Question

我想用多个模型计算出我的数据具有不同概率的所有可能的预测。结果是一个列表。

public function getId(){
    return $this->id;
}

public function getTodo(){
    return $this->todo;
}

然后我创建一个物流模型：

df<-iris
df$y<-sample(0:1,nrow(df),replace=TRUE)
set.seed(101)
#Now Selecting 80% of data as sample from total 'n' rows of the data  
sample <- sample.int(n = nrow(df), size = floor(.8*nrow(df)), replace = F)
train <- df[sample, ]
test  <- df[-sample, ]

首先，我有带有预测的表格。然后我创建了一个具有所有可能概率的向量。

 full <- glm(y~., data = train, family = "binomial")
 min <- glm( y~ 1, data = train, family = "binomial")

 backward <- step(full,direction = "backward",trace=0)
 forward <- step(min,scope=list(lower=min, upper=full),direction = "forward",trace=0)
model2<- glm(y~Sepal.Length+Sepal.Width , data = train, family = "binomial")

  models<-list(backward,forward,model2)
prediction<- lapply(models, function(x){predict(x,newdata=test,type="response")})

问题是我想应用差异中断点。我尝试了purrr包的map2函数，但不起作用。

    p <- seq(from = 0.1, to = 0.9, by = 0.5)

问题是：错误：pred = map2(prediction,p, function(x,pi){ifelse(x > pi, 1, 0)})（3）和.x（2）的长度不同

有人可以帮忙吗？

我认为最好将更改应用于sapply，然后我将有一个data.frame。

.y

那我可以使用pmap函数吗？谢谢

编辑：我更新了所有代码。

Answer 1

看看这是否有意义：

df<-iris
df$y<-sample(0:1,nrow(df),replace=TRUE)
set.seed(101)
#Now Selecting 80% of data as sample from total 'n' rows of the data  
sample <- sample.int(n = nrow(df), size = floor(.8*nrow(df)), replace = F)
train <- df[sample, ]
test  <- df[-sample, ]

full <- glm(y~., data = train, family = "binomial")
min <- glm( y~ 1, data = train, family = "binomial")

backward <- step(full,direction = "backward",trace=0)
forward <- step(min,scope=list(lower=min, upper=full),direction = "forward",trace=0)
model2<- glm(y~Sepal.Length+Sepal.Width , data = train, family = "binomial")

models<-list(backward,forward,model2)
prediction<- lapply(models, function(x){predict(x,newdata=test,type="response")})

p <- seq(from = 0.1, to = 0.9, by = 0.5) 

combn = cross2(prediction, p)

pred <- map(combn, 
            function(combination) {
              x <- combination[[1]]
              pi <- combination[[2]]
              ifelse(x > pi, 1, 0)
            }
)

使用map2迭代两个参数（purrr函数）

1 个答案: