gpt4 book ai didi

python - 双三次插值 Python

转载 作者:太空狗 更新时间:2023-10-29 22:28:33 26 4
gpt4 key购买 nike

我开发了双三次插值,用于向一些使用 Python 编程语言的本科生进行演示。

方法如 wikipedia 中所述。 ,代码工作正常,除了我得到的结果与使用 scipy 库时获得的结果略有不同。

插值代码如下所示,在函数bicubic_interpolation中。

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits import mplot3d
from scipy import interpolate
import sympy as syp
import pandas as pd
pd.options.display.max_colwidth = 200
%matplotlib inline

def bicubic_interpolation(xi, yi, zi, xnew, ynew):

# check sorting
if np.any(np.diff(xi) < 0) and np.any(np.diff(yi) < 0) and\
np.any(np.diff(xnew) < 0) and np.any(np.diff(ynew) < 0):
raise ValueError('data are not sorted')

if zi.shape != (xi.size, yi.size):
raise ValueError('zi is not set properly use np.meshgrid(xi, yi)')

z = np.zeros((xnew.size, ynew.size))

deltax = xi[1] - xi[0]
deltay = yi[1] - yi[0]
for n, x in enumerate(xnew):
for m, y in enumerate(ynew):

if xi.min() <= x <= xi.max() and yi.min() <= y <= yi.max():

i = np.searchsorted(xi, x) - 1
j = np.searchsorted(yi, y) - 1

x0 = xi[i-1]
x1 = xi[i]
x2 = xi[i+1]
x3 = x1+2*deltax

y0 = yi[j-1]
y1 = yi[j]
y2 = yi[j+1]
y3 = y1+2*deltay

px = (x-x1)/(x2-x1)
py = (y-y1)/(y2-y1)

f00 = zi[i-1, j-1] #row0 col0 >> x0,y0
f01 = zi[i-1, j] #row0 col1 >> x1,y0
f02 = zi[i-1, j+1] #row0 col2 >> x2,y0

f10 = zi[i, j-1] #row1 col0 >> x0,y1
f11 = p00 = zi[i, j] #row1 col1 >> x1,y1
f12 = p01 = zi[i, j+1] #row1 col2 >> x2,y1

f20 = zi[i+1,j-1] #row2 col0 >> x0,y2
f21 = p10 = zi[i+1,j] #row2 col1 >> x1,y2
f22 = p11 = zi[i+1,j+1] #row2 col2 >> x2,y2

if 0 < i < xi.size-2 and 0 < j < yi.size-2:

f03 = zi[i-1, j+2] #row0 col3 >> x3,y0

f13 = zi[i,j+2] #row1 col3 >> x3,y1

f23 = zi[i+1,j+2] #row2 col3 >> x3,y2

f30 = zi[i+2,j-1] #row3 col0 >> x0,y3
f31 = zi[i+2,j] #row3 col1 >> x1,y3
f32 = zi[i+2,j+1] #row3 col2 >> x2,y3
f33 = zi[i+2,j+2] #row3 col3 >> x3,y3

elif i<=0:

f03 = f02 #row0 col3 >> x3,y0

f13 = f12 #row1 col3 >> x3,y1

f23 = f22 #row2 col3 >> x3,y2

f30 = zi[i+2,j-1] #row3 col0 >> x0,y3
f31 = zi[i+2,j] #row3 col1 >> x1,y3
f32 = zi[i+2,j+1] #row3 col2 >> x2,y3
f33 = f32 #row3 col3 >> x3,y3

elif j<=0:

f03 = zi[i-1, j+2] #row0 col3 >> x3,y0

f13 = zi[i,j+2] #row1 col3 >> x3,y1

f23 = zi[i+1,j+2] #row2 col3 >> x3,y2

f30 = f20 #row3 col0 >> x0,y3
f31 = f21 #row3 col1 >> x1,y3
f32 = f22 #row3 col2 >> x2,y3
f33 = f23 #row3 col3 >> x3,y3


elif i == xi.size-2 or j == yi.size-2:

f03 = f02 #row0 col3 >> x3,y0

f13 = f12 #row1 col3 >> x3,y1

f23 = f22 #row2 col3 >> x3,y2

f30 = f20 #row3 col0 >> x0,y3
f31 = f21 #row3 col1 >> x1,y3
f32 = f22 #row3 col2 >> x2,y3
f33 = f23 #row3 col3 >> x3,y3

px00 = (f12 - f10)/2*deltax
px01 = (f22 - f20)/2*deltax
px10 = (f13 - f11)/2*deltax
px11 = (f23 - f21)/2*deltax

