迭代二维回归依赖截断正态/拉普拉斯分布

Question

我需要您的帮助来创建一个正态或更好的截断拉普拉斯分布，该分布线性依赖于另一个拉普拉斯分布变量。

这是我迄今为止在派生的 y 分布中使用 NaN 所取得的成果：

import numpy as np
from matplotlib import pyplot as plt
from scipy.stats import gaussian_kde, truncnorm

slope = 0.2237
intercept = 1.066
spread = 4.8719

def dependency(x):
    y_lin = slope * x + intercept
    lower = slope / spread * 3 * x
    upper = slope * spread / 3 * x + 2 * intercept

    y_lin_noise = np.random.laplace(loc=0, scale=spread, size=len(y_lin)) + y_lin

    y_lin_noise[y_lin_noise < lower] = np.nan  # This is the desperate solution where
    y_lin_noise[y_lin_noise > upper] = np.nan  # NaNs are introduced

    return y_lin_noise

max = 100
min = 1
mean = 40
sigma = 25

x = truncnorm((min-mean)/sigma, (max-mean)/sigma, loc=mean, scale=sigma).rvs(5000)
y = dependency(x)

# Plotting
xx = np.linspace(np.nanmin(x), np.nanmax(x), 100)
yy = slope * xx + intercept
lower = slope/spread*3*xx
upper = slope*spread/3*xx + 2*intercept

mask = ~np.isnan(y) & ~np.isnan(x)
x = x[mask]
y = y[mask]

xy = np.vstack([x, y])
z = gaussian_kde(xy)(xy)
idz = z.argsort()
x, y, z = x[idz], y[idz], z[idz]

fig, ax = plt.subplots(figsize=(5, 5))
plt.plot(xx, upper, 'r-.', label='upper constraint')
plt.plot(xx, lower, 'r--', label='lower constraint')

ax.scatter(x, y, c=z, s=3)
plt.xlabel(r'$\bf X_{laplace}$')
plt.ylabel(r'$\bf Y_{{derived}}$')
plt.plot(xx, yy, 'r', label='regression model')
plt.legend()
plt.tight_layout()
plt.show()

我最终想要得到的是没有 NaN 的 y 分布，因此对于每个 x，在上限/下限范围内都有一个对应的 y。也就是说，回归线周围的下限/上截断分布。

我期待想法！

谢谢你和最好的问候。

Answer 1

您想要做的基本上是重试超出您界限的值。您可以定义一个对每个值都执行此操作的函数，或者您可以定义初始值并循环遍历答案以“更正”超出范围的值，如下所示：

import numpy as np
from matplotlib import pyplot as plt
from scipy.stats import gaussian_kde, truncnorm

slope = 0.2237
intercept = 1.066
spread = 4.8719


#slow but effective
def truncated_noise(y, lower, upper):
    while True:
        y_noise = np.random.laplace(loc=0, scale=spread, size=1) + y
        if upper > y_noise > lower:
            return y_noise


def refine_noise(y, y_lin_noise, lower, upper):
    for i in range(len(y_lin_noise)):
        if upper[i] < y_lin_noise[i] or lower[i] > y_lin_noise[i]:
            y_lin_noise[i] = truncated_noise(y[i], lower[i], upper[i])
    return y_lin_noise


def dependency(x):
    y_lin = slope * x + intercept
    lower = slope / spread * 3 * x
    upper = slope * spread / 3 * x + 2 * intercept

    y_lin_noise = np.random.laplace(loc=0, scale=spread, size=len(y_lin)) + y_lin

    y_lin_noise = refine_noise(y_lin, y_lin_noise, lower, upper)

    return y_lin_noise

max = 100
min = 1
mean = 40
sigma = 25

x = truncnorm((min-mean)/sigma, (max-mean)/sigma, loc=mean, scale=sigma).rvs(5000)
y = dependency(x)

# Plotting
xx = np.linspace(np.nanmin(x), np.nanmax(x), 100)
yy = slope * xx + intercept
lower = slope/spread*3*xx
upper = slope*spread/3*xx + 2*intercept


xy = np.vstack([x, y])
z = gaussian_kde(xy)(xy)
idz = z.argsort()
x, y, z = x[idz], y[idz], z[idz]

fig, ax = plt.subplots(figsize=(5, 5))
plt.plot(xx, upper, 'r-.', label='upper constraint')
plt.plot(xx, lower, 'r--', label='lower constraint')

ax.scatter(x, y, c=z, s=3)
plt.xlabel(r'$\bf X_{laplace}$')
plt.ylabel(r'$\bf Y_{{derived}}$')
plt.plot(xx, yy, 'r', label='regression model')
plt.legend()
plt.tight_layout()
plt.show()

我不知道你想用它做什么，但我知道这不再是一个随机的拉普拉斯分布，所以如果你的函数需要它，就不要这样做。