正态分布的峰度为3。随着分布中离群值的增加,尾巴变成“脂肪”,峰度增加到3以上。
如何在峰度大于3(最好是5-7)的两个数字之间生成随机正态分布?
进口
import numpy as np
import scipy.stats import kurtosis
0.01-0.10之间的随机制服
# Random Uniform Distribution
runif = np.random.uniform(0.01, 0.10, 10000)
kurtosis(runif, fisher=False)
1.8124891901330156
随机法线介于0.01-0.10之间
lower = 0.01
upper = 0.10
mu = (upper)/2
sigma = 0.01
N = 10000
retstats = scipy.stats.truncnorm.rvs((lower-mu)/sigma,(upper-mu)/sigma,loc=mu,scale=sigma,size=N)
mean = .05
stdev = .01 # 99.73% chance the sample will fall in your desired range
values = [gauss(mean, stdev) for _ in range(10000)]
kurtosis(values, fisher=False)
3.015004351756201
随机正态,尾巴在0.01-0.10之间
???
上面的链接给出了使用R(叹息)代码的示例,但我认为它很简单,可以让您用Python编写等效代码。这是我所知道的几个扩展(即功能分层)之一,可以使您实现这一目标。
不幸的是,我知道没有简单的解决方案。
但是,通过剪掉尾巴,可以减少峰度。为了生成范围有限且峰度较高的分布,您需要确保切口对尾部的影响最小。通俗地说,您需要一个非常尖刻的分发。
import numpy as np
from scipy.stats import kurtosis
min_range = 0.01
max_range = 0.10
midpoint = (max_range + min_range)/2
samples = 10000
def filter_tails(x):
return x[(x >= min_range) & (x <= max_range)]
runif = np.random.uniform(min_range, max_range, samples)
value = kurtosis(filter_tails(runif), fisher=False)
print(f"uniform kurtosis = {value}")
sigma = 0.01
runif = np.random.normal(midpoint, sigma, samples)
value = kurtosis(filter_tails(runif), fisher=False)
print(f"gaussian kurtosis = {value}")
exponential_decay = 0.001
runif = np.random.laplace(midpoint, exponential_decay, samples)
value = kurtosis(filter_tails(runif), fisher=False)
print(f"laplace kurtosis = {value}")
运行脚本,我得到:
uniform kurtosis = 1.8011863970680828 gaussian kurtosis = 3.0335178694177785 laplace kurtosis = 5.76290423111418