py00 = (f21 - f01)/2*deltay
py01 = (f22 - f02)/2*deltay
py10 = (f31 - f11)/2*deltay
py11 = (f32 - f12)/2*deltay

pxy00 = ((f22-f20) - (f02-f00))/4*deltax*deltay
pxy01 = ((f32-f30) - (f12-f10))/4*deltax*deltay
pxy10 = ((f23-f21) - (f03-f01))/4*deltax*deltay
pxy11 = ((f33-f31) - (f13-f11))/4*deltax*deltay


f = np.array([p00, p01, p10, p11,
px00, px01, px10, px11,
py00, py01, py10, py11,
pxy00, pxy01, pxy10, pxy11])

a = A@f

a = a.reshape(4,4).transpose()
z[n,m] = np.array([1, px, px**2, px**3]) @ a @ np.array([1, py, py**2, py**3])

return z

在函数 bicubic_interpolation 中,输入是 xi= 旧的 x 数据范围,yi= 旧的 y 范围,zi= 网格点 (x,y) 处的旧值,xnewynew 是新的水平数据范围。所有输入都是一维 numpy 数组,除了 zi 是二维 numpy 数组。

我正在测试功能的数据如下所示。我还将结果与 scipy 和真实模型(函数 f)进行比较。

def f(x,y):
return np.sin(np.sqrt(x ** 2 + y ** 2))

x = np.linspace(-6, 6, 11)
y = np.linspace(-6, 6, 11)

xx, yy = np.meshgrid(x, y)

z = f(xx, yy)

x_new = np.linspace(-6, 6, 100)
y_new = np.linspace(-6, 6, 100)

xx_new, yy_new = np.meshgrid(x_new, y_new)

z_new = bicubic_interpolation(x, y, z, x_new, y_new)

z_true = f(xx_new, yy_new)

f_scipy = interpolate.interp2d(x, y, z, kind='cubic')

z_scipy = f_scipy(x_new, y_new)

fig, ax = plt.subplots(2, 2, sharey=True, figsize=(16,12))

img0 = ax[0, 0].scatter(xx, yy, c=z, s=100)
ax[0, 0].set_title('original points')
fig.colorbar(img0, ax=ax[0, 0], orientation='vertical', shrink=1, pad=0.01)

img1 = ax[0, 1].imshow(z_new, vmin=z_new.min(), vmax=z_new.max(), origin='lower',
extent=[x_new.min(), x_new.max(), y_new.max(), y_new.min()])
ax[0, 1].set_title('bicubic our code')
fig.colorbar(img1, ax=ax[0, 1], orientation='vertical', shrink=1, pad=0.01)


img2 = ax[1, 0].imshow(z_scipy, vmin=z_scipy.min(), vmax=z_scipy.max(), origin='lower',
extent=[x_new.min(), x_new.max(), y_new.max(), y_new.min()])
ax[1, 0].set_title('bicubic scipy')
fig.colorbar(img2, ax=ax[1, 0], orientation='vertical', shrink=1, pad=0.01)


img3 = ax[1, 1].imshow(z_true, vmin=z_true.min(), vmax=z_true.max(), origin='lower',
extent=[x_new.min(), x_new.max(), y_new.max(), y_new.min()])
ax[1, 1].set_title('true model')
fig.colorbar(img3, ax=ax[1, 1], orientation='vertical', shrink=1, pad=0.01)

plt.subplots_adjust(wspace=0.05, hspace=0.15)

plt.show()

结果如下图所示:

enter image description here

矩阵 A(内部函数 bicubic_interpolation)如维基百科网站中所述,可以使用以下代码简单地获得:

x = syp.Symbol('x')
y = syp.Symbol('y')
a00, a01, a02, a03, a10, a11, a12, a13 = syp.symbols('a00 a01 a02 a03 a10 a11 a12 a13')
a20, a21, a22, a23, a30, a31, a32, a33 = syp.symbols('a20 a21 a22 a23 a30 a31 a32 a33')

p = a00 + a01*y + a02*y**2 + a03*y**3\
+ a10*x + a11*x*y + a12*x*y**2 + a13*x*y**3\
+ a20*x**2 + a21*x**2*y + a22*x**2*y**2 + a23*x**2*y**3\
+ a30*x**3 + a31*x**3*y + a32*x**3*y**2 + a33*x**3*y**3

px = syp.diff(p, x)
py = syp.diff(p, y)
pxy = syp.diff(p, x, y)

df = pd.DataFrame(columns=['function', 'evaluation'])

