我考虑过要找到每个矩阵的船体图形会有多困难，并且甚至不清楚矩阵中是否有孔。但是，为什么不让numpy / matplotlib完成所有工作呢？

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

M1 = ([5,      3,      6,      7], 
      [2,      np.nan, 3,      6], 
      [6,      7,      8,      np.nan], 
      [9,      10,     np.nan, np.nan], 
      [11,     12,     np.nan, np.nan])

M2 = ([np.nan, np.nan, np.nan, np.nan],
      [np.nan, np.nan, 1,      2], 
      [np.nan, 4,      8,      5], 
      [np.nan, np.nan, 6,      9], 
      [np.nan, np.nan, np.nan, np.nan])

#convert arrays into truth values regarding the presence of NaNs
M1arr = ~np.isnan(M1)
M2arr = ~np.isnan(M2)

#combine arrays based on False = 0, True = 1
M1M2arr = np.sum([M1arr, 2 * M2arr], axis=0) 

#define color scale for the plot
cmapM1M2 = colors.ListedColormap(["white", "tab:blue", "tab:orange", "tab:red"])

cb = plt.imshow(M1M2arr, cmap=cmapM1M2)
cbt= plt.colorbar(cb, ticks=np.linspace(0, 3, 9)[1::2])
cbt.ax.set_yticklabels(["M1 & M2 NaN", "only M1 values", "only M2 values", "M1 & M2 values"])
plt.xlabel("a[i]")
plt.ylabel("b[i]")

plt.tight_layout()
plt.show()

样本输出：

我一直保持方向，imshow因为这是打印输出时如何读取矩阵条目的方式。你可以通过更改以下行将此图像转换为通常的坐标表示形式：

cb = plt.imshow(M1M2arr, cmap=cmapM1M2, origin="lower")