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max： z = 4x1 + 3x2

st:      2x1 + 3x2<=10

           x1 + x2 <=8

           x2 <= 7

           x1,x2 > 0

from scipy.optimize import linprog

c = [4,3]        #默認linprog求解的是最小值，若求最大值，此處c取反即可得到最大值的相反數。

A = [[2,3],[1,1]]

b = [10,8]

x1_bounds = [0,None]

x2_bounds =[0,7]

res = linprog(c,A,b,bounds=(x1_bounds,x2_bounds))

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import numpy as np

import matplotlib.pyplot as plt

x = np.arange(1990,1997,1)

y = np.array([70 ,122 ,144 ,152, 174, 196, 202])

z1 = ployfit(x,y,1)  #之前畫過原始數據，數據走向為ax+b類型。故采用一次多項式擬合

p1 = np.ploy1d(z1)

yvalue = p1(x)

plt.plot(x,y,'*',label = '原始數據')

plt.plot(z1,yvalue,label = '擬合曲線')

plt.xlabel('x axis')

plt.ylabel('y axis')

plt.legend(loc = 4 )

plt.tittle('多項式擬合')

plt.show()

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from __future__ import print_function

from __future__ import division

import numpy as np

import scipy as sp

import scipy.misc

def f(x): return 2*x*x + 3*x + 1

print(sp.misc.derivative(f, 2))</pre>

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from __future__ import print_function

from __future__ import division

import numpy as np

import scipy as sp

import scipy.integrate

f = lambda x : x**2

print(sp.integrate.quad(f, 0, 2))

print(sp.integrate.fixed_quad(f, 0, 2))

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from __future__ import print_function

from __future__ import division

import numpy as np

import scipy as sp

import scipy.optimize

def f(x):

    return [5*x[1] + 3, 4*x[0]*x[0], x[1]*x[2] - 1.5]

ans = sp.optimize.fsolve(f, [0, 0, 0])

print(ans)

print(f(ans))

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from __future__ import print_function

from __future__ import division

import numpy as np

import scipy as sp

import matplotlib.pylab as plt

import scipy.linalg

a = np.array([[1, 3, 5], [2, 5, 1], [2, 3, 8]])

b = np.array([10, 8, 3])

print(sp.linalg.solve(a, b))

# print(sp.linalg.inv(a).dot(b))