Python调用Gurobi基本操作_gurobi python

接上篇学会了如何用python调用gurobipy之后，这篇总结一些学到的基本操作。

tuplelist、tupledict、multidict、创建list、

1.tuplelist方法：

tuplelist是Python list的扩展对象，使用tuplelist()不能忘记from gurobipy import *，tuplelist增加了快速筛选select功能，比传统的if...else...筛选速度快。

from gurobipy import *
import time
 
T1 = time.time()
Cities=[("A","B"),("A","C"),("B","C"),("B","D"),("C","D")]
Routes=tuplelist(Cities)
print(Routes.select("A","*"))
T2 =time.time()
 
print('程序运行时间:%s毫秒' % ((T2 - T1)*1000))

tuplelist运行结果：

C:\Users\xzr\.conda\envs\py310gurobi\python.exe F:\PycharmProjects\workspace\untitled\jizulunban\caogao.py 
<gurobi.tuplelist (2 tuples, 2 values each):
 ( A , B )
 ( A , C )
>
程序运行时间:1.9958019256591797毫秒
 
进程已结束,退出代码0

for...else...方法：

from gurobipy import *
import time
 
T1 = time.time()
Cities=[("A","B"),("A","C"),("B","C"),("B","D"),("C","D")]
Result=[]
for i,j in Cities:
    if i=="A":
        Result.append((i,j))
print(Result)
T2 =time.time()
 
print('程序运行时间:%s毫秒' % ((T2 - T1)*1000))

for...else...运行结果：

C:\Users\xzr\.conda\envs\py310gurobi\python.exe F:\PycharmProjects\workspace\untitled\jizulunban\caogao.py 
[('A', 'B'), ('A', 'C')]
程序运行时间:0.0毫秒
 
进程已结束,退出代码0

尴尬了，竟然是for...else...的运行速度更快，原因是数据量太少了体现不出效果。

2.tupledict方法：

tupledict是Python Dictionary的扩展对象，键值是tuple（元组），可以使用select、sum、prob函数。用于变量和约束。后面有详细介绍。

3.multidict方法：

multidict()创建tuplelist和tupledict的便捷方法。代码示例：

from gurobipy import *
import time
 
# T1 = time.time()
cities,supply,demand = multidict({
    "A":[100,20],
    "B":[150,50],
    "C":[20,300],
    "D":[10,200]})
print(cities)
print(supply)
print(demand)
# T2 =time.time()

运行结果：

C:\Users\xzr\.conda\envs\py310gurobi\python.exe F:\PycharmProjects\workspace\untitled\jizulunban\caogao.py 
['A', 'B', 'C', 'D']
{'A': 100, 'B': 150, 'C': 20, 'D': 10}
{'A': 20, 'B': 50, 'C': 300, 'D': 200}
 
进程已结束,退出代码0

运行结果第一行是list，第二三行是dictionary。

4.创建list：

python有多种创建list的方法：

a=[]
a.append("A")

b=[i**2 for i in range(6)] #[0,1,4,9,16,25]

c=[(i,j)for j in range(4) for i in range(j)] #[(0,1),(0,2),(1,2),(0,3),(1,3),(2,3)]

d=[i for i in range(10) if i not in b]  #[2,3,5,6,7,8]

Pairs=[]
for j in range(4):
    for i in range(j):
        Pairs.append((i,j))

对于求和，Python的Generator（生成器）：

SumSquares=sum(i**2 for i in range(6)) #55

Gurobi中采用quicksum，效率更高：

obj=quicksum(cost[i,j]*x[i,j] for i,j in arcs)

5.tupledict建模示例

tupledict（Gurobi变量一般都是tupledict类型）有sum函数

from gurobipy import *
import time
 
# T1 = time.time()
m=Model()
x=m.addVars(3,4,vtype=GRB.BINARY,name="x")
m.addConstrs((x.sum(i,"*")<=1 for i in range(3)),name="con")
m.update()
m.write("test.lp")
# T2 =time.time()
 
# print('程序运行时间:%s毫秒' % ((T2 - T1)*1000))

运行后，打开“test.lp”文件查看写入的模型

\ LP format - for model browsing. Use MPS format to capture full model detail.
Minimize
 
