Replacing zeros with a desired value

Question

I wish to remove all 0 values that have 4 or more consecutive values of 0 with NaN. Your help would be appreciated.

I have the following data :

 A B C
 1 2 0
 0 8 0
 0 0 0
 0 0 0
 0 0 8
 0 4 9

What I am looking to get :

  A B C
  1 2 Na
 Na 8 Na
 Na 0 Na
 Na 0 Na
 Na 0 8
 Na 4 9

Answer 1

We can use rle to do this. Loop over the columns with lapply, use rle on a logical vector (x==0) to return the adjacent elements that are same ('values') and its lengths), assign the 'values' that are not having lengths greater than or equal to 4 to FALSE, unlist the list with inverse.rle and use that for replaceing the values of 'x' to NA

df1[] <- lapply(df1, function(x) {
     i1 <- inverse.rle(within.list(rle(x == 0),
                 values[!(values & lengths >=4)] <- FALSE))
     replace(x, i1, NA)
   })

df1
#   A B  C
#1  1 2 NA
#2 NA 8 NA
#3 NA 0 NA
#4 NA 0 NA
#5 NA 0  8
#6 NA 4  9

---

Or an option with rleid

library(data.table)
library(dplyr)
df1 %>% 
   mutate(across(everything(), ~ replace(., . == 0 &
            ave(., rleid(.), FUN = length) >= 4, NA) ))

data

df1 <- structure(list(A = c(1L, 0L, 0L, 0L, 0L, 0L), B = c(2L, 8L, 0L, 
0L, 0L, 4L), C = c(0L, 0L, 0L, 0L, 8L, 9L)), class = "data.frame", 
row.names = c(NA, 
-6L))