11

可以在线条之间填充颜色：

http://matplotlib.sourceforge.net/examples/pylab_examples/fill_between_demo.html

也可以对一条线使用连续的颜色图：

http://matplotlib.sourceforge.net/examples/pylab_examples/multicolored_line.html

是否可以（并且相当容易）为两行之间的彩色填充使用连续的颜色图？例如，颜色填充可以基于 x 处两条线之间的差异（或基于另一组数据）沿 x 变化。

4

2 回答 2

11

我找到了解决这个问题的方法。它建立在@Hooked的出色但老套的解决方案之上。您从许多小盒子中创建了一个 2D 网格。这不是最快的解决方案，但它应该非常灵活（比将 imshow 应用于补丁的解决方案更灵活）。

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()

结果是， 在此处输入图像描述

于 2015-04-15T17:19:51.907 回答
7

您的解决方案很棒而且很灵活！特别是 2D 案例非常好。fill_between如果函数的颜色 kwargs 可以接受 x 和 y 长度相同的数组，则可以添加这样的功能？

fill_between这是使用该函数的一维案例的更简单案例。它的作用相同，但由于它使用梯形而不是矩形，因此结果更平滑。

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()

在此处输入图像描述

于 2016-08-31T13:09:40.880 回答