Question

我有三个模型概率预测mod.p1, mod.p2, mod.p3

require(Runuran)
mod.p1<- urnorm(n = 1000, mean = 0.2, sd = 0.35, lb = 0, ub = 1)
mod.p2<- urnorm(n = 1000, mean = 0.23, sd = 0.37, lb = 0, ub = 1)
mod.p3 = urnorm(n = 1000, mean = 0.19, sd = 0.39, lb = 0, ub = 1)

最终结果以判决（是，否）

给出

Verdict <- sample( c("yes","No"), 1000, replace=TRUE, prob=c(0.2,0.8) )

每个模型我有三个权重

i1=0.3
i2=0.3
i3=0.4

创建集合预测并计算auc

  Ensemble=i1*mod.p1+i2*mod.p2+i3*mod.p3
  require(ROCR)    
  Ensemble.pred = prediction(Ensemble, Verdict)
  Ensemble.auc = as.numeric(performance(Ensemble.pred, "auc")@y.values)
  Ensemble.auc # 0.52

现在我如何编写for循环来尝试i1 , i2 and i3的不同值，并返回i1 , i2 and i3的最佳值，这样可以获得最大AUC。

Answer 1

修改此版本以应用约束并以0.1为增量进行探索。这使用了一种稍微强制的方式来限制参数，但使其更加通用

require(Runuran)
require(ROCR)    

mod.p1<- urnorm(n = 1000, mean = 0.2, sd = 0.35, lb = 0, ub = 1)
mod.p2<- urnorm(n = 1000, mean = 0.23, sd = 0.37, lb = 0, ub = 1)
mod.p3 = urnorm(n = 1000, mean = 0.19, sd = 0.39, lb = 0, ub = 1)

## creating ensemble predictions and calculating auc

# make the examples reproducible
set.seed = 123  

# expand.grid, in base R, returns a dataframe
grid <- expand.grid(
  i1 = seq(0.1, 1, by=0.1),
  i2 = seq(0.1, 1, by=0.1),
  i3 = seq(0.1, 1, by=0.1))

# apply contraints to the grid
grid <- grid[(grid$i1 + grid$i2 + grid$i3) == 1,] 

Ensembles <- list()
grid$auc <- NA

for (i in 1:nrow(grid)) {
  # final outcomes is given in verdict ( Yes , No )
  Verdict <- sample( c("yes","No"), 1000, replace=TRUE, prob=c(0.2,0.8) )
  this_set <- grid[i,]
  Ensemble=(this_set[["i1"]]*mod.p1) + (this_set[["i2"]]*mod.p2) + (this_set[["i3"]]*mod.p3)
  Ensemble.pred = prediction(Ensemble, Verdict)
  Ensemble.auc = as.numeric(performance(Ensemble.pred, "auc")@y.values)
  grid$auc[i] <- Ensemble.auc
  Ensembles[i] <- list(Ensemble)
}

# print the best value (note there is some other sourc
# of randomness so
# result svary form run to run)
grid[grid$auc == max(grid$auc),]

#    i1  i2  i3      auc
# 8 0.8 0.1 0.1 0.554802

Answer 2

知道了，这就是我所看到的 -

# creation of Dummy data
mod.p1<- urnorm(n = 1000, mean = 0.2, sd = 0.35, lb = 0, ub = 1)
mod.p2<- urnorm(n = 1000, mean = 0.23, sd = 0.37, lb = 0, ub = 1)
mod.p3 = urnorm(n = 1000, mean = 0.19, sd = 0.39, lb = 0, ub = 1)
Verdict <- sample( c("yes","No"), 1000, replace=TRUE, prob=c(0.2,0.8) )

#loop for optimizing the weights in ensemble
auc = 0
i1 = 0
i2 = 0
i3 = 0

for(i in seq(0,1,0.05))  {
  for (j in seq(0,1-i,0.05))  {
    k = 1-i-j
    e = i1*mod.p1 + i2*mod.p2 + i3*mod.p3
    Ensemble.pred = prediction(e, Verdict)
    Ensemble.auc = as.numeric(performance(Ensemble.pred, "auc")@y.values)
    if (Ensemble.auc>auc)  {
      auc = Ensemble.auc
      i1 = i
      i2 = j
      i3 = k
      pred = e
    }
  }
}

# get final values
auc #0.524
i1  #0
i2  #0.1
i3  #0.9

Answer 3

我认为你自己提出的解决方案实际上非常好。我唯一不理解的是你不能替换i,j,k行中的e = i1*mod.p1 + i2*mod.p2 + i3*mod.p3值。我会像下面一样调整，这很好。

for(i in seq(0,1,0.03))  {
    for (j in seq(0,1-i,0.03))  {
        k = 1-i-j
        e = i*mod.p1 + j*mod.p2 + k*mod.p3
        Ensemble.pred = prediction(e, Verdict)
        Ensemble.auc = as.numeric(performance(Ensemble.pred, "auc")@y.values)
        if (Ensemble.auc>auc)  {
            auc = Ensemble.auc
            i1 = i
            i2 = j
            i3 = k
            pred = e
        }
    }
}

# get final values
auc #0.543
i1  #0.84
i2  #0.15
i3  #0.01

使用for循环优化集合模型权重

3 个答案: