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[英]How to generate correlated random numbers for truncated normal distribution in Python?
[英]How to generate a random normal distribution of integers
如何使用np.random.randint()
生成随机数 integer,但正态分布在 0 左右。
np.random.randint(-10, 10)
返回具有离散均匀分布的整数np.random.normal(0, 0.1, 1)
返回具有正态分布的浮点数
我想要的是两种功能之间的一种组合。
获得看起来像<\/em>正态分布的离散分布的另一种方法是从多项分布中提取,其中概率是根据正态分布计算的。
import scipy.stats as ss
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(-10, 11)
xU, xL = x + 0.5, x - 0.5
prob = ss.norm.cdf(xU, scale = 3) - ss.norm.cdf(xL, scale = 3)
prob = prob / prob.sum() # normalize the probabilities so their sum is 1
nums = np.random.choice(x, size = 10000, p = prob)
plt.hist(nums, bins = len(x))
可以从四舍五入的截断正态分布<\/em><\/a>生成类似的分布。 这是 scipy 的truncnorm()<\/a>的示例。
import numpy as np
from scipy.stats import truncnorm
import matplotlib.pyplot as plt
scale = 3.
range = 10
size = 100000
X = truncnorm(a=-range/scale, b=+range/scale, scale=scale).rvs(size=size)
X = X.round().astype(int)
代码:
#--------*---------*---------*---------*---------*---------*---------*---------*
# Desc: Discretize a normal distribution centered at 0
#--------*---------*---------*---------*---------*---------*---------*---------*
import sys
import random
from math import sqrt, pi
import numpy as np
import matplotlib.pyplot as plt
def gaussian(x, var):
k1 = np.power(x, 2)
k2 = -k1/(2*var)
return (1./(sqrt(2. * pi * var))) * np.exp(k2)
#--------*---------*---------*---------*---------*---------*---------*---------#
while 1:# M A I N L I N E #
#--------*---------*---------*---------*---------*---------*---------*---------#
# # probability density function
# # for discrete normal RV
pdf_DGV = []
pdf_DGW = []
var = 9
tot = 0
# # create 'rough' gaussian
for i in range(-var - 1, var + 2):
if i == -var - 1:
r_pdf = + gaussian(i, 9) + gaussian(i - 1, 9) + gaussian(i - 2, 9)
elif i == var + 1:
r_pdf = + gaussian(i, 9) + gaussian(i + 1, 9) + gaussian(i + 2, 9)
else:
r_pdf = gaussian(i, 9)
tot = tot + r_pdf
pdf_DGV.append(i)
pdf_DGW.append(r_pdf)
print(i, r_pdf)
# # amusing how close tot is to 1!
print('\nRough total = ', tot)
# # no need to normalize with Python 3.6,
# # but can't help ourselves
for i in range(0,len(pdf_DGW)):
pdf_DGW[i] = pdf_DGW[i]/tot
# # print out pdf weights
# # for out discrte gaussian
print('\npdf:\n')
print(pdf_DGW)
# # plot random variable action
rv_samples = random.choices(pdf_DGV, pdf_DGW, k=10000)
plt.hist(rv_samples, bins = 100)
plt.show()
sys.exit()
这个版本在数学上是不正确的(因为你剪掉了铃铛),但如果不需要那么精确,它会快速且容易理解地完成这项工作:
def draw_random_normal_int(low:int, high:int):
# generate a random normal number (float)
normal = np.random.normal(loc=0, scale=1, size=1)
# clip to -3, 3 (where the bell with mean 0 and std 1 is very close to zero
normal = -3 if normal < -3 else normal
normal = 3 if normal > 3 else normal
# scale range of 6 (-3..3) to range of low-high
scaling_factor = (high-low) / 6
normal_scaled = normal * scaling_factor
# center around mean of range of low high
normal_scaled += low + (high-low)/2
# then round and return
return np.round(normal_scaled)
老问题,新答案:
对于整数 {-10, -9, ..., 9, 10} 的钟形分布,您可以使用 n=20 和 p=0.5 的二项式分布<\/a>,并从样本中减去 10。
例如,
In [167]: import numpy as np
In [168]: import matplotlib.pyplot as plt
In [169]: N = 5000000 # Number of samples to generate
In [170]: samples = rng.binomial(n=20, p=0.5, size=N) - 10
In [171]: samples.min(), samples.max()
Out[171]: (-10, 10)
我不确定是否有(在 scipy 生成器中)要生成的 var 类型选项,但是常见的生成可以是 scipy.stats
# Generate pseudodata from a single normal distribution
import scipy
from scipy import stats
import numpy as np
import matplotlib.pyplot as plt
dist_mean = 0.0
dist_std = 0.5
n_events = 500
toy_data = scipy.stats.norm.rvs(dist_mean, dist_std, size=n_events)
toy_data2 = [[i, j] for i, j in enumerate(toy_data )]
arr = np.array(toy_data2)
print("sample:\n", arr[1:500, 0])
print("bin:\n",arr[1:500, 1])
plt.scatter(arr[1:501, 1], arr[1:501, 0])
plt.xlabel("bin")
plt.ylabel("sample")
plt.show()
或者以这种方式(也没有 dtype 选择的选项):
import matplotlib.pyplot as plt
mu, sigma = 0, 0.1 # mean and standard deviation
s = np.random.normal(mu, sigma, 500)
count, bins, ignored = plt.hist(s, 30, density=True)
plt.show()
print(bins) # <<<<<<<<<<
plt.plot(bins, 1/(sigma * np.sqrt(2 * np.pi)) * np.exp( - (bins - mu)**2 / (2 * sigma**2) ),
linewidth=2, color='r')
plt.show()
没有可视化最常见的方式(也没有可能指出 var 类型)
bins = np.random.normal(3, 2.5, size=(10, 1))
可以完成一个包装类来实例化具有给定 vars-dtype 的容器(例如,通过将浮点数舍入为整数,如上所述)...
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