Python/Scipy 2D 插值(非均匀数据) [英] Python/Scipy 2D Interpolation (Non-uniform Data)
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)
假设我想在一个二维矩形区域上进行插值.我的变量z"包含如下所示的数据.每一列都是一个常数值，但是，数组的每一行可能是不同的值，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)
作为另一种可视化方式，我知道的值数组是:
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，它们应该被插入.我尝试用 None 替换它们，但随后对所有结果都使用了 '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.
我想我需要使用griddata"或interp2".griddata 似乎产生了我期望的结果，但 'interp2' 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 '
'
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
看起来你明白了.
在您的上层代码示例和之前的(linked)问题中，您有结构化数据.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(网格上的所有点都有一个已知值).网格不一定必须具有相同的 dx 和 dy.(如果所有 dx 和 dy 都相等，则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，您当前的问题是问如果不是所有的点都知道该怎么办.这称为非结构化数据.您所拥有的只是一个字段中的点选择.您不一定能构造所有顶点都具有已知值的矩形.对于这种类型的数据，您可以使用(正如您所拥有的)griddata，或一种 BivariateSpline.
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.如果您想使用样条线插入数据，请使用这些.这使您的函数平滑且可微，但您可以获得在 Z.max() 或 Z.min() 之外摆动的曲面.
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().
因为您正在设置 ky=1 和 kx=1 并且得到的结果我很确定只是 结构化 上的线性插值数据，我个人只是从 RectBivariateSpline 样条方案切换到 interp2d 结构化网格插值方案.我知道文档说它是用于 常规网格，但是 __doc__ 中的示例 本身只是结构化，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，我会很好奇您是否发现这些方法之间存在任何显着差异.欢迎使用 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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