Subject To
 con[0]: x[0,0] + x[0,1] + x[0,2] + x[0,3] <= 1
 con[1]: x[1,0] + x[1,1] + x[1,2] + x[1,3] <= 1
 con[2]: x[2,0] + x[2,1] + x[2,2] + x[2,3] <= 1
Bounds
Binaries
 x[0,0] x[0,1] x[0,2] x[0,3] x[1,0] x[1,1] x[1,2] x[1,3] x[2,0] x[2,1]
 x[2,2] x[2,3]
End

太妙了太妙了

tupledict（Gurobi变量一般都是tupledict类型）还有prob函数，用于变量和系数相乘后累加

以下表达式等效

obj=quicksum(cost[i,j]*x[i,j] for i,j in arcs)

obj=x.prod(cost)

建模建议，尽量采用稀疏方式，采用tuplelists筛选和指定合适的下标组合关系，基于这些组合关系建立变量和数据字典，利用tuplelist.select()以及tupledict.select()，tupledict.sum()，tupledict.prob()来对下标进行组合处理。

 

  

 

 

 

 Gurobi建模举例1

 ​​​​​​​

from gurobipy import *
 
# T1 = time.time()
 
try:
    #Create a new model
    m=Model("mip1")
 
    #Create variables
    x=m.addVar(vtype=GRB.BINARY,name="x")
    y=m.addVar(vtype=GRB.BINARY,name="y")
    z=m.addVar(vtype=GRB.BINARY,name="z")
 
    #Set objective
    m.setObjective(x+y+2*z,GRB.MAXIMIZE)
 
    #Add constraint:x+2y+3z<=4
    m.addConstr(x+2*y+3*z<=4,"c0")
 
    #Add constraint:x+y>=1
    m.addConstr(x+y>=1,"c1")
 
    m.optimize()
 
    for v in m.getVars():      #getVars获取所有变量
        print("%s %g" % (v.varName,v.x)) #(v.varName,v.x)是（变量名字，优化结果）
    print("Obj: %g" % m.objVal)
 
except GurobiError as e:
    print("Error code" + str(e.errno)+":"+str(e))
 
except AttributeError:
    print("Encountered an attribute error")
 
 
# T2 =time.time()
# print('程序运行时间:%s毫秒' % ((T2 - T1)*1000))

 运行结果

C:\Users\xzr\.conda\envs\py310gurobi\python.exe F:\PycharmProjects\workspace\untitled\jizulunban\caogao.py 
Gurobi Optimizer version 10.0.0 build v10.0.0rc2 (win64)
 
CPU model: Intel(R) Core(TM) i7-7500U CPU @ 2.70GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 2 physical cores, 4 logical processors, using up to 4 threads
 
Optimize a model with 2 rows, 3 columns and 5 nonzeros
Model fingerprint: 0x98886187
Variable types: 0 continuous, 3 integer (3 binary)
Coefficient statistics:
  Matrix range     [1e+00, 3e+00]
  Objective range  [1e+00, 2e+00]
  Bounds range     [1e+00, 1e+00]
  RHS range        [1e+00, 4e+00]
Found heuristic solution: objective 2.0000000
Presolve removed 2 rows and 3 columns
Presolve time: 0.02s
Presolve: All rows and columns removed
 
Explored 0 nodes (0 simplex iterations) in 0.03 seconds (0.00 work units)
Thread count was 1 (of 4 available processors)
 
Solution count 2: 3 2 
 
Optimal solution found (tolerance 1.00e-04)
Best objective 3.000000000000e+00, best bound 3.000000000000e+00, gap 0.0000%
x 1
y 0
z 1
Obj: 3
 
进程已结束,退出代码0

Gurobi建模举例2

 

 

from gurobipy import *
 
categories,minNutrition,maxNutrition=multidict({
    "calories":[1800,2200],
    "protein":[91,GRB.INFINITY],
    "fat":[0,65],
    "sodium":[0,1779]
})
 
foods,cost=multidict({
    "hamburger":2.49,
    "chicken":2.89,
    "hot dog":1.50,
    "fries":1.89,
    "macaroni":2.09,
    "pizza":1.99,
    "salad":2.49,
    "milk":0.89,
    "ice cream":1.59
})
 
