我读过这篇文章http://www.r-bloggers.com/comparing-hist-and-cut-r-functions/并且在我的电脑上测试hist()
比cut()
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,我尝试切换到更快的功能,但很难按照cut()
获得准确的输出。
来自以下示例代码:
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cut
的确切输出:
[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:如何使用hist()
输出的元素(即中断,计数,密度,中等等)或findInterval
来达到我的目标? < / p>
另外,我使用findInterval
从https://stackoverflow.com/questions/12379128/r-switch-statement-on-comparisons找到了一个示例,但这需要我事先创建间隔标签,这不是我想要的。
任何帮助将不胜感激。提前谢谢!
答案 0 :(得分:6)
以下是基于findInterval
建议的实施,比经典cut
快5-6倍:
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
答案 1 :(得分:4)
hist
函数按照与table
和cut
组合类似的方式创建分类计数。例如,
set.seed(1)
x <- rnorm(100)
hist(x, plot = FALSE)
## $breaks
## [1] -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5
##
## $counts
## [1] 1 3 7 14 21 20 19 9 4 2
table(cut(x, seq.int(-2.5, 2.5, 0.5)))
## (-2.5,-2] (-2,-1.5] (-1.5,-1] (-1,-0.5] (-0.5,0] (0,0.5] (0.5,1]
## 1 3 7 14 21 20 19
## (1,1.5] (1.5,2] (2,2.5]
## 9 4 2
如果您想要cut
的原始输出,则无法使用hist
。
但是,如果cut
的速度有问题(并且您可能想要仔细检查它确实是分析的缓慢部分;请参阅premature optimization is the root of all evil),那么您可以使用较低的1级}}。这忽略了.bincode
。
cut