列举几个python解决数学建模的例子

# max： z = 4x1 + 3x2
# st:      -2x1 - 3x2<=-10
#          x1 + x2 <=8
#          x2 <= 7
#          x1,x2 > 0

from scipy.optimize import linprog
c = [4,3]        #默认linprog求解的是最小值，若求最大值，此处c取反即可得到最大值的相反数。
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))
print(res)

     con: array([], dtype=float64)
     fun: 10.000000000157014
 message: ‘Optimization terminated successfully.‘
     nit: 4
   slack: array([-1.24439126e-10,  4.66666667e+00])
  status: 0
 success: True
       x: array([1.40727180e-10, 3.33333333e+00])

import numpy as np
import matplotlib.pyplot as plt
x = np.arange(1, 6, 1)
y = np.array([2, 6, 12, 20, 30])  # y=x^2+x
#拟合
print(np.polyfit(x, y, 2))   

#可视化
f = np.polyfit(x, y, 2)
p = np.poly1d(f)
yvals = p(x)
plt.figure(figsize=(10,8))
plt.plot(x, y, ‘.‘)
plt.plot(x, yvals)
plt.xlabel(‘x轴‘)   #可以用来标注
plt.ylabel(‘y轴‘)
plt.show()

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))

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))

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])  #实际用起来问题很大，很不准
print(ans)
#print(ans[0]*ans[1]*ans[2])
print(f(ans))

import numpy as np
import scipy as sp
import scipy.optimize
import scipy.linalg


a = np.array([[1, 1, 1], [2, 1, 1], [1, -1, 1]])
b = np.array([2, 3, 6])
print(sp.linalg.solve(a, b))
#print(sp.linalg.inv(a).dot(b))  #或利用矩阵运算求解

# coding=utf-8

from pulp import LpProblem, LpVariable, LpConstraint, LpConstraintLE, LpConstraintGE, LpMaximize, LpBinary, LpStatus

# Create a new model
m = LpProblem(name="MIP Model", sense=LpMaximize)

# Create variables
x = LpVariable(cat=LpBinary, name="x")
y = LpVariable(cat=LpBinary, name="y")
z = LpVariable(cat=LpBinary, name="z")

# Add constraint: x + 2y + 3z <= 4
m += LpConstraint(e=(x + 2*y + 3*z), sense=LpConstraintLE, rhs=4, name=‘c0‘)

# Add constraint: x + y >= 1
m += LpConstraint(e=(x + y), sense=LpConstraintGE, rhs=1, name=‘c1‘)

# Set objective
m.setObjective(x + y + 2*z)

# Calculate with the default CBC optimizer
status = m.solve()

if LpStatus[status] == ‘Optimal‘:
    for v in m.variables():
        print(‘%s %g‘ % (v.name, v.varValue))
    print(‘Obj: %g‘ % m.objective.value())