教你如何利用python进行数值分析

import numpy as np
x=np.random.uniform(a,b,100) #产生长度为100的U(a,b)

import numpy as np
x=np.random.normal(mu, sig, 100) #产生长度为100的N(mu, sqart(sig))

def spline_fit():
	size = 20
    x = np.linspace(-10, 10, size)
    y = np.sin(x) + np.random.normal(0, 0.1, size)
    y2 = [0] * len(y)
    # for y_i in y:

    pp.plot(x, y)
    cs = CubicSpline(x, y)
    x2 = x = np.linspace(-10, 10, size * 100)
    pp.plot(x2, cs(x2))
    pp.show()

def least_square():
    f = lambda p0, xx: p0[0] * np.sin(xx * p0[1]) + p0[2]
    LEN = 100
    x = np.linspace(-1, 1, LEN)
    y = x ** 2 + 5
    # 默认情况，param只会返回求得的参数和返回的错误码，1-4为成功，5-8为失败，如果想输出更多参数，可以指定full_out=1，可以看到出错原因和其他参数
    param = leastsq(lambda p0, xx, yy: f(p0, xx) - yy, (1, 1, 1), args=(x, y)) #初值的选择比较重要，如果选取不当，容易陷入局部最优
    print(param)
    pp.scatter(x, y)
    p0 = param[0]
    pp.plot(x, f(p0, x))
    pp.show()

def lagrange()
	from scipy.optimize import minimize
    import numpy as np
    e = 1e-10
    fun = lambda x: 8 * (x[0] * x[1] * x[2])  # f(x,y,z) =8 *x*y*z
    cons = ({'type': 'eq', 'fun': lambda x: x[0] ** 2 + x[1] ** 2 + x[2] ** 2 - 1},  # x^2 + y^2 + z^2=1
            {'type': 'ineq', 'fun': lambda x: x[0] - e},  # x>=e等价于 x > 0
            {'type': 'ineq', 'fun': lambda x: x[1] - e},
            {'type': 'ineq', 'fun': lambda x: x[2] - e}
            )
    x0 = np.array((1.0, 1.0, 1.0))  # 设置初始值
    res = minimize(fun, x0, method='SLSQP', constraints=cons)
    print('最大值：', res.fun)
    print('最优解：', res.x)
    print('迭代终止是否成功：', res.success)
    print('迭代终止原因：', res.message)