I read this article http://www.r-bloggers.com/comparing-hist-and-cut-r-functions/ and tested hist() to be faster than cut() by ~4 times on my PC. My script loops through cut() many times so the time-saving would be significant. I thus tried to switch to the speedier function but am having difficulties getting the exact output as per cut().
From the sample code below:
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)
I wish to get a vector comprising levels that each element of data is assigned to using my breakpoints, i.e. the exact output of 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]
>
My question: How do I use elements of the hist() output (i.e. breaks, counts, density, mids, etc) or findInterval to reach my objective?
Separately, I found an example from https://stackoverflow.com/questions/12379128/r-switch-statement-on-comparisons using findInterval, but this requires me to create the interval labels beforehand, which is not what I want.
Any help would be appreciated. Thanks in advance!
Here is an implementation based on your findInterval suggestion which is 5-6 times faster than classical 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