获取连续数字而不重复

Question

从一个随机数字序列 {"1", "2", "3", "4", "5"} 中，要求确定 3 个连续数字组出现的次数，这意味着，使用 base = c ("1", "2", "3", "4", "5") 生成以下任何组的次数 {"123", "234" , "345" }.

# I undertand that I have generate a sample with 5 numbers
a<-c(sample(1:5,5))
a
#I generated the list, as you can see I didn't fix a seed because I know that in every single sequence I will have differents grupos of 3 consecutive numbers, so I should obtain something like this

b<-c(2,3,4,5,1) #this example gives me just one that it would be {2,3,4}
b
#answer expected
1
#Then, I don't know how to obtain the sequence I have tried with permutations and combinations but I don't get it.

Answer 1

这将统计出现的三项连续增加的次数（例如 123）。

countsequence <- function(x){
  if (length(x)<3){
    return(0)
  }
  count <- 0
  for(i in 3:length(x)){
    if(x[i-1]==x[i]-1 && x[i-2] == x[i]-2){
      count <- count + 1
    }
  }
  return(count)
} 

countsequence(1:5)
countsequence(c(2, 3, 4, 1, 5))

Answer 2

参数k是子序列长度（在你的例子中是3）：

f <- function(b, k = 3){

  if(k > length(b)) return(0)
  
  check_list <- lapply(k:length(b), function(x) all(diff(b[(x-k+1):x]) == 1)) 
  sum(unlist(check_list))
}

如果你想计算增加或减少的子序列，用适当的关系和数字替换 == 和 1。

获取连续数字而不重复

Getting consecutive numbers without repetition

random

numbers

r