如何将因子更改为数字变量或以其他方式处理我在线性回归中遇到的这个错误

Question

Trying to run a linear regression model with this dataset, mro.csv, but when I run lm() it gives the error message:

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

Not sure what parts of the dataset are factors and not numeric, all the data is numbers except column names.Also unsure what the '-' not meaningful for factors part is about because there are no -'s in the dataset either.

Not sure how to share the dataset, but here's the csv in a google sheet: mro.csv

> raw <- read.csv("/Users/cpt.jack/Downloads/mro.csv",header<-F,sep<-",") 
> colnames(raw)<- c("inlf","hours","kidslt6","kidsge6","age", "educ",  "wage", "repwage",             "hushrs", "husage", "huseduc","huswage",  "faminc",  "mtr",  "motheduc",  "fatheduc",    "unem","city", "exper",  "nwifeinc",  "lwage",  "expersq")  
> 
> 
> dim(raw)
[1] 753  22
> 
> set.seed(88)
> raw  <- raw[sample(nrow(raw)),]
> 
> 
> raw1<-raw[raw$inlf==1,]
> dim(raw)
[1] 753  22
> dim(raw1)
[1] 428  22
> 
> 
> reg1 <- lm(wage~ hours + kidslt6 + kidsge6 + age + educ + hushrs + husage + huseduc + huswage
+mtr+motheduc+fatheduc+unem
+exper+nwifeinc, data=raw1)

Warning messages:

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
> reg1 <- lm(wage~ hours,data=raw1)

Answer 1

wage and lwage are being read as factor s because they contain the value "." which can't be parsed as numeric. This value can be handled manually.

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))

Now the lm should run without issues.

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
)