基于python对密立根油滴实验数据处理改进基于python对密立根油滴实验数据处理改进

基于python对密立根油滴实验数据处理改进

近似最大公约数

import math
ac=1.34
def accuracy(x, y):
    return math.fabs(x - y) < ac
def gcd(x, y):
    if accuracy(x, 0):
        return y
    if accuracy(y, 0):
        return x
    return gcd(y, x%y)

b=[3.2,3.2,6.4,4.8,8.0,9.6]
a=[1.60217663410,2*1.60217663410,3*1.60217663410,4*1.60217663410,5*1.60217663410,6*1.60217663410]
c=[7.88,6.25,3.11,3.18,4.57,6.29]
d=[9.36,3.16,3.41,10.78,4.90,3.11]
x=[7.88,6.25,3.11,3.18,4.57,6.29]
f=[2*1.60217663410,100,3*1.60217663410,4*1.60217663410,5*1.60217663410,6*1.60217663410]
n=6
e=2*1.60217663410
for j in range(0,10):
    for i in range(1,n):
        e=gcd(e, a[i])
    for i in range(1,n):
        if(a[i]%e>10):
            del a[i]
            n=n-1  
print(e)

梯度下降拟合“倒过来验证”方法

import numpy as np
import matplotlib.pyplot as plt

N = 6 
x = np.array([5,4,2,2,3,4]).reshape(N,1)       
y = np.array([7.88,6.25,3.11,3.18,4.57,6.29]).reshape(N,1) 
plt.scatter(x, y)
ones = np.ones((N, 1))
x = np.hstack((x, ones))

def CostGradient(theta, x, y):
    diff = np.dot(x, theta) - y
    gradient = (1./N)*(np.dot(np.transpose(x), diff))
    return gradient

def Iteration():
    alpha = 0.001
    theta = np.array([1.6, 0]).reshape(2, 1)         
    gradient = CostGradient(theta, x, y)
    epsilon = 0.001
    while np.linalg.norm(gradient) > epsilon:
        theta = theta - alpha * gradient
        gradient = CostGradient(theta, x, y)
    return theta

Right_theta = Iteration()
print('下降后e值为：')
print('e=',Right_theta[0][0])
print('拟合函数为:')
print('y=',Right_theta[0][0],'x')

x1 = np.linspace(1, 10, 100)
y1 = Right_theta[0]*x1 + Right_theta[1]
plt.plot(x1, y1)
plt.show()

最后

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