问题描述
这是我上一篇文章的后续问题: Python/Scipy插值(map_coordinates)
This is a follow-up question to my previous post: Python/Scipy Interpolation (map_coordinates)
假设我要在2d矩形区域内插值.我的变量"z"包含如下所示的数据.每列都是一个恒定值，但是，数组的每一行都可能具有不同的值，如下面的注释所示.
Let's say I want to interpolate over a 2d rectangular area. My variable 'z' contains the data as shown below. Each column is at a constant value, however, each row of the array may be at a different value as shown in the comment below.
from scipy import interpolate
from numpy import array
import numpy as np
# # 0.0000, 0.1750, 0.8170, 1.0000
z = array([[-2.2818,-2.2818,-0.9309,-0.9309], # 0.0000, 0.0000, 0.0000, 0.0000
[-2.2818,-2.2818,-0.9309,-0.9309], # 0.2620, 0.2784, 0.3379, 0.3526
[-1.4891,-1.4891,-0.5531,-0.5531], # 0.6121, 0.6351, 0.7118, 0.7309
[-1.4891,-1.4891,-0.5531,-0.5531]]) # 1.0000, 1.0000, 1.0000, 1.0000
# Rows, Columns = z.shape
cols = array([0.0000, 0.1750, 0.8170, 1.0000])
rows = array([0.0000, 0.2620, 0.6121, 1.0000])
sp = interpolate.RectBivariateSpline(rows, cols, z, kx=1, ky=1, s=0)
xi = np.array([0.00000, 0.26200, 0.27840, 0.33790, 0.35260, 0.61210, 0.63510,
0.71180, 0.73090, 1.00000], dtype=np.float)
yi = np.array([0.000, 0.167, 0.815, 1.000], dtype=np.float)
print sp(xi, yi)
另一种可视化方法，我知道的值数组为:
As another way of visualizing this, the array of values I KNOW would be:
rows = array([0.0000, 0.2620, 0.2784, 0.3379, 0.3526,
0.6121, 0.6351, 0.7118, 0.7309, 1.0000])
# # 0.0000, 0.1750, 0.8170, 1.0000
z = array([[-2.2818,-2.2818,-0.9309,-0.9309], # 0.0000
[-2.2818, ?, ?, ?], # 0.2620,
[ ?,-2.2818, ?, ?], # 0.2784
[ ?, ?,-0.9309, ?], # 0.3379
[ ? ,?, ?,-0.9309], # 0.3526
[-1.4891, ?, ?, ?], # 0.6121
[ ?,-1.4891, ?, ?], # 0.6351
[ ?, ?,-0.5531, ?], # 0.7118
[ ?, ?, ?,-0.5531], # 0.7309
[-1.4891,-1.4891,-0.5531,-0.5531]]) # 1.0000
我不知道'?'值，并且应该对其进行插值.我尝试将它们替换为无"，但是所有结果都得到"nan".
I do not know the '?' values, and they should be interpolated. I tried replacing them with None, but then get 'nan' for all of my results.
我认为我需要使用'griddata'或'interp2'. griddata似乎产生了我期望的结果，但是'interp2'却没有.
I think I need to use either 'griddata' or 'interp2'. griddata seems to produce the result I expect, but 'interp2' does not.
