在python中使用不同维数的数组进行二维插值

Question

我有三个二维数组，形状为(100,100)。每个数组看起来像:

x =
[[-104.09417725 -104.08866882 -104.0831604  ..., -103.8795166  -103.87399292
-103.86849976]
..., 
[-104.11058044 -104.10507202 -104.09954834 ..., -103.89535522
-103.88983154 -103.88430786]
[-104.11141968 -104.10591125 -104.10038757 ..., -103.89614868 -103.890625
-103.88513184]]
y = 
[[ 40.81712341  40.81744385  40.81776428 ...,  40.82929611  40.82960129
40.82990646]
 ..., 
[ 40.98789597  40.9882164   40.98854065 ...,  41.00011063  41.00041199
41.00072479]]
z =
[[ 1605.58544922  1615.62341309  1624.33911133 ...,  1479.11254883
1478.328125    1476.13378906]
 ..., 
[ 1596.03857422  1600.5690918   1606.30712891 ...,  1598.56982422
1594.90454102  1594.07763672]]

我还有两个一维数组x1和y1。这些x1和y1分别在x和y的范围内，例如:

x1 = [ 104.07794  104.03169  104.03352  104.03584  104.03835  104.04085
104.04334  104.07315  104.07133  104.07635  104.07916  104.0321
104.03481  104.03741  104.04002  104.04366  104.04572  104.04787
...................................................................   
103.92937  103.89825  103.90027  103.90253  103.90352  103.90375
103.89922  103.89931  103.90145  103.90482  103.90885  103.91058
103.91243  103.91525  103.91785  103.92078  103.97814]
y1 = [ 40.9542   40.96922  40.96733  40.96557  40.96377  40.96218  40.96043
40.95446  40.95686  40.95296  40.95184  40.94984  40.94834  40.9469
40.94538  40.94287  40.94154  40.94008  40.93824  40.93705  40.93579
.........................................................................  
40.89675  40.9015   40.90044  40.89948  40.89766  40.89513  40.88374
40.88118  40.87915  40.87933  40.87917  40.878    40.87675  40.87598
40.87515  40.87421  40.91258]

x1和y1对应于(104.07794,40.9542)，(104.03169,40.96922)等索引下的(x1,y1)。这里我想要得到的是z1对应于(x1,y1)内插x,y,z。为此，我编写了如下代码:

x1,y1 = np.meshgrid(x1,y1)
f = interpolate.interp2d(x,y,z,kind='linear')
or
f = interpolate.Rbf(x,y,z,function='linear')
z1 = f(x1,y1)

然而，我不想将x1, y1转换为二维网格，因为这个函数填充了我不喜欢填充的网格点。所以，我想在不转换为二维网格的情况下插值x1,y1，但是这些二维插值方法似乎要求x,y和x1,y1具有相同的维度。有没有办法在不使(x1,y1)和(x,y)维数相等的情况下进行插值?谢谢你！艾萨克

Answer 1

我不太清楚你说的x,y and x1,y1 have same dimesion是什么意思。

我可以构造一个输入数据集:

In [294]: x,y=np.meshgrid(np.arange(10),np.arange(8))    
In [295]: z=x+y
In [296]: f=interpolate.Rbf(x,y,z,kind='linear')
In [297]: x.shape
Out[297]: (8, 10)

我使用meshgrid只是因为它是最简单的方法来生成一对二维数组，使一个合理的表面。因为interp2d不喜欢在这个表面上工作。

我可以定义另一组点作为2个1d数组。点的数量与定义曲面的点的数量或点的布局无关。我只需要给出一个(x1,y1)对它对应于定义曲面的(x,y,z)三元组

In [298]: x1=np.linspace(0,10,15)
In [299]: y1=np.linspace(0,10,15)    
In [300]: f(x1,y1)
Out[300]: 
array([  1.78745907e-13,   1.42752327e+00,   2.85761392e+00,
         4.28560518e+00,   5.71422460e+00,   7.14293770e+00,
         8.57139192e+00,   1.00000000e+01,   1.14285329e+01,
         1.28573610e+01,   1.42852315e+01,   1.57040689e+01,
         1.70924026e+01,   1.84049594e+01,   1.95946258e+01])

x为2d无关;我可以使输入变平。interp2d的文档特别提到了对多维输入执行此操作。

 f1=interpolate.Rbf(x.flatten(), y.flatten(), z.flatten(), kind='linear')

插值点也可以排列成二维形状

f(x1.reshape((3,5)), y1.reshape((3,5)))

相同的插值值，只是排列在(3,5)数组中。

---

interp2d的操作有点不同。"立方"似乎比"线性"更受欢迎(我还没有深入研究为什么):

In [326]: f2=interpolate.interp2d(x,y,z,kind='cubic')    
In [327]: z1=f2(x1,y1)
In [328]: z1.shape
Out[328]: (15, 15)

结果是(x1.shape, y1.shape)——它将x1,y1视为定义了一个类似meshgrid的表面。

但我可以提取对角线，并获得本质上与Rbf相同的值(除了两端):

In [329]: z1.diagonal()
Out[329]: 
array([  7.61803576e-18,   1.42857137e+00,   2.85714294e+00,
         4.28571419e+00,   5.71428543e+00,   7.14285905e+00,
         8.57142306e+00,   1.00000000e+01,   1.14287955e+01,
         1.28551995e+01,   1.41778962e+01,   1.54127554e+01,
         8.99313361e+00,   1.60000000e+01,   1.60000000e+01])

因此，在Rbf中，您指定了要插入值的确切位置，而interp2d则指定了2d空间的x,y坐标。