【发布时间】：2020-06-01 06:26:24
【问题描述】：
我想最小化以下功能
cost=function(x){
events=sum(x==1)
rev = sum(df$rev[which(x==0)])*0.8
costcomp1=tail(comp1$cost[which(comp1$Nevents<events)],1)
costcomp2=tail(comp2$percentuale[which(comp2$k<=rev)],1)*rev+50000
cost=costcomp1+costcomp2
return(cost)
}
x 应该是一个二元向量，因为对于每件商品我都想选择要购买的供应商
head(comp2)
k percentuale
1 800000 0.0500
2 1800000 0.0325
3 88888888888888 0.0200
head(comp1)
Nevents cost
1 1500 13000
2 1750 17000
3 2000 21000
4 2500 22500
5 3000 26000
6 3500 29000
我试过optimize(cost,x),x=sample(c(0,1),nrow(df),replace = T)，但它不起作用。它给了我错误invalid function value in 'optimize'
【问题讨论】：
-
optim(cost, x)怎么样？这篇文章看起来和你想要的差不多：stackoverflow.com/questions/17013612/…. -
Error in optim(cost, x) : cannot coerce type 'closure' to vector of type 'double' Inoltre: Warning message: In optim(cost, x) : one-dimensional optimization by Nelder-Mead is unreliable: use "Brent" or optimize() directly
标签： r optimization minimize minimization