获取一定长度的连续整数的运行并从第一个值中采样

Question

I am trying to create a function that will return the first integer of a subset of a vector such that the values of the subset are discrete, increasing by 1, and of a specified length.

For example, using the input data 'v' and a specified length 'l' of 3:

v <- c(3, 4, 5, 6, 15, 16, 25, 26, 27)
l <- 3

The possible sub-vectors of consecutive values of length 3 would be:

c(3, 4, 5)
c(4, 5, 6)
c(25, 26, 27)

Then I want to randomly choose one of these vectors and return the first/lowest number, ie 3, 4, or 25.

Answer 1

Here's an approach with base R :

First, we create all possible sub-vectors of length length . Next, we subset that list of vectors based on the cumsum of their difference equalling 1 . The is.na test ensures the last vectors which contain NA are also filtered out. Then we just bind the remaining vectors into a matrix and sample the first column.

SampleSequencialVectors <- function(vec, length){
  all.vecs <- lapply(seq_along(vec),function(x)vec[x:(x+(length-1))])
  seq.vec <- all.vecs[sapply(all.vecs,function(x) all(diff(x) == 1 & !is.na(diff(x))))]
  sample(do.call(rbind,seq.vec)[,1],1)
}

replicate(10, SampleSequencialVectors(v, 3))
# [1]  3  4  3  3  4  4 25 25  3 25

---

Or if you prefer a tidyverse type approach:

SampleSequencialVectorsPurrr <- function(vec, length){
  vec %>%
    seq_along %>%
    purrr::map(~vec[.x:(.x+(length-1))]) %>%
    purrr::keep(~ all(diff(.x) == 1 & !is.na(diff(.x)))) %>%
    purrr::invoke(rbind,.) %>%
    {sample(.[,1],size = 1)}
}
replicate(10, SampleSequencialVectorsPurrr(v, 3))
 [1]  4 25 25  3 25  4  4  3  4 25

Answer 2

1. Split the vector into runs of consecutive values*: split(v, cumsum(c(1L, diff(v) != 1)))
2. Select runs of length above or equal to the limit: runs[lengths(runs) >= lim]
3. From each run, select the possible first values ( x[1:(length(x) - lim + 1)] ).

4. From all possible first values, sample 1.

    runs = split(v, cumsum(c(1L, diff(v),= 1))) first = lapply(runs[lengths(runs) >= lim]: function(x) x[1,(length(x) - lim + 1)]) sample(unlist(first), 1)

---

Here we loop over runs of sufficient length, and not all individual values (see the other answers), thus it may be faster on larger vectors (haven't tested).

---

Slightly more compact using data.table :

 sample(data.table(v)[ , if(.N >= 3) v[1:(length(v) - lim + 1)],
                       by = .(cumsum(c(1L, diff(v) != 1)))]$V1, 1)

---

*Credits to the nice canonical: How to split a vector into groups of consecutive sequences? .

Answer 3

Base R two lines: Please note this solution assumes v is sorted.

consec_seq <- sapply(seq_along(v), function(i)split(v, abs(v - v[i]) > 1)[1])
consec_seq[lengths(consec_seq) == l][sample.int(l, 1)]

As a reusable function (not assuming sorted v):

conseq_split_sample <- function(vec, n){ 
  v <- sort(vec)
  consec_seq <- sapply(seq_along(v), function(i)split(v, abs(v - v[i]) > 1)["FALSE"])
  consec_seq[lengths(consec_seq) == n][sample.int(n, 1)]
}
conseq_split_sample(v, l)

Data:

 l <- 3
 v <- c(3, 4, 5, 6, 15, 16, 25, 26, 27)

Answer 4

Tooting my own horn -- cgwtools::seqle is like rle but you can specify the desired increment in a run. seqle(x, incr = 0,..) is the same as rle(x)

Then just grab the run lengths and starting values from the result.