查找二进制序列中的连续出现项

Question

我有一个二进制序列，看起来像这样：

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

如何找到连续正好有 2 个、连续 3 个等的次数？例如，我可以使用

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

Answer 1

我们可以创建一个具有运行长度编码的函数(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

---

或者另一种选择是从data.tablerleid

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 

如果我们想一次有多个'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 

Answer 2

其他基本 R 方法

table(scan(text=gsub("0+",";",paste0(x,collapse="")),sep=";",quiet = T))
1        11       111      1111     11111    111111 111111111 
114        63        34        19         7         3         1 

甚至：

table(strsplit(gsub("0+",";",paste0(x,collapse="")),";")[[1]])

Answer 3

您还可以使用rle来设置可以通过不同方式查询的数据帧，例如：

library(dplyr)
rle_x = rle(x)
results = data.frame(
x = x,
run_length = rep(rle_x$lengths, times = rle_x$lengths),
group = rep(1:length(rle_x$lengths), times = rle_x$lengths)
)
# Output:
#   x run_length group
# 1 0          2     1
# 2 0          2     1
# 3 1          2     2
# 4 1          2     2
# 5 0          1     3
# Find runs of 1 with length exactly == 2
results %>%
filter(x == 1, run_length == 2) %>%
summarize(groups = n_distinct(group))
# Output:
#   groups
# 1     63
# Runs of '1' of at least length 2:
results %>%
filter(x == 1, run_length >= 2) %>%
summarize(groups = n_distinct(group))
#      groups
# 1    127