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Find the "previous" (lag()) or "next" (lead()) values in a vector. Useful for comparing values behind of or ahead of the current values.

Usage

lag(x, n = 1L, default = NULL, order_by = NULL, ...)

lead(x, n = 1L, default = NULL, order_by = NULL, ...)

Arguments

x

A vector

n

Positive integer of length 1, giving the number of positions to lag or lead by

default

The value used to pad x back to its original size after the lag or lead has been applied. The default, NULL, pads with a missing value. If supplied, this must be a vector with size 1, which will be cast to the type of x.

order_by

An optional secondary vector that defines the ordering to use when applying the lag or lead to x. If supplied, this must be the same size as x.

...

Not used.

Value

A vector with the same type and size as x.

Examples

lag(1:5)
#> [1] NA  1  2  3  4
lead(1:5)
#> [1]  2  3  4  5 NA

x <- 1:5
tibble(behind = lag(x), x, ahead = lead(x))
#> # A tibble: 5 × 3
#>   behind     x ahead
#>    <int> <int> <int>
#> 1     NA     1     2
#> 2      1     2     3
#> 3      2     3     4
#> 4      3     4     5
#> 5      4     5    NA

# If you want to look more rows behind or ahead, use `n`
lag(1:5, n = 1)
#> [1] NA  1  2  3  4
lag(1:5, n = 2)
#> [1] NA NA  1  2  3

lead(1:5, n = 1)
#> [1]  2  3  4  5 NA
lead(1:5, n = 2)
#> [1]  3  4  5 NA NA

# If you want to define a value to pad with, use `default`
lag(1:5)
#> [1] NA  1  2  3  4
lag(1:5, default = 0)
#> [1] 0 1 2 3 4

lead(1:5)
#> [1]  2  3  4  5 NA
lead(1:5, default = 6)
#> [1] 2 3 4 5 6

# If the data are not already ordered, use `order_by`
scrambled <- slice_sample(
  tibble(year = 2000:2005, value = (0:5) ^ 2),
  prop = 1
)

wrong <- mutate(scrambled, previous_year_value = lag(value))
arrange(wrong, year)
#> # A tibble: 6 × 3
#>    year value previous_year_value
#>   <int> <dbl>               <dbl>
#> 1  2000     0                   1
#> 2  2001     1                  16
#> 3  2002     4                  NA
#> 4  2003     9                   4
#> 5  2004    16                  25
#> 6  2005    25                   9

right <- mutate(scrambled, previous_year_value = lag(value, order_by = year))
arrange(right, year)
#> # A tibble: 6 × 3
#>    year value previous_year_value
#>   <int> <dbl>               <dbl>
#> 1  2000     0                  NA
#> 2  2001     1                   0
#> 3  2002     4                   1
#> 4  2003     9                   4
#> 5  2004    16                   9
#> 6  2005    25                  16