R 子集数据。帧并集多个间隔

考虑以下

data.frame

d <- data.frame(x = seq(0, 10, length=100), value = rnorm(100))

现在，测试功能似乎是多余的，因为它看起来非常像内置的

findInterval

，但我找不到一种优雅的方法来使用它

condition <- Reduce(`|`, lapply(lapply(intervals,  findInterval,
                                       x=d$x, all.inside=FALSE), `==`, 1))

d[condition, ]

条件
编辑：使用
findInterval

d[findInterval(d$x,unlist(intervals))%%2==1,]

下面是一个使用
interval
包的解决方案

d <- data.frame(x = seq(0, 10, length=100), value = rnorm(100))
intervals <- list(c(0.2, 0.8), c(1, 2), c(8, 8.2))
library(intervals)
intervals <- Intervals( do.call( rbind, intervals ) )
intervals <- reduce( intervals )  # Simplify, if they overlap
condition <- distance_to_nearest(d$x, intervals) == 0
# The following would allow for non-closed intervals,
# but it is awfully slow.
condition <- sapply( d$x, function(u) 
  any(!empty(interval_intersection( Intervals(c(u,u)), intervals ))))
d[condition,]

@baptiste OK，添加了一个
findInterval
solutionnice，它还消除了
for
循环在
Reduce
d[findInterval(d$x,unlist(intervals))%%2==1,]

d <- data.frame(x = seq(0, 10, length=100), value = rnorm(100))
intervals <- list(c(0.2, 0.8), c(1, 2), c(8, 8.2))
library(intervals)
intervals <- Intervals( do.call( rbind, intervals ) )
intervals <- reduce( intervals )  # Simplify, if they overlap
condition <- distance_to_nearest(d$x, intervals) == 0
# The following would allow for non-closed intervals,
# but it is awfully slow.
condition <- sapply( d$x, function(u) 
  any(!empty(interval_intersection( Intervals(c(u,u)), intervals ))))
d[condition,]

intervals <- list(c(0.2, 0.8), c(1, 2), c(8, 8.2))
condition <- findInterval( d$x, unlist(intervals) ) %% 2 == 1
d[condition,]