r - 查找二进制序列中的连续出现

Question

我有一个看起来像这样的二进制序列:

set.seed(1)
n <- 1000
x <- sample(c(0,1), n, rep = TRUE)

如何找到恰好连续 2 个、连续 3 个等的次数？例如，我可以使用

找到连续 至少 2 个的次数

length(which((x[-1] == 1) & (diff(x) == 0)))

Answer 1

我们可以创建一个带有游程编码的函数 (rle)

with(rle(x), sum(values == 1 & lengths == 2))

即

fn_len <- function(vec, val, n) {
            with(rle(vec), sum(values == val & lengths == n))
 }

fn_len(x, 1, 2)
#[1] 63
fn_len(x, 1, 3)
#[1] 34

如果我们需要获取多个元素的长度

sapply(2:5, fn_len, vec = x, val = 1)
#[1] 63 34 19  7

---

或者另一个选项是 data.table 中的 rleid

library(data.table)
data.table(x)[, .N, .(x, rleid(x))][x==1, sum(N==2)]
#[1] 63

基准

set.seed(1)
n <- 1e7
x <- sample(c(0, 1), n, replace = TRUE)

system.time(out1 <- table(scan(text=gsub("0+",";",paste0(x,collapse="")),
       sep=";",quiet = T))[2])
#   user  system elapsed 
# 11.818   0.152  11.976 

system.time(out2 <- table(strsplit(gsub("0+",";",paste0(x,collapse="")),
              ";")[[1]])[3])
#   user  system elapsed 
#10.708   0.200  10.913 

system.time(fn_len(x, 1, 2))
#  user  system elapsed 
#  0.671   0.399   1.073 

如果我们想同时拥有多个'n'，data.table 方法会更快

system.time(data.table(x)[, .N, .(x, rleid(x))][x==1, .N, N])
#   user  system elapsed 
#   2.246   0.285   2.561 

system.time(sapply(2:21, fn_len, vec = x, val = 1))
#   user  system elapsed 
#  14.171   6.103  20.323 

system.time(table(strsplit(gsub("0+",";",paste0(x,collapse="")),";")[[1]]))
#   user  system elapsed 
# 10.570   0.192  10.770