matplotlib 具有交替单元宽度的pcolormesh

of1yzvn4 于 2023-10-24 发布在其他

关注(0)|答案(1)|浏览(146)

我想做一个matplotlib colourmesh，它被不规则地缩放，使一些行和列比其他的宽。
请记住，我对matplotlib不是很熟悉，很可能忽略了一个明显的解决方案。
假设我有以下代码：

plt.axes().set_aspect("equal")
maze = plt.pcolormesh(arr)
plt.show()

其中arr是一个2d位数组，生成一个colourmesh，如下所示：

我想减少每一行和每一列的宽度来压缩上面迷宫的墙壁。
任何帮助将不胜感激，谢谢。

编辑：这里是一个数据样本。它基本上只是一个1和0的二维数组，但为了可读性，我已经将其转换为字符串。

0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   
0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   
0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   1   0   
0   1   0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   1   1   0   
0   1   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   1   0   1   0   
0   1   0   1   0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   1   0   0   0   0   0   0   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   0   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   0   0   0   0   1   0   1   0   1   0   1   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   0   1   1   1   1   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   0   0   0   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   0   1   1   1   0   1   1   1   0   1   0   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   1   0   0   0   0   0   0   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   0   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   1   1   0   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   1   1   0   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   0   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   1   0   1   1   1   0   
0   1   0   1   0   1   0   0   0   0   0   0   0   0   0   1   0   1   0   0   0   0   0   0   0   0   0   1   0   1   0   1   0   
0   1   1   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   0   1   1   1   0   
0   0   0   0   0   0   0   1   0   1   0   1   0   1   0   0   0   0   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   1   1   1   1   1   1   0   1   1   1   0   1   1   1   1   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   
0   1   0   0   0   0   0   1   0   1   0   1   0   1   0   0   0   0   0   1   0   1   0   1   0   1   0   0   0   1   0   1   0   
0   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   0   1   1   1   1   1   0   
0   1   0   1   0   1   0   0   0   0   0   0   0   0   0   1   0   1   0   0   0   0   0   0   0   0   0   1   0   1   0   1   0   
0   1   0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   
0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   1   0   
0   1   0   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   0   
0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0

matplotlib

来源：https://stackoverflow.com/questions/77182518/pcolormesh-with-alternating-cell-widths

1条答案

按热度按时间

shstlldc1#

plt.pcolormesh()可以用1个或3个未命名的参数调用。当只有1个参数时，该参数应该是网格数据。当有3个参数时，前2个参数或x和y方向上网格线的位置。
您可以通过平铺宽单元格和窄单元格的宽度来计算这些位置，然后计算cumulative sum。请注意，网格位置需要比单元格多一个。因此，33单元格的网格线（或2*17）。如果x和y方向上的单元格数量不相等，您可以单独计算网格位置。
这里有一个例子。宽度的选择使得宽单元格和窄单元格占用2个单位，这将使轴坐标与示例相似。如果这些坐标不重要，您可以使用任何宽度。

import matplotlib.pyplot as plt
import numpy as np

arr = np.array([[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0],[0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0],[0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0],[0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0],[0,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0],[0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0],[0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0],[0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0],[0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0],[0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0],[0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0],[0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0],[0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0],[0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]])

dists = np.tile([1.6, 0.4], (arr.shape[0] + 1) // 2).cumsum()

plt.pcolormesh(dists, dists, arr)
plt.axis('equal')
plt.axis('off')
plt.show()

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matplotlib 具有交替单元宽度的pcolormesh

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