如果x是dataframe dat的列名，x的类型是factor，使用:

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

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)))

虽然其他人已经很好地讨论了这个话题，但我想补充一个额外的快速思考/提示。可以使用regexp提前检查字符是否可能仅由数字组成。

for(i in seq_along(names(df)){
     potential_numcol[i] <- all(!grepl("[a-zA-Z]",d[,i]))
}
# and now just convert only the numeric ones
d <- sapply(d[,potential_numcol],as.numeric)

想要了解更多复杂的正则表达式，以及为什么要学习/体验它们的力量，请访问这个非常好的网站:http://regexr.com/

我会加一条评论(不能低评级)

只需要添加user276042和pangratz

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

这将覆盖现有列x的值

虽然你的问题严格是关于数字的，但在开始r时，有许多转换是难以理解的。我将致力于解决帮助的方法。这个问题和这个问题类似。

在R中，类型转换可能是一种痛苦，因为(1)因子不能直接转换为数字，它们需要首先转换为字符类，(2)日期是一种特殊情况，通常需要单独处理，(3)跨数据帧列的循环可能很棘手。幸运的是，“潮流宇宙”已经解决了大部分问题。

This solution uses mutate_each() to apply a function to all columns in a data frame. In this case, we want to apply the type.convert() function, which converts strings to numeric where it can. Because R loves factors (not sure why) character columns that should stay character get changed to factor. To fix this, the mutate_if() function is used to detect columns that are factors and change to character. Last, I wanted to show how lubridate can be used to change a timestamp in character class to date-time because this is also often a sticking block for beginners.

library(tidyverse) 
library(lubridate)

# Recreate data that needs converted to numeric, date-time, etc
data_df
#> # A tibble: 5 × 9
#>             TIMESTAMP SYMBOL    EX  PRICE  SIZE  COND   BID BIDSIZ   OFR
#>                 <chr>  <chr> <chr>  <chr> <chr> <chr> <chr>  <chr> <chr>
#> 1 2012-05-04 09:30:00    BAC     T 7.8900 38538     F  7.89    523  7.90
#> 2 2012-05-04 09:30:01    BAC     Z 7.8850   288     @  7.88  61033  7.90
#> 3 2012-05-04 09:30:03    BAC     X 7.8900  1000     @  7.88   1974  7.89
#> 4 2012-05-04 09:30:07    BAC     T 7.8900 19052     F  7.88   1058  7.89
#> 5 2012-05-04 09:30:08    BAC     Y 7.8900 85053     F  7.88 108101  7.90

# Converting columns to numeric using "tidyverse"
data_df %>%
    mutate_all(type.convert) %>%
    mutate_if(is.factor, as.character) %>%
    mutate(TIMESTAMP = as_datetime(TIMESTAMP, tz = Sys.timezone()))
#> # A tibble: 5 × 9
#>             TIMESTAMP SYMBOL    EX PRICE  SIZE  COND   BID BIDSIZ   OFR
#>                <dttm>  <chr> <chr> <dbl> <int> <chr> <dbl>  <int> <dbl>
#> 1 2012-05-04 09:30:00    BAC     T 7.890 38538     F  7.89    523  7.90
#> 2 2012-05-04 09:30:01    BAC     Z 7.885   288     @  7.88  61033  7.90
#> 3 2012-05-04 09:30:03    BAC     X 7.890  1000     @  7.88   1974  7.89
#> 4 2012-05-04 09:30:07    BAC     T 7.890 19052     F  7.88   1058  7.89
#> 5 2012-05-04 09:30:08    BAC     Y 7.890 85053     F  7.88 108101  7.90

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)