Question

无法为以下逻辑回归

修复以下错误

training=(IBM$Serial<625)
data=IBM[!training,]
dim(data)
stock.direction <- data$Direction
training_model=glm(stock.direction~data$lag2,data=data,family=binomial)
###Error### ----  Error in eval(family$initialize) : y values must be 0 <= y <= 1

我正在使用的数据中的几行

X   Date    Open    High    Low Close   Adj.Close   Volume  Return  lag1    lag2    lag3    Direction   Serial
1   28-11-2012  190.979996  192.039993  189.270004  191.979996  165.107727  3603600 0.004010855 0.004010855 -0.001198021    -0.006354834    Up  1
2   29-11-2012  192.75  192.899994  190.199997  191.529999  164.720734  4077900 0.00114865  0.00114865  -0.004020279    -0.009502386    Up  2
3   30-11-2012  191.75  192 189.5   190.070007  163.465073  4936400 0.003630178 0.003630178 -0.001894039    -0.005576956    Up  3
4   03-12-2012  190.759995  191.300003  188.360001  189.479996  162.957703  3349600 0.001213907 0.001213907 -0.002480478    -0.001636046    Up  4

Answer 1

要求y值介于0和1之间的原因是因为数据中的分类特征（如“方向”）属于“字符”类型。您需要将它们转换为使用as.factor(data$Direction)键入'factor'。所以：glm(Direction ~ lag2, data=...)不需要声明stock.direction。

您可以使用命令class(variable)检查变量类，如果它们是字符，则可以转换为factor并在同一数据框中创建新列。它应该工作。

Answer 2

我遇到了同样的错误“ eval（family $ initialize）中的错误：y值必须为0 <= y <= 1”，并通过在red.csv函数中添加“ stringsAsFactors = T”来解决。

之前：gene.train = read.csv（“ gene.train.csv”，标头= T）＃错误

之后：gene.train = read.csv（“ gene.train.csv”，标头= T，stringsAsFactors = T）＃没有错误。

Answer 3

如果不了解数据，你应该这样做

library(dplyr)
df <- read.table(header = T, stringsAsFactors = F,  text ="X   Date    Open    High    Low Close   Adj.Close   Volume  Return  lag1    lag2    lag3    Direction   Serial
1   28-11-2012  190.979996  192.039993  189.270004  191.979996  165.107727  3603600 0.004010855 0.004010855 -0.001198021    -0.006354834    Up  1
2   29-11-2012  192.75  192.899994  190.199997  191.529999  164.720734  4077900 0.00114865  0.00114865  -0.004020279    -0.009502386    Up  2
3   30-11-2012  191.75  192 189.5   190.070007  163.465073  4936400 0.003630178 0.003630178 -0.001894039    -0.005576956    Up  3
4   03-12-2012  190.759995  191.300003  188.360001  189.479996  162.957703  3349600 0.001213907 0.001213907 -0.002480478    -0.001636046    Up  4
1   28-11-2012  190.979996  192.039993  189.270004  191.979996  165.107727  3603600 0.004010855 0.004010855 -0.001198021    -0.006354834    Up  1
2   29-11-2012  192.75  192.899994  190.199997  191.529999  164.720734  4077900 0.00114865  0.00114865  -0.004020279    -0.009502386    Down  2
3   30-11-2012  191.75  192 189.5   190.070007  163.465073  4936400 0.003630178 0.003630178 -0.001894039    -0.005576956    Up  3
4   03-12-2012  190.759995  191.300003  188.360001  189.479996  162.957703  3349600 0.001213907 0.001213907 -0.002480478    -0.001636046    Down  4
") %>%
  mutate(bin = ifelse(Direction == "Up", 1, 0))

glm(bin ~ High, family = "binomial", data = df)

因子的Logistic回归：eval中的错误（family $ initialize）：y值必须为0＆lt; = y＆lt; = 1

3 个答案: