双三次插值Python
我开发了Bicubic插值法，以使用Python编程语言向一些本科生进行演示。
方法如wikipedia中所述， 该代码工作正常，除了我得到的结果与使用scipy库时获得的结果略有不同。
插值代码显示在函数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，输入为xi =旧的x数据范围，yi =旧的y范围，zi =网格点（x，y的旧值） ），xnew和ynew是新的水平数据范围。除了zi是2D numpy数组之外，所有输入都是一维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()
结果如下所示：
矩阵A（内部函数bicubic_interpolation）如Wikipedia网站中所述，可以使用以下代码轻松获得：
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)
我想知道bicubic_interpolation函数的问题在哪里，为什么它与scipy得到的结果略有不同？
任何帮助将不胜感激！
3 个答案:
答案 0 :(得分：6)
不确定为什么Wikipedia的实施无法正常工作。可能的原因是，这些值可能以不同于其所在位置说明的方式近似。
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文档的实现方式有所不同，与Wikipedia相比，它得到了很好的解释和理解。我使用此实现获得的结果与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()
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()
答案 1 :(得分：1)
Khalil Al Hooti 的升级解决方案。我希望它会更好。
def BiCubicInterp(X, Y, Z, h = 0.01):
new_Z = []
new_X = []
new_Y = []
new_n_x = int((X[1] - X[0]) / h)
new_n_y = int((Y[1] - Y[0]) / h)
count_X = len(X)
count_Y = len(Y)
X_m = np.array([[-1, 0, 1, 8],
[1, 0, 1, 4],
[-1, 0, 1, 2],
[1, 1, 1, 1]])
Y_m = np.array([[-1, 1, -1, 1],
[0, 0, 0, 1],
[1, 1, 1, 1],
[8, 4, 2, 1]])
X_m = np.linalg.inv(X_m)
Y_m = np.linalg.inv(Y_m)
for i in range(1, count_X):
px = X[i - 1]
k = i - 1
for s in range(new_n_x):
for j in range(1, count_Y):
py = Y[j - 1]
l = j - 1
for r in range(new_n_y):
x1 = X[k]
x2 = X[k+1]
y1 = Y[l]
y2 = Y[l+1]
P_x = (px - x1)/(x2 - x1)
P_y = (py - y1)/(y2 - y1)
f00 = Z[(count_Y + l-1) % count_Y, (count_X + k - 1) % count_X]
f01 = Z[(count_Y + l-1) % count_Y, (count_X + k) % count_X]
f02 = Z[(count_Y + l-1) % count_Y, (count_X + k + 1) % count_X]
f03 = Z[(count_Y + l-1) % count_Y, (count_X + k + 2) % count_X]
f10 = Z[(count_Y + l) % count_Y, (count_X + k - 1) % count_X]
f11 = Z[(count_Y + l) % count_Y, (count_X + k) % count_X]
f12 = Z[(count_Y + l) % count_Y, (count_X + k + 1) % count_X]
f13 = Z[(count_Y + l) % count_Y, (count_X + k + 2) % count_X]
f20 = Z[(count_Y + l + 1) % count_Y, (count_X + k - 1) % count_X]
f21 = Z[(count_Y + l + 1) % count_Y, (count_X + k) % count_X]
f22 = Z[(count_Y + l + 1) % count_Y, (count_X + k + 1) % count_X]
f23 = Z[(count_Y + l + 1) % count_Y, (count_X + k + 2) % count_X]
f30 = Z[(count_Y + l + 2) % count_Y, (count_X + k - 1) % count_X]
f31 = Z[(count_Y + l + 2) % count_Y, (count_X + k) % count_X]
f32 = Z[(count_Y + l + 2) % count_Y, (count_X + k + 1) % count_X]
f33 = Z[(count_Y + l + 2) % count_Y, (count_X + k + 2) % count_X]
Z_m = np.array([[f00, f01, f02, f03],
[f10, f11, f12, f13],
[f20, f21, f22, f23],
[f30, f31, f32, f33]])
Cr = np.dot(Z_m, X_m)
R = np.dot(Cr, np.array([P_x**3, P_x**2, P_x, 1]).T)
Cc = np.dot(Y_m, R)
new_Z.append((np.dot(np.array([P_y**3, P_y**2, P_y, 1]), Cc)))
new_X.append(px)
new_Y.append(py)
py += h
py = round(py, 2)
px += h
px = round(px, 2)
return new_X, new_Y, new_Z
答案 2 :(得分：0)
对于将来的通知，我认为问题在于维基百科上详细介绍的算法用于单位平方上的三次三次插值。 相反，如果要在直线网格上进行插值，则需要稍微修改矢量x。 请参阅“扩展到直线网格”部分，该部分现已包含在Wikipedia页面上。 https://en.wikipedia.org/wiki/Bicubic_interpolation#Extension_to_rectilinear_grids
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