将Matlab interp2移植到SciPy interp2d的问题。

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.

我正在用Python语言重写Matlab的代码。在Matlab代码中，我有这个函数:gt= interp2(I(:， I)'，xi,yi，'立方体');在这里我是RGB图像，xi和yi是二维矩阵，具有相同的形状，定义了x和y坐标。当我设置gt(isnan(gt))= 0;对于边界之外的值。这个函数在Matlab中很好地运行。

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.

在Python中，我编写了以下代码:gt=插值。interp2d(x,y,img. t,kind='cubic'，fill_value=0)，其中x和y与Matlab中的xi和yi相同，img是灰度图像。无论如何，我得到了如下例外:“非矩形网格输入z的长度无效”)ValueError:输入z的长度为非矩形网格的无效长度。是什么错了吗?非常感谢。

1 个解决方案

#1

Looking at the documentation you should note a few things:

看看这些文档，你应该注意到以下几点:

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

如果x和y表示一个规则网格，考虑使用RectBivariateSpline。
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.

x和y应该是一维的，不像在Matlab中，它们与Matlab的x和y匹配，而不是与xi和yi。这些将来的。
SciPy will return a function to use to interpolate later. Matlab retuns the interpolated vector immediately.

SciPy将返回一个函数，用于稍后插入。用Matlab对插补矢量进行了快速的调整。

You've used it in Matlab but implicitly assuming that

你在Matlab中已经用过了，但隐含的假设。

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

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

这是你需要传递给插值的X和Y，而不是xi和yi，它们会在后面出现。

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):

在SciPy中，你可以跳过网格的步骤，只需要使用一个正常的范围，但它也会输出一个函数，而不是在点上执行插值(注意，在我的x和y定义中可能有小错误，我不太熟悉arange):

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

and then when you want to do the interpolation:

当你想做插值的时候

gt = f(xi, yi)

#1