使用matplotlib cmap填充子图之间

编辑

针对已编辑的问题，这是一种替代方法，该方法垂直进行渐变，但不使用imshow.

import matplotlib.pyplot as plt
from  matplotlib import colors, patches
import numpy as np
import pandas as pd

n = 100
nc = 100

x = np.linspace(0, np.pi*5, n)
y1 = [-50.0]
y2 = [50.0]
for ii in range(1, n):
    y1.append(y1[ii-1] + (np.random.random()-0.3)*3)
    y2.append(y2[ii-1] + (np.random.random()-0.5)*3)
y1 = np.array(y1)
y2 = np.array(y2)
z = np.linspace(0, 10, nc)
normalize = colors.Normalize(vmin=z.min(), vmax=z.max())
cmap = plt.cm.get_cmap('winter')

fig, ax = plt.subplots(1)
for ii in range(len(df['x'].values)-1):
    y = np.linspace(y1[ii], y2[ii], nc)
    yn = np.linspace(y1[ii+1], y2[ii+1], nc)
    for kk in range(nc - 1):
        p = patches.Polygon([[x[ii], y[kk]], 
                             [x[ii+1], yn[kk]], 
                             [x[ii+1], yn[kk+1]], 
                             [x[ii], y[kk+1]]], color=cmap(normalize(z[kk])))
        ax.add_patch(p)

plt.plot(x, y1, 'k-', lw=1)
plt.plot(x, y2, 'k-', lw=1)
plt.show()

这里的想法与我最初的答案类似，只是将梯形分为nc个部分，并且每个部分分别着色.如比较所示，这样做的优点是可以正确缩放以适应y1[ii]，y2[ii]距离，

The idea here being similar to that in my original answer, except the trapezoids are divided into nc pieces and each piece is colored separately. This has the advantage of scaling correctly for varying y1[ii], y2[ii] distances, as shown in this comparison,

但是，它的缺点是比imshow或水平梯度方法慢很多，很多，并且无法正确处理交叉".

It does, however, have the disadvantages of being much, much slower than imshow or the horizontal gradient method and of being unable to handle 'crossing' correctly.

上面比较中生成第二张图像的代码:

The code to generate the second image in the above comparison:

import matplotlib.pyplot as plt
import numpy as np
from matplotlib import patches
from matplotlib.path import Path

x = np.linspace(0, 10, n)
y1 = [-50.0]
y2 = [50.0]
for ii in range(1, n):
    y1.append(y1[ii-1] + (np.random.random()-0.2)*3)
    y2.append(y2[ii-1] + (np.random.random()-0.5)*3)
y1 = np.array(y1)
y2 = np.array(y2)

verts = np.vstack([np.stack([x, y1], 1), np.stack([np.flip(x), np.flip(y2)], 1)])
path = Path(verts)

patch = patches.PathPatch(path, facecolor='k', lw=2, alpha=0.0)
plt.gca().add_patch(patch)

plt.imshow(np.arange(10).reshape(10,-1), cmap=plt.cm.winter, interpolation="bicubic",
             origin='upper', extent=[0,10,-60,60], aspect='auto', clip_path=patch, 
             clip_on=True)
plt.show()

原始

这有点骇人听闻，部分是基于这个问题.它看起来确实工作得很好，但是在沿x轴的较高密度下效果最佳.想法是为与x对[x[ii], x[ii+1]]对相对应的每个梯形分别调用fill_between.这是一个使用一些生成数据的完整示例

Original

This is a bit of a hack, partly based on the answers in this question. It does seem to work fairly well but works best with higher density along the x axis. The idea is to call fill_between separately for each trapezoid corresponding to x pairs, [x[ii], x[ii+1]]. Here is a complete example using some generated data

import matplotlib.pyplot as plt
from  matplotlib import colors
import numpy as np
import pandas as pd

n = 1000

X = np.linspace(0, np.pi*5, n)
Y1 = np.sin(X)
Y2 = np.cos(X)
Z = np.linspace(0, 10, n)
normalize = colors.Normalize(vmin=Z.min(), vmax=Z.max())
cmap = plt.cm.get_cmap('winter')

df = pd.DataFrame({'x': X, 'y1': Y1, 'y2': Y2, 'z': Z})
x = df['x'].values
y1 = df['y1'].values
y2 = df['y2'].values
z = df['z'].values

for ii in range(len(df['x'].values)-1):
    plt.fill_between([x[ii], x[ii+1]], [y1[ii], y1[ii+1]], 
                     [y2[ii], y2[ii+1]], color=cmap(normalize(z[ii])))

plt.plot(x, y1, 'k-', x, y2, 'k-')
plt.show()

这可以推广到二维彩色网格，但需要进行一些简单的修改

This can be generalized to a 2 dimensional color grid but would require non-trivial modification