插入符号预测目标变量nrow为空

问题描述

df:

library(caret)

a = c("aa","bb","cc","aa","bb") 
b = c("aa","bb") 
c = c("aa","bb") 
d = c("aa","bb") 
e = c(1,1,1)

#df1
df1 = data.frame(a,b,c,d,e)
#df2
df2 = data.frame(a,e)

Caret Log-red模型:

df1$e <- as.factor(df1$e)
df2$e <- as.factor(df2$e)

# define training control
train_control <- trainControl(method = "cv",number = 5)

# train the model on training set
model <- train(e ~ .,data = df1,trControl = train_control,method = "glm",family=binomial())

# logistic <- glm(WonLost ~ . -PANum,data=train,family="binomial")
df2$predict <- caret::predict.train(model,newdata=df2,type = "prob")


nrow(df2$predict)
nrow(df2$e)

为什么nrow(df2 $ e)为零?我根据先前遇到的以下错误将目标变量更改为一个因数,但这似乎是造成当前问题的原因。

警告消息:1:在train.default(x,y,weights = w,...)中:您 正在尝试进行回归,您的结果只有两种可能 值您是否要进行分类?如果是这样,请使用2级 因素作为结果列。

解决方法

有时caret对变量很敏感,即使您的glm logit模型存在回归或分类方面的麻烦也有其影响。我学到的一个建议是将目标变量重新编码为是/否。另外,请注意,将插入符号的预测作为新数据帧添加到df2中,这就是nrow()起作用而e只是一个向量的原因,因此您必须使用length()NROW()。这里的代码:

library(caret)
#Vectors
a = c("aa","bb","cc","aa","bb") 
b = c("aa","bb") 
c = c("aa","bb") 
d = c("aa","bb") 
e = c(1,1,1)

#df1
df1 = data.frame(a,b,c,d,e)
#df2
df2 = data.frame(a,e)
#Format
df1$e[df1$e==1] <- 'Yes'
df1$e[df1$e==0] <- 'No'
df2$e[df2$e==1] <- 'Yes'
df2$e[df2$e==0] <- 'No'

# define training control
train_control <- trainControl(method = "cv",number = 5)

# train the model on training set
model <- train(e ~ .,data = df1,trControl = train_control,method = "glm",family=binomial())

#Predict
df2$predict <- caret::predict.train(model,newdata=df2,type = "prob")
#Checks
nrow(df2$predict)
NROW(df2$e)
length(df2$e)

输出:

df2
    a  b  c  d   e   predict.No predict.Yes
1  aa aa aa aa Yes 7.500000e-01        0.25
2  bb bb bb bb  No 2.500000e-01        0.75
3  cc cc cc cc Yes 8.646869e-09        1.00
4  aa aa aa aa  No 7.500000e-01        0.25
5  aa aa aa aa  No 7.500000e-01        0.25
6  aa aa aa aa  No 7.500000e-01        0.25
7  bb bb bb bb Yes 2.500000e-01        0.75
8  cc cc cc cc Yes 8.646869e-09        1.00
9  bb bb bb bb Yes 2.500000e-01        0.75
10 bb bb bb bb Yes 2.500000e-01        0.75

nrow(df2$predict)
[1] 10
NROW(df2$e)
[1] 10
length(df2$e)
[1] 10