今天在使用R语言对Weekly进行交叉验证时，发生如下报错：

错误于eval(family$initialize): y值必需满足0 <= y <= 1

错误代码为：

Weekly<-read.csv("Weekly.csv")
set.seed(1)
attach(Weekly)
glm.fit1 = glm(Direction~Lag1+Lag2, data=Weekly, family=binomial)
summary(glm.fit1)

现在让我们来解决这个问题。

首先打开数据：

fix(Weekly)

我们发现Weekly数据集中的变量Direction为分类变量，为此需要对他进行0-1变换。

 Direction 0-1编码，0表示Down，1表示Up 

修改一：

Direction.1<-as.factor(Direction)
Weekly<-data.frame(Weekly,Direction.1)
Direction<-Direction.1

Weekly<-resd.csv('Weekly.csv')
attach(Weekly)
Direction.1<-as.factor(Direction)
Weekly<-data.frame(Weekly,Direction.1)
Direction<-Direction.1
fix(Weekly)
glm.fit<-glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+colume,data=Weekly,family=binomial)
summary(glm.fit)

 上面修改方式过于繁琐，为此下面对修改结果进行简化：

修改二：

Weekly$Direction=as.factor(Direction)

Weekly<-read.csv("Weekly.csv")
set.seed(1)
attach(Weekly)
Weekly$Direction=as.factor(Direction)
summary(Weekly)

glm.fit1 = glm(Direction~Lag1+Lag2, data=Weekly, family=binomial)
summary(glm.fit1)

Call:
glm(formula = Direction ~ Lag1 + Lag2, family = binomial, data = Weekly)

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)  0.22122    0.06147   3.599 0.000319 ***
Lag1        -0.03872    0.02622  -1.477 0.139672    
Lag2         0.06025    0.02655   2.270 0.023232 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 1496.2  on 1088  degrees of freedom
Residual deviance: 1488.2  on 1086  degrees of freedom
AIC: 1494.2

Number of Fisher Scoring iterations: 4

代码成功运行，问题解决。