如何将 python 中的二维插值函数显示为矩阵?
How to display a 2d interpolation function in python as a matrix?
我环顾四周,但很难找到答案。基本上,当一个插值 v -> w 时,您通常会使用许多插值函数之一。但是我想得到对应的矩阵Av = w.
在我的例子中,w 是一个 200x200 矩阵,其中 v 是 w 的随机子集,具有一半的点数。我真的不喜欢花哨的数学,它可以像按距离平方加权已知点一样简单。我已经尝试过用一些 for 循环来实现它,但它只适用于小值。但也许它有助于解释我的问题。
from random import sample
def testScatter(xbig, ybig):
NumberOfPoints = int(xbig * ybig / 2) #half as many points as in full Sample
#choose random coordinates
Index = sample(range(xbig * ybig),NumberOfPoints)
IndexYScatter = np.remainder(Index, xbig)
IndexXScatter = np.array((Index - IndexYScatter) / xbig, dtype=int)
InterpolationMatrix = np.zeros((xbig * ybig , NumberOfPoints), dtype=np.float32)
WeightingSum = np.zeros(xbig * ybig )
coordsSamplePoints = []
for i in range(NumberOfPoints): #first set all the given points (no need to interpolate)
coordsSamplePoints.append(IndexYScatter[i] + xbig * IndexXScatter[i])
InterpolationMatrix[coordsSamplePoints[i], i] = 1
WeightingSum[coordsSamplePoints[i]] = 1
for x in range(xbig * ybig): #now comes the interpolation
if x not in coordsSamplePoints:
YIndexInterpol = x % xbig #xcoord in interpolated matrix
XIndexInterpol = (x - YIndexInterpol) / xbig #ycoord in interp. matrix
for y in range(NumberOfPoints):
XIndexScatter = IndexXScatter[y]
YIndexScatter = IndexYScatter[y]
distanceSquared = (np.float32(YIndexInterpol) - np.float32(YIndexScatter))**2+(np.float32(XIndexInterpol) - np.float32(XIndexScatter))**2
InterpolationMatrix[x,y] = 1/distanceSquared
WeightingSum[x] += InterpolationMatrix[x,y]
return InterpolationMatrix/ WeightingSum[:,None] , IndexXScatter, IndexYScatter
您需要花一些时间阅读 Numpy 文档,从 this page 的顶部开始,然后逐步阅读。 研究 此处关于 SO 的答案,询问在使用 Numpy 数组时如何对操作进行矢量化的问题会对您有所帮助。如果您发现您正在迭代索引并使用 Numpy 数组执行计算,那么可能有更好的方法。
第一次剪...
第一个for循环可以替换为:
coordsSamplePoints = IndexYScatter + (xbig * IndexXScatter)
InterpolationMatrix[coordsSamplePoints,np.arange(coordsSamplePoints.shape[0])] = 1
WeightingSum[coordsSamplePoints] = 1
这主要利用了 elementwise arithmetic and Index arrays - 应该阅读完整的索引教程
您可以通过增强功能并执行 for 循环以及 Numpy 方式 然后比较结果来测试它。
...
IM = InterpolationMatrix.copy()
WS = WeightingSum.copy()
for i in range(NumberOfPoints): #first set all the given points (no need to interpolate)
coordsSamplePoints.append(IndexYScatter[i] + xbig * IndexXScatter[i])
InterpolationMatrix[coordsSamplePoints[i], i] = 1
WeightingSum[coordsSamplePoints[i]] = 1
cSS = IndexYScatter + (xbig * IndexXScatter)
IM[cSS,np.arange(cSS.shape[0])] = 1
WS[cSS] = 1
# TEST Validity
print((cSS == coordsSamplePoints).all(),
(IM == InterpolationMatrix).all(),
(WS == WeightingSum).all())
...
外循环:
...
for x in range(xbig * ybig): #now comes the interpolation
if x not in coordsSamplePoints:
YIndexInterpol = x % xbig #xcoord in interpolated matrix
XIndexInterpol = (x - YIndexInterpol) / xbig #ycoord in interp. matrix
...
