R 如何将因子更改为数值变量或以其他方式处理此错误I'；我在做线性回归

尝试使用此数据集mro.csv运行线性回归模型，但运行lm（）时会显示错误消息：

1: In model.response(mf, "numeric") :
  using type = "numeric" with a factor response will be ignored
2: In Ops.factor(y, z$residuals) : ‘-’ not meaningful for factors

不确定数据集的哪些部分是因子而不是数字，除列名外，所有数据都是数字。也不确定对于因子部分没有意义的“-”，因为数据集中也没有“-”

不确定如何共享数据集，但以下是谷歌表单中的csv：

原始尺寸（原始） [1] 753 22 > >种子（88） >生的 > >raw1尺寸（原始） [1] 753 22 >dim（raw1） [1] 428 22 > >

---

>reg1 reg1

工资

和

lwage

被解读为

系数

s，因为它们包含的值

“

无法解析为数字。此值可以手动处理

df <- read.csv(
  "~/Downloads/mro.csv",
  header = FALSE,
  stringsAsFactors = FALSE,
  col.names = c(
    "inlf", "hours", "kidslt6", "kidsge6", "age", "educ",  "wage",
    "repwage", "hushrs", "husage", "huseduc", "huswage",  "faminc",
    "mtr",  "motheduc",  "fatheduc", "unem", "city", "exper",
    "nwifeinc", "lwage", "expersq"
  )
)

df$wage <- as.numeric(ifelse(df$wage == ".", 0, df$wage))
df$lwage <- as.numeric(ifelse(df$lwage == ".", 0, df$lwage))

---

这条消息说，响应“工资”是一个因素。你能分享你的数据吗（

dput（raw）

）？其他一些事情：对于参数，你应该使用

=

，而不是

，我不知道如何分享数据@Paul，但我刚刚添加了一个到数据集的链接
df <- read.csv(
  "~/Downloads/mro.csv",
  header = FALSE,
  stringsAsFactors = FALSE,
  col.names = c(
    "inlf", "hours", "kidslt6", "kidsge6", "age", "educ",  "wage",
    "repwage", "hushrs", "husage", "huseduc", "huswage",  "faminc",
    "mtr",  "motheduc",  "fatheduc", "unem", "city", "exper",
    "nwifeinc", "lwage", "expersq"
  )
)

df$wage <- as.numeric(ifelse(df$wage == ".", 0, df$wage))
df$lwage <- as.numeric(ifelse(df$lwage == ".", 0, df$lwage))

df <- df[sample(nrow(df)), ]
df1 <- df[df$inlf == 1, ]

reg1 <- lm(
  wage ~ hours + kidslt6 + kidsge6 + age + educ + hushrs + husage + huseduc +
         huswage + mtr + motheduc + fatheduc + unem + exper + nwifeinc,
  data = df1
)