Find if certain value appears more than n-times subsequently [R]-CodePudding

I have a list of vectors, for instance:

vec1 <- c(rep(0,5), 1, rep(0,11), rep(1,4), rep(0,6))
vec2 <- c(rep(0,11), 1, rep(0,18))
vec3 <- c(rep(0,3), rep(1,5), rep(0,21))
vec4 <- c(rep(0,23))
  
test_list <- list(vec1, vec2, vec3, vec4)

I would like to filter this list based on 2 conditions:

1 is present within the vector.
1 appears consecutively (in a row) more than 3 times.

My output should contain vec1 and vec3.

I wrote a following function:

filter_ones <- test_list[sapply(test_list,function(vec) 1 %in% vec )]

And it returns vec1, vec2, and vec3.

How to apply the second condition? I probably shall use rle() but have no idea, how to do so. I will be grateful for help.

CodePudding user response：

We could add a second condition using rle short-circuiting with the OP's first logical expression (1 %in% vec) in Filter to filter the elements of the list.

The rle on the logical converted binary values is converted to a second logical based on whether the lengths (from rle) is greater than threshold 'n' and it is a 1 (TRUE), wrap with any to return a single TRUE/FALSE

n <- 3
Filter(function(x) 1 %in% x && any(with(rle(as.logical(x)), 
      lengths > n & values)), test_list)

-output

[[1]]
 [1] 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0

[[2]]
 [1] 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Or using the OP's sapply

test_list[sapply(test_list,function(vec) 1 %in% vec && 
      any(with(rle(as.logical(vec)), 
      lengths > n & values)))]
[[1]]
 [1] 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0

[[2]]
 [1] 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

CodePudding user response：

I condensed the vectors to a string and used grepl to get those vectors that match both conditions.

test_list[
  # Get non-empty results
  vapply(
    # Find vectors where conditions apply
    sapply(test_list, function(x) {
      # condense vector to string and do grepl
      # find those vectors where there is a 1 and a sequence of at least 3 1s and vice versa
      if(grepl("1.*111|111.*1", paste(x, collapse = ""))) x
    # Get non-empty results  
    }), Negate(is.null), NA
  )
]