英文:
Mean, variance of normal distribution for each seed in Python
问题
我正在使用numpy.random.seed()来获得不同的正态分布。但我希望对于每个种子，mean,var保持恒定。我该如何实现这一目标？以下是当前和期望的输出。
当前输出为：
Mean = 49.99989488440971
var = 1.0009790234390512
Mean = 50.00911322245587
var = 0.8709906144821978
期望的输出是：
Mean = 50.000
var = 1.000
Mean = 50.000
var = 1.000
英文:
I am using numpy.random.seed() to get different normal distribution. But I want to have the mean,var to remain constant for each seed. How can I achieve this? I present the current and expected output.
from scipy.stats import truncnorm
import numpy as np
import os
import csv
import pandas as pd
import random
import matplotlib.pyplot as plt
for i in range(0,2):
np.random.seed(i)
mu, sigma = 49.925, 0.996 # mean and standard deviation
Nodes=220
r = (1e-6)*np.random.normal(mu, sigma, Nodes)
sort_r = np.sort(r)
r1=sort_r[::-1]
r1=r1.reshape(1,Nodes)
r2 = r.copy()
np.random.shuffle(r2.ravel()[1:])
r2=r2.reshape(1,Nodes)
maximum = r2.max()
indice1 = np.where(r2 == maximum)
r2[indice1] = r2[0][0]
r2[0][0] = maximum
r2[0][Nodes-1] = maximum
mean=np.mean(r)
print("Mean =",mean*1e6)
var=np.var(r)
print("var =",var*1e12)
r=pd.Series(r*1e6)
r.hist()
plt.xlabel("Pore throat size [\u03BCm]",size=15)
plt.ylabel("Count",size=15)
plt.title(rf"var={round(var*1e12,1)}",size=25)
plt.savefig(rf"C:\Users\USER\OneDrive - Technion\Research_Technion\Python_PNM\Surfactant A-D0 nodes_1\var_1\Histogram_0.png")
The current output is
Mean = 49.99989488440971
var = 1.0009790234390512
Mean = 50.00911322245587
var = 0.8709906144821978
The expected output is
Mean = 50.000
var = 1.000
Mean = 50.000
var = 1.000
答案1
得分: 1
IIUC，您可以使用标准归一化（z-分数）：
for i in range(0, 2):
np.random.seed(i)
mu, sigma = 49.925, 2.996 # 均值和标准差
Nodes = 220
r = (1e-6) * np.random.normal(mu, sigma, Nodes)
# 目标
t_mean = 50
t_std = 3
# 标准缩放器
a = (1e-6) * (t_mean + (r - r.mean()) / (r.std() / t_std))
print("Mean =", a.mean() * 1e6)
print("var =", a.std() * 1e6)
输出：
Mean = 50.000000000000014
var = 2.999999999999999
Mean = 49.99999999999999
var = 3.000000000000001
英文:
IIUC, you can use the standard normalization (z-score):
for i in range(0,2):
np.random.seed(i)
mu, sigma = 49.925, 2.996 # mean and standard deviation
Nodes = 220
r = (1e-6) * np.random.normal(mu, sigma, Nodes)
# target
t_mean = 50
t_std = 3
# standard scaler
a = (1e-6) * (t_mean + (r - r.mean()) / (r.std() / t_std))
print("Mean =", a.mean() * 1e6)
print("var =", a.std() * 1e6)
Output:
Mean = 50.000000000000014
var = 2.999999999999999
Mean = 49.99999999999999
var = 3.000000000000001
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