3

我正在尝试编写一段非常简单的代码，该代码将从一组现有数据中进行插值，以创建值的综合分布。

到目前为止，我的代码如下所示：

import pandas as pd
import numpy as np
import scipy
from scipy.interpolate import griddata
import matplotlib

CRN_data=pd.read_table('disequilibrium data.dat',sep=',')
kzz=CRN_data['Kzz']
temperature=CRN_data['Temperature']
degree=CRN_data['Mean Degree']
points=np.ndarray(shape=(len(kzz),2),dtype='float')
for i in range(len(kzz)):
    points[i][0]=kzz[i]
    points[i][1]=temperature[i]
gridx,gridy= np.mgrid[0:1:100j,0:1:200j]
grid=griddata(points,degree,(gridx,gridy),method='cubic')
print grid

我从中插值的数据集如下所示：

Kzz,Temperature,Mean Degree,   
1.00E+06,400,7.41E+18
1.00E+06,500,4.48E+23
...
1.00E+08,400,4.67E+18
1.00E+08,500,6.88E+23
1.00E+08,750,1.88E+34
...
1.00E+10,750,2.73E+33
1.00E+10,900,2.82E+37
1.00E+10,1000,1.19E+39
...

但是，当代码运行时，我得到的主要输出是

[[ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]
 ..., 
 [ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]]

这显然不是很有帮助。这是 Scipy 中的错误，还是（更有可能）我做错了什么？

4

2 回答 2

2

您正在接收nan值，因为您请求的点包含在输入点的凸包中gridx并且位于 中的凸包之外。您可以指定 a用于外推点，但您可能会考虑重新指定分配的限制并产生有意义的结果，例如：gridypointsfill_valuegridxgridy

import pandas as pd
import numpy as np
from scipy.interpolate import griddata
import matplotlib.pyplot as plt

CRN_data = pd.DataFrame([
[1.00E+06,400,7.41E+18],
[1.00E+06,500,4.48E+23],
[1.00E+08,400,4.67E+18],
[1.00E+08,500,6.88E+23],
[1.00E+08,750,1.88E+34],
[1.00E+10,750,2.73E+33],
[1.00E+10,900,2.82E+37],
[1.00E+10,1000,1.19E+39]],
columns=['Kzz','Temperature','Mean Degree'])

kzz = CRN_data['Kzz']
temperature = CRN_data['Temperature']
degree = CRN_data['Mean Degree']

points = np.matrix([[kzz[i], temperature[i]] for i in range(len(kzz))])

gridx, gridy = np.mgrid[kzz.min():kzz.max():100j,temperature.min():temperature.max():200j]

grid = griddata(points, degree, (gridx, gridy), method='cubic')

产量：

[[7.41000000e+18 1.35147259e+22 2.70220418e+22 ...            nan
             nan            nan]
 [           nan 1.07878728e+33 1.26216288e+33 ...            nan
             nan            nan]
 [           nan            nan 1.38255505e+35 ...            nan
             nan            nan]
 ...
 [           nan            nan            nan ... 1.16569048e+39
             nan            nan]
 [           nan            nan            nan ... 1.16394396e+39
  1.17798560e+39            nan]
 [           nan            nan            nan ... 1.16129655e+39
  1.17564827e+39 1.19000000e+39]]

并绘制：

在此处输入图像描述

于 2018-10-29T20:18:49.940 回答
0

将源数据的范围与您的目标网格进行比较。从我看到的源范围 x=1e6:1e10 和 y=400:1000 而你的网格是 0-1（x 和 y）。在这种情况下，“目标”在源数据之外，使用“线性”或“立方”会给你 NaN - 尝试“最近”，Nan 会消失。

于 2018-10-29T19:53:51.937 回答