考虑以下data.frame
,
d <- data.frame(x = seq(0, 10, length=100), value = rnorm(100))
我希望基于属于以下任意间隔的
x
进行子集化,intervals <- list(c(0.2, 0.8), c(1, 2), c(8, 8.2))
test <- function(range, x){
which(x >= range[1] & x <= range[2])
}
d[Reduce(`union`, lapply(intervals, test, x=d$x)), ]
现在,测试功能似乎是多余的,因为它看起来很像内置的
findInterval
,但是我找不到一种优雅的使用方式。condition <- Reduce(`|`, lapply(lapply(intervals, findInterval,
x=d$x, all.inside=FALSE), `==`, 1))
d[condition, ]
你能建议更好吗?
最佳答案
d[unlist(sapply(intervals, function(x) which(!is.na(cut(d$x,x))))),]
x value
3 0.2020202 0.15488314
4 0.3030303 -0.06891842
5 0.4040404 1.59909655
6 0.5050505 0.31006866
7 0.6060606 1.68986821
8 0.7070707 0.18500635
11 1.0101010 0.18721091
12 1.1111111 0.32485063
13 1.2121212 -0.42728405
14 1.3131313 0.84220081
15 1.4141414 -1.30745237
16 1.5151515 -1.90335389
17 1.6161616 -0.47139683
18 1.7171717 0.01622827
19 1.8181818 0.76362918
20 1.9191919 -0.37827765
81 8.0808081 0.46672521
82 8.1818182 1.27038641
编辑:使用
findInterval
的相同结果d[findInterval(d$x,unlist(intervals))%%2==1,]