文章目录
（一）简单陈述本文章的内容
python建模会持续更新，用途是只作为个人笔记。我博客中的所有资料都可通过我提供的链接永久获取，最后希望大家一起相互促进，相互努力。
本文章只介绍如何用python中的库和模块求解相对应的建模，不介绍相应的理论知识，理论知识可以查看链接下载：《数学建模算法与应用》 提取码：fx4q
本文章所有（源码、文件）获取：链接：https://pan.baidu.com/s/1SFazXZ5uqJ2A6rIYXiG0cA
提取码：kozt
（二）常用导入文件方式
- 为了方便初学者使用，下面是把几种常用的导入文件方法写入到一个定义函数中。
books文件下载：books提取码：ouvw--来自百度网盘的分享
# 数据导入例程
import pandas as pd
# 读取数据文件
def readDataFile(readPath): # readPath: 数据文件的地址和文件名
# readPath = "../data/youcansxupt.csv" # 文件路径也可以直接在此输入
try:
if (readPath[-4:] == ".csv"):
dfFile = pd.read_csv(readPath, header=0, sep=",") # 间隔符为逗号，首行为标题行
# dfFile = pd.read_csv(filePath, header=None, sep=",") # sep: 间隔符，无标题行
elif (readPath[-4:] == ".xls") or (readPath[-5:] == ".xlsx"): # sheet_name 默认为 0
dfFile = pd.read_excel(readPath, header=0) # 首行为标题行
# dfFile = pd.read_excel(filePath, header=None) # 无标题行
elif (readPath[-4:] == ".dat"): # sep: 间隔符，header：首行是否为标题行
dfFile = pd.read_table(readPath, sep=" ", header=0) # 间隔符为空格，首行为标题行
# dfFile = pd.read_table(filePath,sep=",",header=None) # 间隔符为逗号，无标题行
else:
print("不支持的文件格式。")
except Exception as e:
print("读取数据文件失败：{}".format(str(e)))
return
return dfFile
# 主程序
def main():
# 读取数据文件
readPath = "books.csv" # 数据文件的地址和文件名
dfFile = readDataFile(readPath) # 调用读取文件子程序
print(type(dfFile)) # 查看 dfFile 数据类型
print(dfFile.shape) # 查看 dfFile 形状（行数，列数）
print(dfFile.head()) # 显示 dfFile 前 5 行数据
return
if __name__ == '__main__':
main()
OUT:
<class 'pandas.core.frame.DataFrame'>
(10000, 23)
id ... small_image_url
0 1 ... https://images.gr-assets.com/books/1447303603s...
1 2 ... https://images.gr-assets.com/books/1474154022s...
2 3 ... https://images.gr-assets.com/books/1361039443s...
3 4 ... https://images.gr-assets.com/books/1361975680s...
4 5 ... https://images.gr-assets.com/books/1490528560s...
