使用speedier hist()或findInterval()获得与cut()相同的输出？

Question

我读了这篇文章http://www.r-bloggers.com/comparing-hist-and-cut-r-functions/,测试hist()速度比cut()我的PC 快了~4倍.我的脚本循环遍历cut()很多次,因此省时很重要.因此我试图切换到更快的功能,但是很难获得准确的输出cut().

从以下示例代码:

data <- rnorm(10, mean=0, sd=1)  #generate data
my_breaks <- seq(-6, 6, by=1)  #create a vector that specifies my break points
cut(data, breaks=my_breaks)

我希望得到一个包含级别的向量,使用我的断点将每个数据元素分配给它,即:的确切输出cut:

 [1] (1,2]   (-1,0]  (0,1]   (1,2]   (0,1]   (-1,0]  (-1,0]  (0,1]   (-2,-1] (0,1]  
Levels: (-6,-5] (-5,-4] (-4,-3] (-3,-2] (-2,-1] (-1,0] (0,1] (1,2] (2,3] (3,4] (4,5] (5,6]
> 

我的问题:我如何使用hist()输出元素(即中断,计数,密度,中等)或findInterval达到我的目标？

另外,我发现从一个例子/sf/ask/866538991/使用findInterval,但是这需要我事先创建的间隔的标签,这是不是我想要的.

任何帮助,将不胜感激.提前致谢!

Answer 1

这是一个基于你的findInterval建议的实现,比经典的快5-6倍cut:

cut2 <- function(x, breaks) {
  labels <- paste0("(",  breaks[-length(breaks)], ",", breaks[-1L], "]")
  return(factor(labels[findInterval(x, breaks)], levels=labels))
}

library(microbenchmark)

set.seed(1)
data <- rnorm(1e4, mean=0, sd=1)

microbenchmark(cut.default(data, my_breaks), cut2(data, my_breaks))

# Unit: microseconds
#                         expr      min        lq    median        uq      max neval
# cut.default(data, my_breaks) 3011.932 3031.1705 3046.5245 3075.3085 4119.147   100
#        cut2(data, my_breaks)  453.761  459.8045  464.0755  469.4605 1462.020   100

identical(cut(data, my_breaks), cut2(data, my_breaks))
# TRUE