可以替换为:
...
space = np.arange(xbig * ybig)
mask = ~(space == cSS[:,None]).any(0)
iP = space[mask] # points to interpolate
yIndices = iP % xbig
xIndices = (iP - yIndices) / xbig
...
完整的解决方案:
import random
import numpy as np
def testScatter(xbig, ybig):
NumberOfPoints = int(xbig * ybig / 2) #half as many points as in full Sample
#choose random coordinates
Index = random.sample(range(xbig * ybig),NumberOfPoints)
IndexYScatter = np.remainder(Index, xbig)
IndexXScatter = np.array((Index - IndexYScatter) / xbig, dtype=int)
InterpolationMatrix = np.zeros((xbig * ybig , NumberOfPoints), dtype=np.float32)
WeightingSum = np.zeros(xbig * ybig )
coordsSamplePoints = IndexYScatter + (xbig * IndexXScatter)
InterpolationMatrix[coordsSamplePoints,np.arange(coordsSamplePoints.shape[0])] = 1
WeightingSum[coordsSamplePoints] = 1
IM = InterpolationMatrix
cSS = coordsSamplePoints
WS = WeightingSum
space = np.arange(xbig * ybig)
mask = ~(space == cSS[:,None]).any(0)
iP = space[mask] # points to interpolate
yIndices = iP % xbig
xIndices = (iP - yIndices) / xbig
dSquared = ((yIndices[:,None] - IndexYScatter) ** 2) + ((xIndices[:,None] - IndexXScatter) ** 2)
IM[iP,:] = 1/dSquared
WS[iP] = IM[iP,:].sum(1)
return IM / WS[:,None], IndexXScatter, IndexYScatter
与你的原始参数相比,我得到了大约 200 倍的改进,参数为 (100,100)。可能还有其他一些小的改进,但它们不会显着影响执行时间。
Broadcasting 是另一个 必须 的 Numpy 技能。
我环顾四周,但很难找到答案。基本上,当一个插值 v -> w 时,您通常会使用许多插值函数之一。但是我想得到对应的矩阵Av = w.
在我的例子中,w 是一个 200x200 矩阵,其中 v 是 w 的随机子集,具有一半的点数。我真的不喜欢花哨的数学,它可以像按距离平方加权已知点一样简单。我已经尝试过用一些 for 循环来实现它,但它只适用于小值。但也许它有助于解释我的问题。
from random import sample
def testScatter(xbig, ybig):
NumberOfPoints = int(xbig * ybig / 2) #half as many points as in full Sample
#choose random coordinates
Index = sample(range(xbig * ybig),NumberOfPoints)
IndexYScatter = np.remainder(Index, xbig)
IndexXScatter = np.array((Index - IndexYScatter) / xbig, dtype=int)
InterpolationMatrix = np.zeros((xbig * ybig , NumberOfPoints), dtype=np.float32)
WeightingSum = np.zeros(xbig * ybig )
coordsSamplePoints = []
for i in range(NumberOfPoints): #first set all the given points (no need to interpolate)
coordsSamplePoints.append(IndexYScatter[i] + xbig * IndexXScatter[i])
InterpolationMatrix[coordsSamplePoints[i], i] = 1
WeightingSum[coordsSamplePoints[i]] = 1
for x in range(xbig * ybig): #now comes the interpolation
if x not in coordsSamplePoints:
YIndexInterpol = x % xbig #xcoord in interpolated matrix
XIndexInterpol = (x - YIndexInterpol) / xbig #ycoord in interp. matrix
for y in range(NumberOfPoints):
XIndexScatter = IndexXScatter[y]
YIndexScatter = IndexYScatter[y]
distanceSquared = (np.float32(YIndexInterpol) - np.float32(YIndexScatter))**2+(np.float32(XIndexInterpol) - np.float32(XIndexScatter))**2
InterpolationMatrix[x,y] = 1/distanceSquared
WeightingSum[x] += InterpolationMatrix[x,y]
return InterpolationMatrix/ WeightingSum[:,None] , IndexXScatter, IndexYScatter
您需要花一些时间阅读 Numpy 文档,从 this page 的顶部开始,然后逐步阅读。 研究 此处关于 SO 的答案,询问在使用 Numpy 数组时如何对操作进行矢量化的问题会对您有所帮助。如果您发现您正在迭代索引并使用 Numpy 数组执行计算,那么可能有更好的方法。
第一次剪...
