挑战：重新编码 data.frame() — 使其更快

重新编码是调查数据的常见做法，但最明显的路线花费的时间比应有的时间要多。

使用提供的示例数据完成相同任务的最快代码system.time()在我的机器上获胜。

## Sample data
dat <- cbind(rep(1:5,50000),rep(5:1,50000),rep(c(1,2,4,5,3),50000))
dat <- cbind(dat,dat,dat,dat,dat,dat,dat,dat,dat,dat,dat,dat)
dat <- as.data.frame(dat)
re.codes <- c("This","That","And","The","Other")

要优化的代码。

for(x in 1:ncol(dat)) { 
    dat[,x] <- factor(dat[,x], labels=re.codes)
    }

Current system.time():

   user  system elapsed 
   4.40    0.10    4.49 

Hint: dat <- lapply(1:ncol(dat), function(x) dat[,x] <- factor(dat[,x],labels=rc)))并没有更快。

组合@DWin 的回答 https://stackoverflow.com/questions/6147119/challenge-recoding-a-data-frame-make-it-faster/6147950#6147950，以及我的回答最有效的列表到 data.frame 方法？ https://stackoverflow.com/q/5942760/271616:

system.time({
  dat3 <- list()
  # define attributes once outside of loop
  attrib <- list(class="factor", levels=re.codes)
  for (i in names(dat)) {              # loop over each column in 'dat'
    dat3[[i]] <- as.integer(dat[[i]])  # convert column to integer
    attributes(dat3[[i]]) <- attrib    # assign factor attributes
  }
  # convert 'dat3' into a data.frame. We can do it like this because:
  # 1) we know 'dat' and 'dat3' have the same number of rows and columns
  # 2) we want 'dat3' to have the same colnames as 'dat'
  # 3) we don't care if 'dat3' has different rownames than 'dat'
  attributes(dat3) <- list(row.names=c(NA_integer_,nrow(dat)),
    class="data.frame", names=names(dat))
})
identical(dat2, dat3)  # 'dat2' is from @Dwin's answer