#Nutrition values for the foods
nutritionValues={
    ("hamburger","calories"):410,
    ("hamburger","protein"):24,
    ("hamburger","fat"):26,
    ("hamburger","sodium"):730,
    ("chicken","calories"):420,
    ("chicken", "protein"):32,
    ("chicken", "fat"):10,
    ("chicken", "sodium"):1190,
    ("hot dog","calories"):560,
    ("hot dog", "protein"):20,
    ("hot dog", "fat"):32,
    ("hot dog", "sodium"):1800,
    ("fries","calories"):380,
    ("fries", "protein"):4,
    ("fries", "fat"):19,
    ("fries", "sodium"):720,
    ("macaroni", "calories"): 320,
    ("macaroni", "protein"): 12,
    ("macaroni", "fat"): 10,
    ("macaroni", "sodium"): 930,
    ("pizza", "calories"): 320,
    ("pizza", "protein"): 15,
    ("pizza", "fat"): 12,
    ("pizza", "sodium"): 820,
    ("salad", "calories"): 320,
    ("salad", "protein"): 31,
    ("salad", "fat"): 12,
    ("salad", "sodium"): 1230,
    ("milk", "calories"): 100,
    ("milk", "protein"): 8,
    ("milk", "fat"): 2.5,
    ("milk", "sodium"): 125,
    ("ice cream", "calories"): 330,
    ("ice cream", "protein"): 8,
    ("ice cream", "fat"): 10,
    ("ice cream", "sodium"): 180
}
 
#Model
m=Model("diet")
 
#Create decision variables for the foods to buy
buy=m.addVars(foods,name="buy")
#也可以是：
# buy=[]
# for f in foods:
#     buy[f]=m.addVar(name=f)
 
#目标函数是最小化cost
m.setObjective(buy.prod(cost),GRB.MINIMIZE)
#如果使用循环结构，应该是：
# m.setObjective(sum(buy[f]*cost[f] for f in foods),GRB.MINIMIZE)
 
#Nutrition constraints
m.addConstrs(
    (quicksum(nutritionValues[f,c]*buy[f] for f in foods)
     == [minNutrition[c],maxNutrition[c]]
     for c in categories),"_")
#如果使用循环结构，应该是：
# for c in categories:
#     m.addRange(
#         sum(nutritionValues[f,c] * buy[f] for f in foods),minNutrition[c],maxNutrition[c],c)
#     )
 
def printSolution():
    if m.status == GRB.Status.OPTIMAL:
        print("\nCost:%g" % m.objval)
        print("\nBuy:")
        buyx=m.getAttr("x",buy)
        for f in foods:
            if buy[f].x>0.0001:
                print("%s %g" % (f,buyx[f]))
    else:
        print("No solution")
 
#solve
m.optimize()
printSolution()
 
 
# T1 = time.time()
 
 
 
 
# T2 =time.time()
# print('程序运行时间:%s毫秒' % ((T2 - T1)*1000))

 运行结果

C:\Users\xzr\.conda\envs\py310gurobi\python.exe F:\PycharmProjects\workspace\untitled\jizulunban\caogao.py 
Gurobi Optimizer version 10.0.0 build v10.0.0rc2 (win64)
 
CPU model: Intel(R) Core(TM) i7-7500U CPU @ 2.70GHz, instruction set [SSE2|AVX|AVX2]
Thread count: 2 physical cores, 4 logical processors, using up to 4 threads
 
Optimize a model with 4 rows, 12 columns and 39 nonzeros
Model fingerprint: 0xed649f3c
Coefficient statistics:
  Matrix range     [1e+00, 2e+03]
  Objective range  [9e-01, 3e+00]
  Bounds range     [7e+01, 2e+03]
  RHS range        [7e+01, 2e+03]
Presolve removed 0 rows and 2 columns
Presolve time: 0.01s
Presolved: 4 rows, 10 columns, 37 nonzeros
 
Iteration    Objective       Primal Inf.    Dual Inf.      Time
       0    0.0000000e+00   1.472500e+02   0.000000e+00      0s
       4    1.1828861e+01   0.000000e+00   0.000000e+00      0s
 
Solved in 4 iterations and 0.03 seconds (0.00 work units)
Optimal objective  1.182886111e+01

 

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注意：

若出现众多warning，删去warning给的路径下的带“~”的文件

若出现ERROR: Could not find a version that satisfies the requirement time (from versions: none)，是找不到适应现有python版本的包