[5 rows x 23 columns]
（三）线性规划
3.1 线性规划的一般模型
max(min) = ∑ j = 1 n \sum_{j=1}^n ∑j=1n c j c_j cj x j x_j xj,
s . t . { ∑ j = 1 n a i j x j ≤ ( ≥ ) b i , i = 1 , 2 , 3 , ⋅ ⋅ ⋅ , m x j > = 0 , j = 1 , 2 , 3 , ⋅ ⋅ ⋅ , m s.t. \begin{cases} \sum_{j=1}^n a_{ij}x_{j} \le(\ge) b_{i}, &i = 1,2,3,···,m\\ x_{j}>=0, &j = 1,2,3,···,m \end{cases} s.t.{
∑j=1naijxj≤(≥)bi,xj>=0,i=1,2,3,⋅⋅⋅,mj=1,2,3,⋅⋅⋅,m
3.2 运用python各种库和模块求解线性方程
min z = c T x c^Tx cTx,
s . t . { A ∙ x ≤ b , A e q ∙ x = b e q , L B ≤ x ≤ U B s.t. \begin{cases} A \bullet x \le b, &\\ Aeq \bullet x = beq, &\\ LB \le x \le UB \end{cases} s.t.⎩⎪⎨⎪⎧A∙x≤b,Aeq∙x=beq,LB≤x≤UB
linprog 的基本调用格式为
from scipy.optimize import linprog
# 默认每个决策变量下界为0，上界为正无穷
res = linprog(c,A,b,Aeq,beq)
res = linprog(c, A = None, b = None, Aeq = None, beq = None, bounds = None, mothod = 'simplex')
print(res.fun) # 显示目标函数的最小值
print(res.x) # 显示最优解
3.2.1 Scipy线性规划模型
max z = 2 x 1 + 3 x 2 − 5 x 3 2x_1 + 3x_2 - 5x_3 2x1+3x2−5x3,
s . t . { x 1 + 3 x 2 + x 3 ≤ 12 , 2 x 1 − 5 x 2 + x 3 ≥ 10 , x 1 + x 2 + x 3 = 7 , x 1 , x 2 , x 3 ≥ 0 s.t. \begin{cases} x_1 + 3x_2 + x_3 \le 12,&\\ 2x_1 - 5x_2 + x_3 \ge 10,\\ x_1 + x_2 +x_3 = 7,\\ x_1, x_2, x_3 \ge 0 \end{cases} s.t.⎩⎪⎪⎪⎨⎪⎪⎪⎧x1+3x2+x3≤12,2x1−5x2+x3≥10,x1+x2+x3=7,x1,x2,x3≥0
# Scipy线性规划模型
from scipy.optimize import linprog
c = [-2,-3,5]; A = [[1,3,1],[-2,5,-1]] # 或者 c = np.array([-2,-3,5])
b = [[12],[-10]]; Aeq = [[1,1,1]]; beq = [7]
LB = [0,0,0]
UB = [None]*len(c) # 生成3个None的列表
bound = tuple(zip(LB,UB)) #生成决策向量界限的元组
res = linprog(c,A,b,Aeq,beq,bound)
print("目标函数的最小值:",res.fun)
print("最优解为:",res.x)
# 目标函数的最小值: -14.571428565645057, 所以最大值为:14.571428565645057
# 最优解为: [6.42857143e+00 5.71428571e-01 2.35900788e-10]
OUT：
目标函数的最小值: -14.571428565645057
最优解为: [6.42857143e+00 5.71428571e-01 2.35900788e-10]
Process finished with exit code 0
3.2.2 pulp线性规划模型
与2.2.1一样的题目，用不同模型求解
max z = 2 x 1 + 3 x 2 − 5 x 3 2x_1 + 3x_2 - 5x_3 2x1+3x2−5x3,
s . t . { x 1 + 3 x 2 + x 3 ≤ 12 , 2 x 1 − 5 x 2 + x 3 ≥ 10 , x 1 + x 2 + x 3 = 7 , x 1 , x 2 , x 3 ≥ 0 s.t. \begin{cases} x_1 + 3x_2 + x_3 \le 12,&\\ 2x_1 - 5x_2 + x_3 \ge 10,\\ x_1 + x_2 +x_3 = 7,\\ x_1, x_2, x_3 \ge 0 \end{cases} s.t.⎩⎪⎪⎪⎨⎪⎪⎪⎧x1+3x2+x3≤12,2x1−5x2+x3≥10,x1+x2+x3=7,x1,x2,x3≥0
