matplotlib 从散点图中获取位图

Question

我有一些需要绘制的点的坐标，然后将绘图转换为黑白位图：

import matplotlib.pyplot as plt
import matplotlib.image as mpimg
from PIL import Image

plt.scatter(x,y)
plt.tight_layout()
fig1 = plt.gcf()
plt.show()

type(fig1)
matplotlib.figure.Figure

如何从这个图中获取黑白位图作为类似于这个的numpy数组：

side = 5
image = np.random.choice([0, 1], size=side*side,  p=[.1, .9])
image = image.reshape(side,side)
image = np.expand_dims(image, axis=-1)
print("image.shape: ",image.shape)
plt.imshow(image, cmap=plt.get_cmap('gray'))

image.shape:  (5, 5, 1)

print(image.reshape(side,side))

[[1 1 1 0 1]
 [1 1 1 1 1]
 [1 0 1 1 0]
 [1 1 1 1 0]
 [1 1 1 1 1]]

更新 1

我还需要将生成的位图作为一个 numpy 数组。如何获得？

如果我使用 Zephyr 给出的解决方案：

fig, ax = plt.subplots(figsize = (5,5))
ax.hist2d(x, y, cmap = 'Greys', cmin = 0, cmax = 1)
plt.show()

我得到的图像与散点图不同。它们应该是相似的：

Answer 1

您可以创建一个网格并使用它来定义地图，其中最近的点将是白色的。我尝试使用随机数据，范围 0 到 1：

import matplotlib.pyplot as plt
import numpy as np

n_points = 10
# create random coordinates
x, y = np.random.rand(n_points,2).T
fig, ax = plt.subplots()
ax.scatter(x,y)
ax.set_xlim([0,1])
ax.set_ylim([0,1])
ax.set_aspect(1.0)

# create a grid
grid_points = 10
grid_x = np.linspace(0,1,grid_points)
grid_y = grid_x.copy()

# initiate array of ones (white)
image = np.ones([grid_points, grid_points])
for xp, yp in zip(x,y):
    # selecing the closest point in grid
    index_x = np.argmin(np.abs(xp - grid_x))
    index_y = np.argmin(np.abs(yp - grid_y))
    # setting to black
    image[index_x,index_y] = 0
    
# you need to transpose it so x is represented
# by the columns and y by the rows
fig, ax = plt.subplots()
ax.imshow(
    image.T,
    origin='lower',
    cmap=plt.get_cmap('gray'))

请注意，最接近的可能并不总是好的。使用更精细的网格会变得更好。

Answer 2

首先，我在(x_min, x_max)和(y_min, y_max)范围内生成随机的N点：

np.random.seed(42)

N = 10
x_min = 0
x_max = 40
y_min = -20
y_max = 20

x = np.random.uniform(x_min, x_max, N)
y = np.random.uniform(y_min, y_max, N)

然后我准备：

• (size, size) 维度的grid（位图）
• 两个向量 x_grid 和 y_grid 在 size + 1 点中重新采样 (x_min, x_max) 和 (y_min, y_max)，所以 size inverval：每个 grid 单元格有一个间隔

size = 10
grid = np.zeros((size, size))
x_grid = np.linspace(x_min, x_max, size + 1)
y_grid = np.linspace(y_min, y_max, size + 1)

然后我循环遍历每个grid 单元格；在每次迭代中，我检查是否至少有 1 个点 (x, y) 保持在该单元格的范围内。如果是这样，我将grid的对应值设置为1：

for i in range(size):
    for j in range(size):
        for x_i, y_i in zip(x, y):
            if (x_grid[i] < x_i <= x_grid[i + 1]) and (y_grid[j] < y_i <= y_grid[j + 1]):
                grid[i, j] = 1
                break

生成的 numpy 矩阵：

[[0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
 [0. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 1. 0. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]]

完整代码

import matplotlib.pyplot as plt
import numpy as np

np.random.seed(42)

N = 10
x_min = 0
x_max = 40
y_min = -20
y_max = 20

x = np.random.uniform(x_min, x_max, N)
y = np.random.uniform(y_min, y_max, N)

size = 10
grid = np.zeros((size, size))
x_grid = np.linspace(x_min, x_max, size + 1)
y_grid = np.linspace(y_min, y_max, size + 1)

for i in range(size):
    for j in range(size):
        for x_i, y_i in zip(x, y):
            if (x_grid[i] < x_i <= x_grid[i + 1]) and (y_grid[j] < y_i <= y_grid[j + 1]):
                grid[i, j] = 1
                break

fig, ax = plt.subplots(1, 2, figsize = (10, 5))

ax[0].scatter(x, y)
ax[0].set_xlim(x_min, x_max)
ax[0].set_ylim(y_min, y_max)
ax[0].grid()
ax[0].set_xticks(x_grid)
ax[0].set_yticks(y_grid)

ax[1].imshow(grid.T, cmap = 'Greys', extent = (x_min, x_max, y_min, y_max))
ax[1].invert_yaxis()

plt.show()

注意

请注意，在ax.imshow 中，您需要转置矩阵 (grid.T)，然后反转 y 轴，以便能够将ax.imshow 与ax.scatter 进行比较。
如果你想让grid矩阵匹配ax.imshow，那么你需要逆时针旋转90°：

grid = np.rot90(grid, k=1, axes=(0, 1))

旋转grid，对应上图：

[[0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 0. 0. 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. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 1. 0. 0.]
 [0. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]]