Y that you need to pass to interpolation.interp2d and NOT xi and yi, they come later.

在SciPy中，您实际上可以跳过meshgrid步骤，仅使用正常范围，但是它会输出一个函数供以后使用，而不是在现场进行插值(请注意，我在x的定义中可能存在小错误和y，我对arange不太熟悉:

In SciPy you can actually skip the meshgrid step and just use a normal range but also it outputs a function for later rather than performing the interpolation on the spot (note there might be small errors in my definitions of x and y, I'm not that familiar with arange):

x = arange(1, img.shape[0]+1)
y = arange(1, img.shape[1]+1)
f = interpolation.interp2d(x, y, img.T)

，然后在想要进行插值时:

and then when you want to do the interpolation:

gt = f(xi, yi)

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