for i in range(2):
for j in range(2):
function = 'p({}, {})'.format(j,i)
df.loc[len(df)] = [function, p.subs({x:j, y:i})]
for i in range(2):
for j in range(2):
function = 'px({}, {})'.format(j,i)
df.loc[len(df)] = [function, px.subs({x:j, y:i})]
for i in range(2):
for j in range(2):
function = 'py({}, {})'.format(j,i)
df.loc[len(df)] = [function, py.subs({x:j, y:i})]
for i in range(2):
for j in range(2):
function = 'pxy({}, {})'.format(j,i)
df.loc[len(df)] = [function, pxy.subs({x:j, y:i})]

eqns = df['evaluation'].tolist()
symbols = [a00,a01,a02,a03,a10,a11,a12,a13,a20,a21,a22,a23,a30,a31,a32,a33]
A = syp.linear_eq_to_matrix(eqns, *symbols)[0]
A = np.array(A.inv()).astype(np.float64)

print(df)

print(A)

enter image description here

enter image description here

我想知道 bicubic_interpolation 函数的问题出在哪里,为什么它与 scipy 得到的结果略有不同?非常感谢任何帮助!

最佳答案

不确定为什么维基百科实现没有按预期工作。原因可能是这些值的近似值可能与其站点中解释的不同。

px00 = (f12 - f10)/2*deltax
px01 = (f22 - f20)/2*deltax
px10 = (f13 - f11)/2*deltax
px11 = (f23 - f21)/2*deltax

py00 = (f21 - f01)/2*deltay
py01 = (f22 - f02)/2*deltay
py10 = (f31 - f11)/2*deltay
py11 = (f32 - f12)/2*deltay

pxy00 = ((f22-f20) - (f02-f00))/4*deltax*deltay
pxy01 = ((f32-f30) - (f12-f10))/4*deltax*deltay
pxy10 = ((f23-f21) - (f03-f01))/4*deltax*deltay
pxy11 = ((f33-f31) - (f13-f11))/4*deltax*deltay

但是,我发现this文档有不同的实现,它比维基百科有很好的解释和理解。我使用此实现获得的结果与通过 SciPy 获得的结果非常相似。

def bicubic_interpolation2(xi, yi, zi, xnew, ynew):

# check sorting
if np.any(np.diff(xi) < 0) and np.any(np.diff(yi) < 0) and\
np.any(np.diff(xnew) < 0) and np.any(np.diff(ynew) < 0):
raise ValueError('data are not sorted')

if zi.shape != (xi.size, yi.size):
raise ValueError('zi is not set properly use np.meshgrid(xi, yi)')

z = np.zeros((xnew.size, ynew.size))

deltax = xi[1] - xi[0]
deltay = yi[1] - yi[0]
for n, x in enumerate(xnew):
for m, y in enumerate(ynew):

if xi.min() <= x <= xi.max() and yi.min() <= y <= yi.max():

i = np.searchsorted(xi, x) - 1
j = np.searchsorted(yi, y) - 1

x1 = xi[i]
x2 = xi[i+1]

y1 = yi[j]
y2 = yi[j+1]

px = (x-x1)/(x2-x1)
py = (y-y1)/(y2-y1)

f00 = zi[i-1, j-1] #row0 col0 >> x0,y0
f01 = zi[i-1, j] #row0 col1 >> x1,y0
f02 = zi[i-1, j+1] #row0 col2 >> x2,y0

f10 = zi[i, j-1] #row1 col0 >> x0,y1
f11 = p00 = zi[i, j] #row1 col1 >> x1,y1
f12 = p01 = zi[i, j+1] #row1 col2 >> x2,y1

f20 = zi[i+1,j-1] #row2 col0 >> x0,y2
f21 = p10 = zi[i+1,j] #row2 col1 >> x1,y2
f22 = p11 = zi[i+1,j+1] #row2 col2 >> x2,y2

if 0 < i < xi.size-2 and 0 < j < yi.size-2:

f03 = zi[i-1, j+2] #row0 col3 >> x3,y0

f13 = zi[i,j+2] #row1 col3 >> x3,y1

f23 = zi[i+1,j+2] #row2 col3 >> x3,y2

f30 = zi[i+2,j-1] #row3 col0 >> x0,y3
f31 = zi[i+2,j] #row3 col1 >> x1,y3
f32 = zi[i+2,j+1] #row3 col2 >> x2,y3
f33 = zi[i+2,j+2] #row3 col3 >> x3,y3

elif i<=0:

f03 = f02 #row0 col3 >> x3,y0

f13 = f12 #row1 col3 >> x3,y1

f23 = f22 #row2 col3 >> x3,y2

f30 = zi[i+2,j-1] #row3 col0 >> x0,y3
f31 = zi[i+2,j] #row3 col1 >> x1,y3
f32 = zi[i+2,j+1] #row3 col2 >> x2,y3
f33 = f32 #row3 col3 >> x3,y3

