如何根据条件在R中将向量拆分为不相等的块？

Question

Let's say I have these two vectors:

x <- c(1,2,4,6,7)
y <- c(3,7)

How can I split x into the elements that are smaller than each element of y ? For example: c(1,2) | c(4,6,7) c(1,2) | c(4,6,7) . .

I guess one option would be to do a double loop and return the smallest element in y that is smaller than the current one in x : c(3,3,3,7,7) . I could then split using this vector.

j <- 1
sapply(x, function(i){
  if (i <= y[j]) {
    y[j]
  } else {
    if (j < length(y)){
      j <- j + 1
    }
    y[j]
  }
})

I feel like there's a cleverer way to do this, but I can't figure it out.

Answer 1

Here is how I would do it:

x <- c(1,2,4,6,7)
y <- c(3,7)
out <- list(x[x < min(y)], x[!x < min(y)])

Here is the result:

> out
[[1]]
[1] 1 2

[[2]]
[1] 4 6 7

Answer 2

Here is a base R method using split and findInterval :

split(x, findInterval(x, y, rightmost.closed=TRUE))

$`0`
[1] 1 2

$`1`
[1] 4 6 7

The findInterval function returns a vector that categorizes the variable values in x along your criteria in y. The split function separates the vector as desired and returns a named list.

Answer 3

Using cut and split in base R:

lapply(y, function(a) split(x, cut(x, c(-Inf, a, Inf))))

# [[1]]
# [[1]]$`(-Inf,3]`
# [1] 1 2

# [[1]]$`(3, Inf]`
# [1] 4 6 7


# [[2]]
# [[2]]$`(-Inf,7]`
# [1] 1 2 4 6 7

# [[2]]$`(7, Inf]`
# numeric(0)

Answer 4

maybe not the best solution but it is quicker:

z <- x < min(y)

end <- x[z]