无法在Python scipy.interp中解释MATLAB的interp2d。

The following code is just to understand the context. My question doesn't require much understanding of this. It need a simple translation of one line of MATLAB code in to Python.

下面的代码只是为了理解上下文。我的问题不需要太多的理解。它需要简单地将一行MATLAB代码转换成Python代码。

us = np.linspace(-(1023)/2,(1023)/2,1024)
vs = np.linspace(-(1023)/2,(1023)/2,1024)
uu,vv = np.meshgrid(-us,vs)
pu = ((((rx*SDD)/(ry+SOD))+us[0])/(-du))+1

xs = np.linspace(-(360-1)/2,(nx-1)/2,360)
ys = np.linspace(-(360-1)/2,(ny-1)/2,360)
zs = np.linspace(-(360-1)/2,(ny-1)/2,360)

xx,yy = np.meshgrid(xs,ys)

angle_rad = np.linspace(0,359,360)
angle_rad = angle_rad*np.pi/180

for i in range(0,360) :
    vol = np.zeros((360,360,360))
    rx = xx*np.cos(angle_rad[i]-np.pi/2) + yy*np.sin(angle_rad[i]-np.pi/2)
    ry = -xx*np.sin(angle_rad[i]-np.pi/2) + yy*np.cos(angle_rad[i]-np.pi/2)
    pu = ((((rx*370)/(ry+9))+us[0])/(-51.2/1024))+1

    for iz in range(0,360) :
        pv = ((((zs[iz]*370)/(ry+9))-vs[0])/(51.2/1024)) +1

So after this step the code should do interpolation and in MATLAB it's like this:

在这一步之后代码应该做插值在MATLAB中它是这样的

vol(:,:,iz) = interp2(uu,vv ,proj',pu,pv,'linear'); This is in MATLAB

My proj, uu and vv are (1024,1024) and pu, pv are (360,360). I need to convert the above line to Python. I tried using scipy.interpolate but it gives the following errors on trying these :

我的proj, uu和vv是(1024,1024)和pu, pv是(360 360)。我需要将上面的代码转换为Python。我试过使用scipy。内插，但它给出了如下错误:

vol[:,:,iz] = Ratio*(interp2d(uu,vv,proj,pu,pv,'cubic'))

TypeError: unhashable type: 'numpy.ndarray'

TypeError:unhashable类型:“numpy.ndarray”

vol[:,:,iz] = Ratio*(interp2d(uu,vv,proj,'cubic'))

OverflowError: Too many data points to interpolate

OverflowError:过多的数据点来插入。

vol[:,:,iz] = Ratio*(interp2d(proj,pu,pv,'cubic'))

ValueError: x and y must have equal lengths for non rectangular grid

ValueError: x和y对非矩形网格的长度必须相等。

vol[:,:,iz] = Ratio*(interp2d(pu,pv,proj,'cubic'))

ValueError: Invalid length for input z for non rectangular grid

ValueError:非矩形网格输入z的长度无效。

I have read all the scipy.interpolate documentations and none seemed to help. Could anyone figure out what's wrong?

我已经阅读了所有的scipy.插值文档，似乎没有任何帮助。谁能弄明白是怎么回事?

1 个解决方案

#1

The problem on a wide scale is that you're looking at the documentation but you're not trying to understand it. If you're using a specific function interp2d in the module scipy.interpolate, then look at the function's documentation, as @pv also suggested in a comment. The fact that you're throwing the arguments all around the place clearly demonstrates that you're trying to use a function based on guesswork. It won't work: a function was implemented with a given syntax, and it will only work like that.

大规模的问题是你在看文档，但你并没有试图去理解它。如果您在模块scipy. insert中使用一个特定的函数interp2d，那么查看函数的文档，就像@pv在注释中建议的那样。事实是，你把所有的论点都抛在了脑后，这清楚地表明你在试图利用一种基于猜测的函数。它不会起作用:一个函数是用给定的语法实现的，它只会这样工作。

So look at the function's signature:

所以看看这个函数的签名:

class scipy.interpolate.interp2d(x, y, z, kind='linear', copy=True, bounds_error=False, fill_value=nan)

类scipy.插值.interp2d(x, y, z, kind='线性'，copy=True, bounds_error=False, fill_value=nan)

The meaning of the parameters is explained afterwards. You can clearly see that there are 3 mandatory parameters: x, y, z. No other array-valued inputs are allowed This is because interp2d only constructs an interpolating function, which you should then use to compute the interpolated values on a mesh (unlike MATLAB, where interp2d gives you the interpolated values directly). So you can call

之后解释参数的含义。你可以清楚地看到有3强制参数:x,y,z。不允许有其他数组值输入这是因为interp2d只构造插值函数,然后你应该使用计算插值网格(不像MATLAB interp2d直接给你插入的值)。所以你可以叫

myfun = interp2(uu,vv,proj,'linear')

to get an interpolating function. You can then substitute at the given values, but note that the function myfun will expect 1d input, and it will construct a mesh internally. So assuming that your output mesh is constructed as

得到一个插值函数。然后可以在给定的值上进行替换，但是注意函数myfun将期望1d输入，并且它将在内部构造一个网格。假设你的输出网格被构造成。

puu,puv = np.meshgrid(puu_vec,puv_vec)

(which is probably not the case, but I'll get back to this later), you need

(很可能不是这样，但我稍后再讲)，你需要。

vol[:,:,iz] = Ratio*myfun(puu_vec,puv_vec)

to obtain the output you need.

获得所需的输出。

However, there are some important points you should note.

然而，你应该注意到一些重要的观点。

In MATLAB you have interp2d(uu,vv,proj',...), but make sure that for scipy the elements of uu, vv, and proj are in the same order. To be on the safe side: in case of an asymmetric mesh size, the shape of uu, vv and proj should all be the same.
在MATLAB中，你有interp2d(uu,vv,proj'，…)，但要确保uu,vv和proj的元素都是相同的顺序。为了安全起见:如果不对称的网目尺寸，uu, vv和proj的形状都应该是相同的。
In MATLAB you're using 'linear' interpolation, while in python you're using 'cubic'. I'm not sure this is really what you want, if you're porting your code to a new language.
在MATLAB中，你使用的是“线性”插值，而在python中，你使用的是“立方”。我不确定这是否是您想要的，如果您将您的代码移植到一种新的语言中。
It seems to me that your output mesh is not defined by a rectangular grid as if from meshgrid, which suggests that interp2d might not be suitable for your case. Anyway, I've had some odd experiences with interp2d, and I wouldn't trust it. So if it's not suitable for your expected output, I'd strongly suggest using scipy.interpolate.griddata instead. This function gives you interpolated points directly: I suggest that you try to figure out its use based on the manual:) My linked answer can also help. You can set the kind of interpolation in the same way, but your output can be any set of scattered points if you like.
在我看来，你的输出网格不是由一个矩形网格定义的，就像网格网格一样，这意味着interp2d可能不适合你的情况。不管怎样，我在interp2d上有过一些奇怪的经历，我不会相信它。因此，如果它不适合您的预期输出，我强烈建议使用scipy. insert .griddata代替。这个函数直接给你插入点:我建议你试着根据手册找出它的用法:)我的链接答案也可以帮助你。你可以用同样的方法来设置插值，但是如果你喜欢的话，你的输出可以是任意的散点集。

#1