elif j<=0:

f03 = zi[i-1, j+2] #row0 col3 >> x3,y0

f13 = zi[i,j+2] #row1 col3 >> x3,y1

f23 = zi[i+1,j+2] #row2 col3 >> x3,y2

f30 = f20 #row3 col0 >> x0,y3
f31 = f21 #row3 col1 >> x1,y3
f32 = f22 #row3 col2 >> x2,y3
f33 = f23 #row3 col3 >> x3,y3


elif i == xi.size-2 or j == yi.size-2:

f03 = f02 #row0 col3 >> x3,y0

f13 = f12 #row1 col3 >> x3,y1

f23 = f22 #row2 col3 >> x3,y2

f30 = f20 #row3 col0 >> x0,y3
f31 = f21 #row3 col1 >> x1,y3
f32 = f22 #row3 col2 >> x2,y3
f33 = f23 #row3 col3 >> x3,y3

Z = np.array([f00, f01, f02, f03,
f10, f11, f12, f13,
f20, f21, f22, f23,
f30, f31, f32, f33]).reshape(4,4).transpose()

X = np.tile(np.array([-1, 0, 1, 2]), (4,1))
X[0,:] = X[0,:]**3
X[1,:] = X[1,:]**2
X[-1,:] = 1

Cr = Z@np.linalg.inv(X)
R = Cr@np.array([px**3, px**2, px, 1])

Y = np.tile(np.array([-1, 0, 1, 2]), (4,1)).transpose()
Y[:,0] = Y[:,0]**3
Y[:,1] = Y[:,1]**2
Y[:,-1] = 1

Cc = np.linalg.inv(Y)@R

z[n,m]=(Cc@np.array([py**3, py**2, py, 1]))


return z

def f(x,y):
return np.sin(np.sqrt(x ** 2 + y ** 2))

x = np.linspace(-6, 6, 11)
y = np.linspace(-6, 6, 11)

xx, yy = np.meshgrid(x, y)

z = f(xx, yy)

x_new = np.linspace(-6, 6, 100)
y_new = np.linspace(-6, 6, 100)

xx_new, yy_new = np.meshgrid(x_new, y_new)

z_new = bicubic_interpolation2(x, y, z, x_new, y_new)

z_true = f(xx_new, yy_new)

f_scipy = interpolate.interp2d(x, y, z, kind='cubic')

z_scipy = f_scipy(x_new, y_new)

fig, ax = plt.subplots(2, 2, sharey=True, figsize=(16,12))

img0 = ax[0, 0].scatter(xx, yy, c=z, s=100)
ax[0, 0].set_title('original points')
fig.colorbar(img0, ax=ax[0, 0], orientation='vertical', shrink=1, pad=0.01)

img1 = ax[0, 1].imshow(z_new, vmin=z_new.min(), vmax=z_new.max(), origin='lower',
extent=[x_new.min(), x_new.max(), y_new.max(), y_new.min()])
ax[0, 1].set_title('bicubic our code')
fig.colorbar(img1, ax=ax[0, 1], orientation='vertical', shrink=1, pad=0.01)


img2 = ax[1, 0].imshow(z_scipy, vmin=z_scipy.min(), vmax=z_scipy.max(), origin='lower',
extent=[x_new.min(), x_new.max(), y_new.max(), y_new.min()])
ax[1, 0].set_title('bicubic scipy')
fig.colorbar(img2, ax=ax[1, 0], orientation='vertical', shrink=1, pad=0.01)


img3 = ax[1, 1].imshow(z_true, vmin=z_true.min(), vmax=z_true.max(), origin='lower',
extent=[x_new.min(), x_new.max(), y_new.max(), y_new.min()])
ax[1, 1].set_title('true model')
fig.colorbar(img3, ax=ax[1, 1], orientation='vertical', shrink=1, pad=0.01)

plt.subplots_adjust(wspace=0.05, hspace=0.15)

plt.show()

enter image description here

fig, ax = plt.subplots(1, 2, sharey=True, figsize=(10, 6))

ax[0].plot(xx[0,:], z[5,:], 'or', label='original')
ax[0].plot(xx_new[0,:], z_true[int(100/10*5),:], label='true')
ax[0].plot(xx_new[0,:], z_new[int(100/10*5), :], label='our interpolation')

ax[1].plot(xx[0,:], z[5,:], 'or', label='original')
ax[1].plot(xx_new[0,:], z_true[int(100/10*5),:], label='true')
ax[1].plot(xx_new[0,:], z_scipy[int(100/10*5), :], label='scipy interpolation')


for axes in ax:
axes.legend()
axes.grid()


plt.show()

enter image description here

关于python - 双三次插值 Python,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52700878/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com