# pulp线性规划模型
import pulp
# 定义一个规划问题
MyProbLP = pulp.LpProblem("LPProbDemo1", sense=pulp.LpMaximize) # 求最大值
# 定义决策变量
x1 = pulp.LpVariable('x1', lowBound=0, upBound=7, cat='Continuous')
x2 = pulp.LpVariable('x2', lowBound=0, upBound=7, cat='Continuous')
x3 = pulp.LpVariable('x3', lowBound=0, upBound=7, cat='Continuous')
# 添加目标函数
MyProbLP += 2*x1 + 3*x2 - 5*x3 # 设置目标函数
# 添加约束条件
MyProbLP += (2*x1 - 5*x2 + x3 >= 10) # 不等式约束
MyProbLP += (x1 + 3*x2 + x3 <= 12) # 不等式约束
MyProbLP += (x1 + x2 + x3 == 7) # 等式约束
MyProbLP.solve()
print("Status:", pulp.LpStatus[MyProbLP.status]) # 输出求解状态
for v in MyProbLP.variables(): # youcans
print(v.name, "=", v.varValue) # 输出每个变量的最优值
print("Max F(x) = ", pulp.value(MyProbLP.objective)) #输出最优解的目标函数值
OUT：
Status: Optimal
x1 = 6.4285714
x2 = 0.57142857
x3 = 0.0
Max F(x) = 14.57142851
Process finished with exit code 0
3.2.3 cvxopt.solvers 模块求解
- cvxopt.solvers 模块求解线性规划模型的标准型如下：
min z = c T x c^Tx cTx,
s . t . { A ∙ x ≤ b , A e q ∙ x = b e q . s.t. \begin{cases} A \bullet x \le b, &\\ Aeq \bullet x = beq. \end{cases} s.t.{ A∙x≤b,Aeq∙x=beq. - 求解线性规划
min z = − 4 x 1 − 5 x 2 -4x_1 - 5x_2 −4x1−5x2,
s . t . { 2 x 1 + x 2 ≤ 3 , x 1 + 2 x 2 ≤ 3 , x 1 ≥ 0 , x 2 ≥ 0. s.t. \begin{cases} 2x_1 + x_2 \le 3,\\ x_1 + 2x_2 \le 3,&\\ x_1 \ge 0, x_2 \ge 0. \end{cases} s.t.⎩⎪⎨⎪⎧2x1+x2≤3,x1+2x2≤3,x1≥0,x2≥0.
# 线性规划 使用cvxopy.solvers模块求解
import numpy
from cvxopt import matrix,solvers
c = matrix([2.,1]); A = matrix([[-1.,1],[-1,-1],[1,-2],[0,-1]]).T # matrix([2.,1]):记住matrix中第一个元素不许是float型
b = matrix([1.,-2,4,0]); Aeq = matrix([1.,2],(1,2)) # Aeq为行向量
beq = matrix(3.5); sol = solvers.lp(c,A,b,Aeq,beq)
print("最优解为:\n",sol['x'])
print("最优值为:\n",sol['primal objective'])
OUT:
D:\ProgramData\Anaconda3\python.exe C:/Users/Administrator/PycharmProjects/pythonProject/数学建模/线性规划/线性规划3.cvxopy.solvers.py
pcost dcost gap pres dres k/t
0: 5.5556e+00 1.2222e+00 1e+01 0e+00 2e+00 1e+00
1: 4.6038e+00 3.7995e+00 2e+00 1e-16 4e-01 2e-01
2: 2.5229e+00 2.4639e+00 2e-01 2e-16 4e-02 4e-02
3: 2.5002e+00 2.4997e+00 2e-03 1e-16 4e-04 4e-04
4: 2.5000e+00 2.5000e+00 2e-05 4e-16 4e-06 4e-06
5: 2.5000e+00 2.5000e+00 2e-07 1e-16 4e-08 4e-08
Optimal solution found.
