如何将散点图转换为轮廓图？

Question

我想将散点图转换为轮廓图。我该如何处理我拥有的数据？这是代码：

import numpy as np
import matplotlib.pyplot as plt
x = np.array([-75.55846, -75.5459 , -75.56686, -75.55276, -75.57951, -75.58955,
   -75.59967, -75.53964, -75.65485, -75.55292, -75.59622, -75.57392,
   -75.47077, -75.58644, -75.68264, -75.56732, -75.59502, -75.37198,
   -75.59585, -75.57081, -75.40989, -75.50928, -75.54841, -75.72734,
   -75.36676, -75.79303, -75.36966, -75.69282, -75.42498, -75.57986,
   -75.58644, -75.64551, -75.45509, -75.47098])
y = np.array([38.07759, 38.07541, 38.06712, 38.09973, 38.06692, 38.09264,
   38.08731, 38.0822 , 38.36981, 38.06027, 38.07962, 38.08531,
   37.93448, 38.07716, 37.71758, 38.08925, 38.0975 , 38.33174,
   38.05731, 38.05515, 38.0547 , 38.1398 , 38.066  , 38.27009,
   37.93415, 38.24889, 38.19691, 38.03272, 38.19954, 37.91286,
   37.97847, 38.29755, 38.01239, 37.93453])
z = np.array([17.526, 21.336, 19.558, 17.78 , 20.828, 20.828, 20.066, 21.082,
   18.542, 20.32 , 19.812, 19.05 , 16.51 , 20.066, 25.654, 16.51 ,
   18.542, 17.018, 20.828, 21.844, 21.59 , 16.764, 20.828, 19.558,
   19.812, 22.606, 25.146, 19.558, 20.574, 24.13 , 35.306, 19.558,
   23.876, 18.796])
fig, ax = plt.subplots(figsize=(8,8))
ax.scatter(x, y, c=z)
plt.show()

Answer 1

这不一定是最好的方法，但是一种实现方法可能是将数据转换为x，y网格中每个点的z值。可以用多种方法完成此操作，得到不同的结果，但是我选择的方法是根据与每个点的距离使用z值的加权平均值。可以调节falloff来改变点的影响，随着距离该点的增加，该影响会改变多少（将其值保持为正）。

import numpy as np
import matplotlib.pyplot as plt

x_arr = np.array([-75.55846, -75.5459, -75.56686, -75.55276, -75.57951, -75.58955,
                  -75.59967, -75.53964, -75.65485, -75.55292, -75.59622, -75.57392,
                  -75.47077, -75.58644, -75.68264, -75.56732, -75.59502, -75.37198,
                  -75.59585, -75.57081, -75.40989, -75.50928, -75.54841, -75.72734,
                  -75.36676, -75.79303, -75.36966, -75.69282, -75.42498, -75.57986,
                  -75.58644, -75.64551, -75.45509, -75.47098])
y_arr = np.array([38.07759, 38.07541, 38.06712, 38.09973, 38.06692, 38.09264,
                  38.08731, 38.0822, 38.36981, 38.06027, 38.07962, 38.08531,
                  37.93448, 38.07716, 37.71758, 38.08925, 38.0975, 38.33174,
                  38.05731, 38.05515, 38.0547, 38.1398, 38.066, 38.27009,
                  37.93415, 38.24889, 38.19691, 38.03272, 38.19954, 37.91286,
                  37.97847, 38.29755, 38.01239, 37.93453])
z_arr = np.array([17.526, 21.336, 19.558, 17.78, 20.828, 20.828, 20.066, 21.082,
                  18.542, 20.32, 19.812, 19.05, 16.51, 20.066, 25.654, 16.51,
                  18.542, 17.018, 20.828, 21.844, 21.59, 16.764, 20.828, 19.558,
                  19.812, 22.606, 25.146, 19.558, 20.574, 24.13, 35.306, 19.558,
                  23.876, 18.796])

resolution = 50

xx, yy = np.meshgrid(
    np.linspace(np.min(x_arr), np.max(x_arr), num=resolution),
    np.linspace(np.min(y_arr), np.max(y_arr), num=resolution)
)

falloff = 2
max_closeness = 100000


def eval_at_point(p_x, p_y):
    closeness = [min(np.hypot(x - p_x, y - p_y) ** -falloff, max_closeness) for x, y in zip(x_arr, y_arr)]

    return sum(close * z for close, z in zip(closeness, z_arr)) / sum(closeness)


zz = [[eval_at_point(x, y) for x, y in zip(x_row, y_row)] for x_row, y_row in zip(xx, yy)]
plt.scatter(x_arr, y_arr, c=z_arr)
plt.contour(xx, yy, zz)
plt.show()

contourf也可以使用。

Answer 2

我想我想出了一个解决方案-参见下文。基本上，x，y和z是一维数组，而X，Y和Z是轮廓接受并起作用的2D数组。我会说我确实环顾四周以解决这个问题。所以它不是我一个人。

def grid(x, y, z, resX=100, resY=100):
    #prepare to create grid
    xi = np.linspace(min(x), max(x), resX)
    yi = np.linspace(min(y), max(y), resY)

    #grid x, y, z
    X, Y = np.meshgrid(xi, yi)
    Z = griddata((x, y), z, (X, Y), method='linear')
    return X, Y, Z