from scipy import interpolate
from numpy import array
import numpy as np
z = array([[-2.2818,-2.2818,-0.9309,-0.9309],
[-2.2818,-2.2818,-0.9309,-0.9309],
[-1.4891,-1.4891,-0.5531,-0.5531],
[-1.4891,-1.4891,-0.5531,-0.5531]])
rows = array([0.0000, 0.0000, 0.0000, 0.0000,
0.2620, 0.2784, 0.3379, 0.3526,
0.6121, 0.6351, 0.7118, 0.7309,
1.0000, 1.0000, 1.0000, 1.0000])
cols = array([0.0000, 0.1750, 0.8180, 1.0000,
0.0000, 0.1750, 0.8180, 1.0000,
0.0000, 0.1750, 0.8180, 1.0000,
0.0000, 0.1750, 0.8180, 1.0000])
xi = array([0.0000, 0.2620, 0.2784, 0.3379, 0.3526, 0.6121, 0.6351, 0.7118,
0.7309, 1.0000], dtype=np.float)
yi = array([0.000, 0.175, 0.818, 1.000], dtype=np.float)
GD = interpolate.griddata((rows, cols), z.ravel(),
(xi[None,:], yi[:,None]), method='linear')
I2 = interpolate.interp2d(rows, cols, z, kind='linear')
print GD.reshape(4, 10).T
print '\n'
print I2(xi, yi).reshape(4, 10).T
import matplotlib.pyplot as plt
import numpy.ma as ma
plt.figure()
GD = interpolate.griddata((rows.ravel(), cols.ravel()), z.ravel(),
(xi[None,:], yi[:,None]), method='linear')
CS = plt.contour(xi,yi,GD,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,GD,15,cmap=plt.cm.jet)
plt.colorbar()
plt.scatter(rows,cols,marker='o',c='b',s=5)
plt.figure()
I2 = I2(xi, yi)
CS = plt.contour(xi,yi,I2,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,I2,15,cmap=plt.cm.jet)
plt.colorbar()
plt.scatter(rows,cols,marker='o',c='b',s=5)
plt.show()
推荐答案
看起来很像.
在您的上层代码示例和上一个(已链接)问题中， 结构化数据.可以使用RectBivariateSpline或interp2d进行插值.这意味着您拥有可以在网格上描述的数据(网格上的所有点都具有已知值).网格不一定必须具有相同的dx和dy. (如果dx和dy都相等，那么您将拥有一个常规网格)
In your upper code example and in your previous (linked) question you have structured data. Which can be interpolated using RectBivariateSpline or interp2d. This means you have data that can be described on a grid (all points on the grid have a known value). The grid doesn't necessarily have to have all the same dx and dy. (if all dx's and dy's were equal, you'd have a Regular Grid)
现在，您当前的问题是如果不是所有要点都知道怎么办.这称为非结构化数据.您所拥有的只是字段中选择的点.您不一定必须构造所有顶点都具有已知值的矩形.对于这种类型的数据，您可以(根据需要)使用griddata或BivariateSpline风格.
Now, your current question asks what to do if not all the points are known. This is known as unstructured data. All you have are a selection of points in a field. You can't necessarily construct rectangles where all vertices have known values. For this type of data, you can use (as you have) griddata, or a flavor of BivariateSpline.
现在选择哪个?
与结构化RectBivariateSpline最接近的类比是非结构化 BivariateSpline 类:SmoothBivariateSpline或LSQBivariateSpline.如果要使用样条线对数据进行插值，请使用这些样条线.这样可以使您的函数平滑且易于区分，但是您可以获得的曲面可以在Z.max()或Z.min()之外摆动.
The nearest analogy to the structured RectBivariateSpline is one of the unstructured BivariateSpline classes: SmoothBivariateSpline or LSQBivariateSpline. If you want to use splines to interpolate the data, go with these. this makes your function smooth and differentiable, but you can get a surface that swings outside Z.max() or Z.min().
由于您正在设置ky=1和kx=1，并且可以确定我只是确定对结构化数据进行线性插值，所以我个人只是从RectBivariateSpline样条线切换 interp2d 结构化网格插值方案.我知道文档说它是用于常规网格的，但是__doc__本身的示例只是结构化的，而不是常规的.
Since you are setting ky=1 and kx=1 and are getting what I am pretty sure is just linear interpolation on the structured data, I'd personally just switch from the RectBivariateSpline spline scheme to the interp2d structured grid interpolation scheme. I know the documentation says it is for regular grids, but the example in the __doc__ itself is only structured, not regular.
如果您最终切换时发现方法之间有任何显着差异，我会很好奇.欢迎来到SciPy.
I'd be curious if you found any significant differences between the methods if you do end up switching. Welcome to SciPy.
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