最优解为:
[ 5.00e-01]
[ 1.50e+00]
最优值为:
2.500000024611047
3.2.4 用cvxpy库求解
P183 . 例题: 已知某种商品 6 个仓库的存货量，8个客户对该商品的需求量，单位商品运价如下表所示。试确定6个仓库到8个客户的商品调运数量，使总的运输费用最小。
| 仓库：单位运价：客户 | V1 | V2 | V3 | V4 | V5 | V6 | V7 | V8 | 存货量 |
|---|---|---|---|---|---|---|---|---|---|
| W1 | 6 | 2 | 6 | 7 | 4 | 2 | 5 | 9 | 60 |
| W2 | 4 | 9 | 5 | 3 | 8 | 5 | 3 | 2 | 55 |
| W3 | 5 | 2 | 1 | 9 | 7 | 4 | 3 | 3 | 51 |
| W4 | 7 | 6 | 7 | 3 | 9 | 2 | 7 | 1 | 43 |
| W5 | 2 | 3 | 9 | 5 | 7 | 2 | 6 | 5 | 41 |
| W6 | 5 | 5 | 2 | 2 | 8 | 1 | 4 | 3 | 52 |
| 需求量 | 35 | 37 | 22 | 32 | 41 | 32 | 43 | 38 |
解 ： 设 x i j x_{ij} xij(i = 1,2,·······,8)表示第 i 个仓库运到第 j 个客户的商品数量， c i j c_{ij} cij表示第 i个仓库到底 j 个客户的单位运价， d j d_j dj 表示第 j 个客户的需求量， e i e_i ei 表示第 i 仓库的存货量，建立如下线性规划模型：
min ∑ i = 1 6 \sum_{i=1}^6 ∑i=16 ∑ j = 1 8 \sum_{j=1}^8 ∑j=18 c i j x i j c_{ij}x_{ij} cijxij,
s . t . { ∑ j = 1 8 x i j ≤ e i , i = 1 , 2 , ⋅ ⋅ ⋅ , 6 ∑ i = 1 6 x i j = d j , j = 1 , 2 , ⋅ ⋅ ⋅ , 8 x i j ≥ 0 , i = 1 , 2 , ⋅ ⋅ ⋅ , 6 ; j = 1 , 2 , ⋅ ⋅ ⋅ , 8. s.t. \begin{cases} \sum_{j=1}^8 x_{ij} \le e_i,&i = 1,2,···,6\\ \sum_{i=1}^6 x_{ij} = d_j,&j = 1,2,···,8\\ x_{ij} \ge 0, &i = 1,2,···,6;j = 1,2,···,8. \end{cases} s.t.⎩⎪⎨⎪⎧∑j=18xij≤ei,∑i=16xij=dj,xij≥0,i=1,2,⋅⋅⋅,6j=1,2,⋅⋅⋅,8i=1,2,⋅⋅⋅,6;j=1,2,⋅⋅⋅,8.
把上表中7行9列的数据保存到Excel文件，并命名为工作簿3.xlsx。
链接：工作簿3.xlsx 提取码：kzmp
import cvxpy as cp
import numpy as np
import pandas as pd
d1 = pd.read_excel("工作簿3.xlsx",header=None)
d2 = d1.values; c = d2[:-1,:-1]
d = d2[-1,:-1].reshape(1,-1)
e = d2[:-1,-1].reshape(-1,1)
x = cp.Variable((6,8))
obj = cp.Minimize(cp.sum(cp.multiply(c,x)))
con = [cp.sum(x,axis=1,keepdims=True)<=e,cp.sum(x,axis=0,keepdims=True)==d,x>=0]
prob = cp.Problem(obj,con)
prob.solve(solver='GLPK_MI',verbose=True)
print("最优值为:\n",prob.value)
print("最优解为:\n",x.value)
OUT:
最优值为:
664.0
最优解为:
[[-0. 19. -0. -0. 41. -0. -0. -0.]
[-0. -0. -0. 32. -0. -0. -0. 1.]
[-0. 12. 22. -0. -0. -0. 17. -0.]
[-0. -0. -0. -0. -0. 6. -0. 37.]
[35. 6. -0. -0. -0. -0. -0. -0.]
[-0. -0. -0. -0. -0. 26. 26. -0.]]
Process finished with exit code 0