考虑到可能存在char列，这是基于@Abdou在获取列类型的excel表格自动回答:

makenumcols<-function(df){
  df<-as.data.frame(df)
  df[] <- lapply(df, as.character)
  cond <- apply(df, 2, function(x) {
    x <- x[!is.na(x)]
    all(suppressWarnings(!is.na(as.numeric(x))))
  })
  numeric_cols <- names(df)[cond]
  df[,numeric_cols] <- sapply(df[,numeric_cols], as.numeric)
  return(df)
}
df<-makenumcols(df)

使用type.convert()和rapply()的通用方式:

convert_types <- function(x) {
    stopifnot(is.list(x))
    x[] <- rapply(x, utils::type.convert, classes = "character",
                  how = "replace", as.is = TRUE)
    return(x)
}
d <- data.frame(char = letters[1:5], 
                fake_char = as.character(1:5), 
                fac = factor(1:5), 
                char_fac = factor(letters[1:5]), 
                num = 1:5, stringsAsFactors = FALSE)
sapply(d, class)
#>        char   fake_char         fac    char_fac         num 
#> "character" "character"    "factor"    "factor"   "integer"
sapply(convert_types(d), class)
#>        char   fake_char         fac    char_fac         num 
#> "character"   "integer"    "factor"    "factor"   "integer"

如果x是dataframe dat的列名，x的类型是factor，使用:

as.numeric(as.character(dat$x))

如果数据帧有多种类型的列，一些字符，一些数字尝试以下转换列包含数值为数值:

for (i in 1:length(data[1,])){
  if(length(as.numeric(data[,i][!is.na(data[,i])])[!is.na(as.numeric(data[,i][!is.na(data[,i])]))])==0){}
  else {
    data[,i]<-as.numeric(data[,i])
  }
}

蒂姆是对的，谢恩有个遗漏。以下是其他例子:

R> df <- data.frame(a = as.character(10:15))
R> df <- data.frame(df, num = as.numeric(df$a), 
                        numchr = as.numeric(as.character(df$a)))
R> df
   a num numchr
1 10   1     10
2 11   2     11
3 12   3     12
4 13   4     13
5 14   5     14
6 15   6     15
R> summary(df)
  a          num           numchr    
 10:1   Min.   :1.00   Min.   :10.0  
 11:1   1st Qu.:2.25   1st Qu.:11.2  
 12:1   Median :3.50   Median :12.5  
 13:1   Mean   :3.50   Mean   :12.5  
 14:1   3rd Qu.:4.75   3rd Qu.:13.8  
 15:1   Max.   :6.00   Max.   :15.0  
R> 

我们的data.frame现在有了因子列的摘要(counts)和as.numeric()的数值摘要(这是错误的，因为它得到了数值因子级别)以及as.numeric(as.character())的(正确的)摘要。

在我的PC (R v.3.2.3)中，apply或sapply给出错误。Lapply效果很好。

dt[,2:4] <- lapply(dt[,2:4], function (x) as.factor(as.numeric(x)))