Roll backwards

[elben10]

Would it be possible to implement backwards rolling. Thus, it would be a very convenient way to implement multi periods condition when handling panel data.

library(tidyverse)
#> + ggplot2 2.2.1          Date: 2017-05-29
#> + tibble  1.3.3             R: 3.3.2
#> + tidyr   0.6.3            OS: Windows 10 x64
#> + readr   1.1.1           GUI: RTerm
#> + purrr   0.2.2.2      Locale: Danish_Denmark.1252
#> + dplyr   0.5.0            TZ: Europe/Paris
#> + stringr 1.2.0        
#> + forcats 0.2.0
#> Warning: package 'ggplot2' was built under R version 3.3.3
#> Warning: package 'tibble' was built under R version 3.3.3
#> Warning: package 'tidyr' was built under R version 3.3.3
#> Warning: package 'readr' was built under R version 3.3.3
#> Warning: package 'purrr' was built under R version 3.3.3
#> Warning: package 'dplyr' was built under R version 3.3.3
#> Warning: package 'forcats' was built under R version 3.3.3
#> -- Conflicts ----------------------------------------------------
#> * filter(),  from dplyr, masks stats::filter()
#> * lag(),     from dplyr, masks stats::lag()
library(RcppRoll)
#> Warning: package 'RcppRoll' was built under R version 3.3.3

df <- tibble(id = rep(1, each = 10), wage = c(980, rep(1000, each = 9)))

df %>%
  mutate(wagecri = if_else(wage >= 1000, 1, 0)) %>%
  mutate(crimean = roll_meanr(wagecri, n = 2))
#> # A tibble: 10 x 4
#>       id  wage wagecri crimean
#>    <dbl> <dbl>   <dbl>   <dbl>
#>  1     1   980       0      NA
#>  2     1  1000       1     0.5
#>  3     1  1000       1     1.0
#>  4     1  1000       1     1.0
#>  5     1  1000       1     1.0
#>  6     1  1000       1     1.0
#>  7     1  1000       1     1.0
#>  8     1  1000       1     1.0
#>  9     1  1000       1     1.0
#> 10     1  1000       1     1.0

# I would like the result to be as follows


tibble(id = rep(1, each = 10), wage = c(980, rep(1000, each = 9))) %>%
  mutate(wagecri = if_else(wage >= 1000, 1, 0)) %>%
  mutate(crimean = c(0.5, rep(1, each = 8), NA_real_))
#> # A tibble: 10 x 4
#>       id  wage wagecri crimean
#>    <dbl> <dbl>   <dbl>   <dbl>
#>  1     1   980       0     0.5
#>  2     1  1000       1     1.0
#>  3     1  1000       1     1.0
#>  4     1  1000       1     1.0
#>  5     1  1000       1     1.0
#>  6     1  1000       1     1.0
#>  7     1  1000       1     1.0
#>  8     1  1000       1     1.0
#>  9     1  1000       1     1.0
#> 10     1  1000       1      NA

Best regards,
Jakob