问题
I am working on a LP model with pyomo,but when I create constraint, it shows a key error of 'can not find a certain combination'. I know list all the combinations can solve this problem. But the real data has many combinations. Is there any easy way to deal with this knid of problem? thanks!Here is a simple example:
This is a similar question for reference: Is there an easier way to avoid this kind of index key error of pyomo?
from pyomo.environ import *
import pandas as pd
data = [['tom','A', 10], ['nick','A', 15], ['juli','B',14]]
df = pd.DataFrame(data, columns = ['Name','Type', 'Age'])
#set
A = set(df['Name'])
B = set(df['Type'])
model = ConcreteModel()
#parameter
C= df.set_index(['Name','Type'])['Age'].to_dict()
#varibale
model.AB = Var(A,B,domain = NonNegativeReals)
#constraint1
def cons1(model,a,b):
return(model.AB[a,b]<=C[a,b])
model.Cons1 = Constraint(A,B,rule = cons1)
#constraint2
def cons2(model,a):
return(sum(model.AB[a,b]<=C[a,b] for b in B)<=1)
model.Cons2 = Constraint(A,rule = cons2)
回答1:
Starting with the solution given in your previous question and fixing what I think is a typo in your second constraint, the trick is to check if the (a,b) pair is in the set IJ
when iterating over B
in the sum:
from pyomo.environ import *
import pandas as pd
data = [['tom','A', 10], ['nick','A', 15], ['juli','B',14], ['juli','A',14]]
df = pd.DataFrame(data, columns = ['Name','Type', 'Age'])
#set
A = set(df['Name'])
B = set(df['Type'])
model = ConcreteModel()
#parameter
C= df.set_index(['Name','Type'])['Age'].to_dict()
#varibale
model.IJ = Set(initialize=C.keys())
model.AB = Var(model.IJ,domain = NonNegativeReals)
#constraint1
def cons1(model,a,b):
return(model.AB[a,b]<=C[a,b])
model.Cons1 = Constraint(model.IJ,rule = cons1)
def cons2(model,a):
return(sum(model.AB[a,b] for b in B if (a,b) in model.IJ)<=1)
model.Cons2 = Constraint(A,rule = cons2)
来源:https://stackoverflow.com/questions/56019266/what-is-the-best-way-to-deal-with-this-kind-of-key-error-in-pyomo