第一个for循环可以替换为:
coordsSamplePoints = IndexYScatter + (xbig * IndexXScatter)
InterpolationMatrix[coordsSamplePoints,np.arange(coordsSamplePoints.shape[0])] = 1
WeightingSum[coordsSamplePoints] = 1
这主要利用了 elementwise arithmetic and Index arrays - 应该阅读完整的索引教程
您可以通过增强功能并执行 for 循环以及 Numpy 方式 然后比较结果来测试它。
...
IM = InterpolationMatrix.copy()
WS = WeightingSum.copy()
for i in range(NumberOfPoints): #first set all the given points (no need to interpolate)
coordsSamplePoints.append(IndexYScatter[i] + xbig * IndexXScatter[i])
InterpolationMatrix[coordsSamplePoints[i], i] = 1
WeightingSum[coordsSamplePoints[i]] = 1
cSS = IndexYScatter + (xbig * IndexXScatter)
IM[cSS,np.arange(cSS.shape[0])] = 1
WS[cSS] = 1
# TEST Validity
print((cSS == coordsSamplePoints).all(),
(IM == InterpolationMatrix).all(),
(WS == WeightingSum).all())
...
外循环:
...
for x in range(xbig * ybig): #now comes the interpolation
if x not in coordsSamplePoints:
YIndexInterpol = x % xbig #xcoord in interpolated matrix
XIndexInterpol = (x - YIndexInterpol) / xbig #ycoord in interp. matrix
...
可以替换为:
...
space = np.arange(xbig * ybig)
mask = ~(space == cSS[:,None]).any(0)
iP = space[mask] # points to interpolate
yIndices = iP % xbig
xIndices = (iP - yIndices) / xbig
...
完整的解决方案:
import random
import numpy as np
def testScatter(xbig, ybig):
NumberOfPoints = int(xbig * ybig / 2) #half as many points as in full Sample
#choose random coordinates
Index = random.sample(range(xbig * ybig),NumberOfPoints)
IndexYScatter = np.remainder(Index, xbig)
IndexXScatter = np.array((Index - IndexYScatter) / xbig, dtype=int)
InterpolationMatrix = np.zeros((xbig * ybig , NumberOfPoints), dtype=np.float32)
WeightingSum = np.zeros(xbig * ybig )
coordsSamplePoints = IndexYScatter + (xbig * IndexXScatter)
InterpolationMatrix[coordsSamplePoints,np.arange(coordsSamplePoints.shape[0])] = 1
WeightingSum[coordsSamplePoints] = 1
IM = InterpolationMatrix
cSS = coordsSamplePoints
WS = WeightingSum
space = np.arange(xbig * ybig)
mask = ~(space == cSS[:,None]).any(0)
iP = space[mask] # points to interpolate
yIndices = iP % xbig
xIndices = (iP - yIndices) / xbig
dSquared = ((yIndices[:,None] - IndexYScatter) ** 2) + ((xIndices[:,None] - IndexXScatter) ** 2)
IM[iP,:] = 1/dSquared
WS[iP] = IM[iP,:].sum(1)
return IM / WS[:,None], IndexXScatter, IndexYScatter
与你的原始参数相比,我得到了大约 200 倍的改进,参数为 (100,100)。可能还有其他一些小的改进,但它们不会显着影响执行时间。
Broadcasting 是另一个 必须 的 Numpy 技能。