# -*- coding: utf-8 -*-
import numpy as np
from scipy.optimize import leastsq
import pylab as pl

def func(x, p):
    """
    数据拟合所用的函数: A*sin(2*pi*k*x + theta)
    """
    A, k, theta = p
    return A*np.sin(2*np.pi*k*x+theta)   

def residuals(p, y, x):
    """
    实验数据x, y和拟合函数之间的差，p为拟合需要找到的系数
    """
    return y - func(x, p)

x = np.linspace(0, -2*np.pi, 100)
A, k, theta = 10, 0.34, np.pi/6 # 真实数据的函数参数
y0 = func(x, [A, k, theta]) # 真实数据
y1 = y0 + 2 * np.random.randn(len(x)) # 加入噪声之后的实验数据    

p0 = [7, 0.2, 0] # 第一次猜测的函数拟合参数

# 调用leastsq进行数据拟合
# residuals为计算误差的函数
# p0为拟合参数的初始值
# args为需要拟合的实验数据
plsq = leastsq(residuals, p0, args=(y1, x))

print u"真实参数:", [A, k, theta] 
print u"拟合参数", plsq[0] # 实验数据拟合后的参数

pl.plot(x, y0, label=u"真实数据")
pl.plot(x, y1, label=u"带噪声的实验数据")
pl.plot(x, func(x, plsq[0]), label=u"拟合数据")
pl.legend()
pl.show()

# -*- coding: utf-8 -*-
# 本程序用各种fmin函数求卷积的逆运算

import scipy.optimize as opt 
import numpy as np 

def test_fmin_convolve(fminfunc, x, h, y, yn, x0): 
    """
    x (*) h = y, (*)表示卷积
    yn为在y的基础上添加一些干扰噪声的结果
    x0为求解x的初始值
    """
    def convolve_func(h):
        """
        计算 yn - x (*) h 的power
        fmin将通过计算使得此power最小
        """ 
        return np.sum((yn - np.convolve(x, h))**2) 

    # 调用fmin函数，以x0为初始值
    h0 = fminfunc(convolve_func, x0) 

    print fminfunc.__name__ 
    print "---------------------" 
    # 输出 x (*) h0 和 y 之间的相对误差
    print "error of y:", np.sum((np.convolve(x, h0)-y)**2)/np.sum(y**2) 
    # 输出 h0 和 h 之间的相对误差
    print "error of h:", np.sum((h0-h)**2)/np.sum(h**2) 
    print 

def test_n(m, n, nscale): 
    """
    随机产生x, h, y, yn, x0等数列，调用各种fmin函数求解b
    m为x的长度, n为h的长度, nscale为干扰的强度
    """
    x = np.random.rand(m) 
    h = np.random.rand(n) 
    y = np.convolve(x, h) 
    yn = y + np.random.rand(len(y)) * nscale
    x0 = np.random.rand(n) 

    test_fmin_convolve(opt.fmin, x, h, y, yn, x0) 
    test_fmin_convolve(opt.fmin_powell, x, h, y, yn, x0) 
    test_fmin_convolve(opt.fmin_cg, x, h, y, yn, x0)
    test_fmin_convolve(opt.fmin_bfgs, x, h, y, yn, x0)

if __name__ == "__main__":
    test_n(200, 20, 0.1) 

from scipy.optimize import fsolve
from math import sin,cos

def f(x):
    x0 = float(x[0])
    x1 = float(x[1])
    x2 = float(x[2])
    return [
        5*x1+3,
        4*x0*x0 - 2*sin(x1*x2),
        x1*x2 - 1.5
    ]

result = fsolve(f, [1,1,1])

print result
print f(result)

# -*- coding: utf-8 -*-
from scipy.optimize import fsolve
from math import sin,cos
def f(x):
    x0 = float(x[0])
    x1 = float(x[1])
    x2 = float(x[2])
    return [
        5*x1+3,
        4*x0*x0 - 2*sin(x1*x2),
        x1*x2 - 1.5
    ]
    
def j(x):
    x0 = float(x[0])
    x1 = float(x[1])
    x2 = float(x[2])    
    return [
        [0, 5, 0],
        [8*x0, -2*x2*cos(x1*x2), -2*x1*cos(x1*x2)],
        [0, x2, x1]
    ]
 
result = fsolve(f, [1,1,1], fprime=j)
print result
print f(result)

# -*- coding: utf-8 -*-
import numpy as np
import pylab as pl
from scipy import interpolate 

x = np.linspace(0, 2*np.pi+np.pi/4, 10)
y = np.sin(x)

x_new = np.linspace(0, 2*np.pi+np.pi/4, 100)
f_linear = interpolate.interp1d(x, y)
tck = interpolate.splrep(x, y)
y_bspline = interpolate.splev(x_new, tck)

pl.plot(x, y, "o",  label=u"原始数据")
pl.plot(x_new, f_linear(x_new), label=u"线性插值")
pl.plot(x_new, y_bspline, label=u"B-spline插值")
pl.legend()
pl.show()

# -*- coding: utf-8 -*-
from scipy.integrate import odeint 
import numpy as np 

def lorenz(w, t, p, r, b): 
    # 给出位置矢量w，和三个参数p, r, b计算出
    # dx/dt, dy/dt, dz/dt的值
    x, y, z = w
    # 直接与lorenz的计算公式对应 
    return np.array([p*(y-x), x*(r-z)-y, x*y-b*z]) 

t = np.arange(0, 30, 0.01) # 创建时间点 
# 调用ode对lorenz进行求解, 用两个不同的初始值 
track1 = odeint(lorenz, (0.0, 1.00, 0.0), t, args=(10.0, 28.0, 3.0)) 
track2 = odeint(lorenz, (0.0, 1.01, 0.0), t, args=(10.0, 28.0, 3.0)) 

# 绘图
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt 

fig = plt.figure()
ax = Axes3D(fig)
ax.plot(track1[:,0], track1[:,1], track1[:,2])
ax.plot(track2[:,0], track2[:,1], track2[:,2])
plt.show()

# -*- coding: utf-8 -*-
import scipy.weave as weave
import numpy as np
import time

def my_sum(a):
    n=int(len(a))
    code="""
    int i;

    double counter;
    counter =0;
    for(i=0;i<n;i++){
        counter=counter+a(i);
    }
    return_val=counter;
    """

    err=weave.inline(
        code,['a','n'],
        type_converters=weave.converters.blitz,
        compiler="gcc"
    )
    return err

a = np.arange(0, 10000000, 1.0)
# 先调用一次my_sum，weave会自动对C语言进行编译，此后直接运行编译之后的代码
my_sum(a)

start = time.clock()
for i in xrange(100):
    my_sum(a)  # 直接运行编译之后的代码
print "my_sum:", (time.clock() - start) / 100.0

start = time.clock()
for i in xrange(100):
    np.sum( a ) # numpy中的sum，其实现也是C语言级别
print "np.sum:", (time.clock() - start) / 100.0

start = time.clock()
print sum(a) # Python内部函数sum通过数组a的迭代接口访问其每个元素，因此速度很慢
print "sum:", time.clock() - start