我找到了解决这个问题的办法。它建立在@Hooked的辉煌但黑客solution之上。您可以从许多小方框中创建二维网格。这不是最快的解决方案，但它应该是相当灵活的（比解决方案，适用于imshow的补丁）。

import numpy as np
import pylab as plt

#Plot a rectangle
def rect(ax, x, y, w, h, c,**kwargs):
    #Varying only in x
    if len(c.shape) is 1:
        rect = plt.Rectangle((x, y), w, h, color=c, ec=c,**kwargs)
        ax.add_patch(rect)
    #Varying in x and y
    else:
        #Split into a number of bins
        N = c.shape[0]
        hb = h/float(N); yl = y
        for i in range(N):
            yl += hb
            rect = plt.Rectangle((x, yl), w, hb, 
                                 color=c[i,:], ec=c[i,:],**kwargs)
            ax.add_patch(rect)

#Fill a contour between two lines
def rainbow_fill_between(ax, X, Y1, Y2, colors=None, 
                         cmap=plt.get_cmap("Reds"),**kwargs):
    plt.plot(X,Y1,lw=0)  # Plot so the axes scale correctly

    dx = X[1]-X[0]
    N  = X.size

    #Pad a float or int to same size as x
    if (type(Y2) is float or type(Y2) is int):
        Y2 = np.array([Y2]*N)

    #No colors -- specify linear
    if colors is None:
        colors = []
        for n in range(N):
            colors.append(cmap(n/float(N)))
    #Varying only in x
    elif len(colors.shape) is 1:
        colors = cmap((colors-colors.min())
                      /(colors.max()-colors.min()))
    #Varying only in x and y
    else:
        cnp = np.array(colors)
        colors = np.empty([colors.shape[0],colors.shape[1],4])
        for i in range(colors.shape[0]):
            for j in range(colors.shape[1]):
                colors[i,j,:] = cmap((cnp[i,j]-cnp[:,:].min())
                                    /(cnp[:,:].max()-cnp[:,:].min()))

    colors = np.array(colors)

    #Create the patch objects
    for (color,x,y1,y2) in zip(colors,X,Y1,Y2):
        rect(ax,x,y2,dx,y1-y2,color,**kwargs)

# Some Test data    
X = np.linspace(0,10,100)
Y1 = .25*X**2 - X
Y2 = X
g = np.exp(-.3*(X-5)**2)

#Plot fill and curves changing in x only
fig, axs =plt.subplots(1,2)
colors = g
rainbow_fill_between(axs[0],X,Y1,Y2,colors=colors)
axs[0].plot(X,Y1,'k-',lw=4)
axs[0].plot(X,Y2,'k-',lw=4)

#Plot fill and curves changing in x and y
colors = np.outer(g,g)
rainbow_fill_between(axs[1],X,Y1,Y2,colors=colors)
axs[1].plot(X,Y1,'k-',lw=4)
axs[1].plot(X,Y2,'k-',lw=4)
plt.show()

结果为x1c 0d1x

您的解决方案是伟大和灵活的！特别是2D的情况真的很好。这样的功能可以添加到fill_between，也许如果函数的颜色kwargs将接受相同长度的x和y的数组？
下面是使用fill_between函数的一维情况的一个更简单的情况。它也做了同样的事情，但由于它使用梯形而不是矩形，结果更平滑。

import matplotlib as mpl
import matplotlib.pyplot as plt

import numpy as np
from scipy.stats import norm

# Select a color map
cmap = mpl.cm.bwr

# Some Test data
npts = 100
x = np.linspace(-4, 4, npts)
y = norm.pdf(x)
z = np.sin(2 * x)
normalize = mpl.colors.Normalize(vmin=z.min(), vmax=z.max())

# The plot
fig = plt.figure()
ax = fig.add_axes([0.12, 0.12, 0.68, 0.78])
plt.plot(x, y, color="gray")
for i in range(npts - 1):
    plt.fill_between([x[i], x[i+1]], [y[i], y[i+1]], color=cmap(normalize(z[i])))

cbax = fig.add_axes([0.85, 0.12, 0.05, 0.78])
cb = mpl.colorbar.ColorbarBase(cbax, cmap=cmap, norm=normalize, orientation='vertical')
cb.set_label("Sin function", rotation=270, labelpad=15)
plt.show()