example:

vector <- round(rnorm(100, mean = 0.5, sd = 0.2), 0)

n <- 3

if there are less than 3 consecutive 1-s, those 1-s = 0.

equal or more than 3 consecutive 1-s remain intact.

example:

vector <- round(rnorm(100, mean = 0.5, sd = 0.2), 0)

n <- 3

if there are less than 3 consecutive 1-s, those 1-s = 0.

equal or more than 3 consecutive 1-s remain intact.

```
vector <- round(rnorm(100, mean = 0.5, sd = 0.2), 0)
n <- 3
library(tidyverse)
(exdf <- enframe(vector))
# https://gist.github.com/brshallo/cadaa40cef6387e28924b6c3756627c9
lag_multiple <- function(x, n_vec){
map(n_vec, ~lag(x=x,n=.,default=0)) %>%
set_names(paste0("lag", n_vec)) %>%
as_tibble()
}
(exdf2 <- mutate(exdf,
lag_multiple(value,seq_len(n)-1)))
(exdf3 <- mutate(rowwise(exdf2),
s=sum(c_across(starts_with("lag")))) %>% ungroup())
(exdf4 <- mutate(exdf3,
value_2 = ifelse(s==n,1,0)) %>% select(starts_with("value")))
print(exdf4,n=100)
```

1 Like

Thank you very much, sir! Meantime I came up with an alternative solution in base R. Meant for only if n = 3.

```
vector <- round(rnorm(100, mean = 0.5, sd = 0.2), 0)
res <- c()
for (i in 2:(length(vector)-1)){
if ((sum(vector[(i-1):(i+1)]) == 3)){
res[i] <- TRUE
} else {res[i] <- FALSE}
}
res2 <- vector(mode = "numeric", length = length(res))
for (i in 2:(length(res)-1)){
if (res[i] == TRUE){
res2[(i-1):(i+1)] <- 1
}
}
res2
```

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