你为什么不试试sitools库？

library(data.table)
dt<-data.table(var = sample(x=1:1e5, size=1e6, replace=T))
library(sitools)
> system.time(dt[, var2 := f2si(var)])
   user  system elapsed 
  10.08    0.09   10.89

**EDIT：**这是一个基于数据表的函数，可以从sitools包中反转f2si：

si2f<-function(x){
  if(is.numeric(x)) return(x)
  require(data.table)
  dt<-data.table(lab=c("y","z","a","f","p","n","µ","m","c","d","", "da", "h", "k","M","G","T","P","E","Z","Y"),
                 mul=c(1e-24, 1e-21, 1e-18, 1e-15, 1e-12, 1e-9, 1e-6, 1e-3, 1e-2, 1e-1, 1L, 10L, 1e2, 1e3, 1e6, 1e9, 1e12, 1e15, 1e18, 1e21, 1e24),
                 key="lab")
  res<-as.numeric(gsub("[^0-9|\\.]","", x))
  x<-gsub("[0-9]|\\s+|\\.","", x)
  .subset2(dt[.(x)], "mul")*res
}

> system.time(dt[, var3 := si2f(var2)])
   user  system elapsed 
  13.18    0.03   13.31 

> dt[, all.equal(var,var3)]
[1] TRUE

这里有一个方法，在我的计算机上大约需要10秒来转换一个具有10M值的向量。您可以将其扩展到涵盖更多的“K”，“M”和“G”

> f_conv <- function(val){
+     # create matrix indexed by name for exponent
+     key <- c(Zero = ""
+          , K = "E3"
+          , M = "E6"
+          , G = "E9"
+          )
+     # extract where the original exponent is 
+     indx <- regexpr("[KMG]", val)
+     # extract the exponent
+     exp <- substring(val, indx)
+     # if there was none, the use "Zero"
+     exp[indx == -1L] <- "Zero"
+     # put fake length
+     indx[indx == -1L] <- 20L
+     # do the conversion
+     as.numeric(paste0(substring(val, 1L, indx - 1L)
+                  , key[exp]
+                  )
+              )
+ }
> 
> # test data
> n <- 10000000
> result <- paste0(sample(1:999, n, TRUE)
+             , sample(c("K", "M", "G", ""), n, TRUE)
+             )
> 
> system.time(x <- f_conv(result))
   user  system elapsed 
   8.48    0.13    8.63 
> cbind(result[1:50], x[1:50])
      [,1]   [,2]          
 [1,] "562K" "562000"      
 [2,] "946"  "946"         
 [3,] "313G" "313000000000"
 [4,] "538M" "538000000"   
 [5,] "697K" "697000"      
 [6,] "486G" "486000000000"
 [7,] "814G" "814000000000"
 [8,] "842"  "842"         
 [9,] "993M" "993000000"   
[10,] "440K" "440000"      
[11,] "435G" "435000000000"
[12,] "407M" "407000000"   
[13,] "919K" "919000"      
[14,] "840"  "840"         
[15,] "766G" "766000000000"
[16,] "977"  "977"         
[17,] "139"  "139"         
[18,] "195G" "195000000000"
[19,] "609M" "609000000"   
[20,] "69"   "69"          
[21,] "147M" "147000000"   
[22,] "104M" "104000000"   
[23,] "509K" "509000"      
[24,] "951M" "951000000"   
[25,] "278"  "278"         
[26,] "797G" "797000000000"
[27,] "106K" "106000"      
[28,] "667K" "667000"      
[29,] "521K" "521000"      
[30,] "9"    "9"           
[31,] "17K"  "17000"       
[32,] "673M" "673000000"

所有其他答案对我来说都不适合NA（或者产生警告，这也不好）。
这是我的解决方案，它重用了其他解决方案中的一些位。

library(stringr)

si2num <- function(x)
{
  conv <- paste0("e", c(seq(-24 ,-3, by=3), -2, -1, seq(3, 24, by=3),3))
  names(conv) <- c("y","z","a","f","p","n","µ","m","c","d","K","M","G","T","P","E","Z","Y","k")
  
  xout <- str_replace_all(x, conv)
  xout <- as.numeric(xout)
  
  return(xout)
}

x <- c(NA,"10", "10.01K",NA,"10.1M", "20K", "21k",NA)

si2num(x)

[1]       NA       10    10010       NA 10100000    20000    21000       NA