import numpy as np

import matplotlib.pyplot as plt

from scipy.optimize import leastsq

plt.figure(figsize=(9,9))

x=np.linspace(0,10,1000)

X = np.array([8.19, 2.72, 6.39, 8.71, 4.7, 2.66, 3.78])

Y = np.array([7.01, 2.78, 6.47, 6.71, 4.1, 4.23, 4.05])

#计算以p为参数的直线和原始数据之间的误差

def f(p):

    k, b = p

    return(Y-(k*X+b))

#leastsq使得f的输出数组的平方和最小，参数初始值为[1,0]

r = leastsq(f, [1,0])

k, b = r[0]

print("k=",k,"b=",b)

plt.scatter(X,Y, s=100, alpha=1.0, marker='o',label=u'数据点')

y=k*x+b

ax = plt.gca()  #gca获取轴这个对象

ax.set_xlabel(..., fontsize=20)

ax.set_ylabel(..., fontsize=20)

#设置坐标轴标签字体大小

plt.plot(x, y, color='r',linewidth=5, linestyle=":",markersize=20, label=u'拟合曲线')

plt.legend(loc=0, numpoints=1)

leg = plt.gca().get_legend()

ltext  = leg.get_texts()

plt.setp(ltext, fontsize='xx-large')

plt.xlabel(u'安培/A')

plt.ylabel(u'伏特/V')

plt.xlim(0, x.max() * 1.1)

plt.ylim(0, y.max() * 1.1)

plt.xticks(fontsize=20)

plt.yticks(fontsize=20)

#刻度字体大小

plt.legend(loc='upper left')

plt.show()

#最小二乘拟合实例

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*cos(2*pi*k*x + theta)

    """

    A, k, theta = p

    return A*np.sin(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, 20, 100)

A, k, theta = 10, 3, 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 = [10, 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, color='r',label=u"真实数据")

pl.plot(x, y1, color='b',label=u"带噪声的实验数据")

pl.plot(x, func(x, plsq[0]), color='g', label=u"拟合数据")

pl.legend()

pl.show()

# -*- coding: utf-8 -*-

"""

Created on Thu Jul 27 16:42:30 2017

@author: Dell

"""

import numpy as np

import pylab as pl

from scipy import interpolate

import matplotlib.pyplot as plt

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)

plt.xlabel(u'安培/A')

plt.ylabel(u'伏特/V')

plt.plot(x, y, "o",  label=u"原始数据")

plt.plot(x_new, f_linear(x_new), label=u"线性插值")

plt.plot(x_new, y_bspline, label=u"B-spline插值")

pl.legend()

pl.show()

# -*- coding: utf-8 -*-

"""

Created on Thu Jul 27 16:53:21 2017

@author: Dell

"""

import numpy as np

from scipy import interpolate

import pylab as pl

#创建数据点集并绘制

pl.figure(figsize=(12,9))

x = np.linspace(0, 10, 11)

y = np.sin(x)

ax=pl.plot()

pl.plot(x,y,'ro')

#建立插值数据点

xnew = np.linspace(0, 10, 101)

for kind in ['nearest', 'zero','linear','quadratic']:

    #根据kind创建插值对象interp1d

    f = interpolate.interp1d(x, y, kind = kind)

    ynew = f(xnew)#计算插值结果

    pl.plot(xnew, ynew, label = str(kind))

pl.xticks(fontsize=20)

pl.yticks(fontsize=20)

pl.legend(loc = 'lower right')

pl.show()

# -*- coding: utf-8 -*-

"""

Created on Thu Jul 27 16:53:21 2017

@author: Dell

"""

import numpy as np

from scipy import interpolate

import pylab as pl

#创建数据点集并绘制

pl.figure(figsize=(12,9))

x = np.linspace(0, 10, 11)

y = np.sin(x)

ax=pl.plot()

pl.plot(x,y,'ro')

#建立插值数据点

xnew = np.linspace(0, 10, 101)

for kind in ['nearest', 'zero','linear','quadratic',5]:

    #根据kind创建插值对象interp1d

    f = interpolate.interp1d(x, y, kind = kind)

    ynew = f(xnew)#计算插值结果

    pl.plot(xnew, ynew, label = str(kind))

pl.xticks(fontsize=20)

pl.yticks(fontsize=20)

pl.legend(loc = 'lower right')

pl.show()

运行得到