将因子的data.frame转换为数字（as.numerics）R的问题

Question

我知道还有其他问题可以解决这个问题，例如How to convert a factor to integer\numeric without loss of information?。但是这些问题提出的解决办法都不适合我。 所以我有这个data.frame，我应该使用as.numeric将其转换为数字，但是在运行循环时，却收到了此错误消息

In Ops.factor(x[i, 1], 15) : ‘<=’ not meaningful for factors

所以我再次尝试使用可以在互联网上找到的答案

as.numeric(as.character(tobacco))

和

as.numeric(paste(tobacco))

但是当我查看它们时，我得到的东西看起来像这样

这是我的data.frame：

structure(list(V1 = structure(c(12L, 13L, 4L, 17L, 4L, 4L, 12L, 
4L, 12L, 8L, 16L, 4L, 12L, 8L, 4L, 8L, 4L, 7L, 6L, 4L, 12L, 4L, 
4L, 1L, 16L, 4L, 4L, 4L, 17L, NA, 1L, 3L, 6L, 6L, 7L, 6L, 15L, 
17L, 6L, 16L, 3L, 15L, NA, 12L, 4L, 7L, 6L, 6L, 11L, 8L, 6L, 
7L, 6L, 6L, 6L, 4L, 12L, 6L, 6L, 12L, 16L, 8L, 4L, 8L, 3L, 10L, 
8L, 12L, 6L, 8L, 14L, 5L, 16L, 12L, 16L, 16L, 8L, 4L, 12L, 4L, 
14L, 15L, 4L, 1L, 12L, 17L, 16L, 12L, 17L, 8L, 17L, 12L, 6L, 
NA, 16L, 6L, 16L, 4L, 6L, 17L, 16L, 10L, 4L, 16L, 12L, 6L, 8L, 
8L, 8L, 16L, 16L, 4L, 4L, 16L, 12L, 12L, 8L, 8L, 8L, 4L, 15L, 
17L, 4L, 12L, 4L, 6L, 12L, 4L, 12L, 16L, 12L, 7L, 17L, 4L, 17L, 
NA, 16L, 4L, 14L, 15L, 16L, 16L, 4L, 16L, 4L, 12L, 16L, 12L, 
4L, 4L, 6L, 14L, 4L, 12L, 8L, 16L, 17L, 16L, 17L, 4L, NA, 4L, 
4L, 12L, 7L, 8L, 4L, 4L, 12L, 17L, 8L, 6L, 16L, 4L, 4L, 13L, 
4L, 4L, 4L, 12L, 3L, 12L, 4L, NA, 16L, 16L, 4L, 4L, 20L, 4L, 
11L, 16L, 15L, 4L, 12L, 17L, 6L, 19L, 14L, 6L, 8L, 12L, 21L, 
16L, 4L, 16L, 18L, 2L, 14L, 17L, 6L, 12L, 6L, 17L, 16L, 6L, 16L, 
12L, 8L, 7L, 8L, 6L, 12L, 4L, 17L, 16L, 1L, 4L, 4L, 4L, 12L, 
8L, 12L, 16L, 12L, 4L, 17L, 18L, 16L, 3L, 17L, 12L, 13L, NA, 
4L, 16L, 8L, 4L, 12L, 12L, 12L, 6L, 10L, 6L, 12L, 4L, 10L, 16L, 
4L, 4L, 12L, 3L, 4L, 17L, 9L, 13L, 17L, 11L, NA, 8L, 12L, 7L, 
12L, 15L), .Label = c("0", "0.5", "1", "10", "120", "15", "2", 
"20", "23", "25", "3", "30", "35", "40", "45", "5", "60", "7", 
"75", "8", "90", "max 15"), class = "factor")), .Names = "V1", row.names = c(NA, 
-274L), class = "data.frame")

有什么想法吗？谢谢 ！