在这类问题中精确地计算闰年是R中的一件痛苦的事情。因此，我用365.25天来近似一年的长度。请查看评论中对步骤的解释。

library(tidyverse)
library(lubridate)

dat <- tibble(Person = c(1, 2, 3, 4, 5, 6),  Neighbourhood = c("XYZ", "XYZ", "XYZ", "XYZ", "ABC", "ABC"),  Date_of_sale =  structure(c(17987, NA, 19275, 17564, 18052, NA), class = "Date"),  Sale = c(1, 0, 1, 1, 1, 0))

sol <- dat |> 
  filter(Sale == 1) |> # keep only persons with sales
  arrange(Neighbourhood, Date_of_sale) |>
  group_by(Neighbourhood) |> 
  mutate(year = ((Date_of_sale - min(Date_of_sale) + 1)/365.25) |> # bring time diff to year level
          unclass() |> 
          ceiling() # round up to full year
         ) |> 
  group_by(Neighbourhood, year) |> 
  summarize(sales = sum(Sale)) |> 
  ungroup() |> 
  complete(Neighbourhood, year = 1:3, fill = list(sales = 0)) |> # Add year without sales
  filter(year <= 3) |> # Remove years greater than 3
  group_by(Neighbourhood) |> 
  mutate(cum_sales = cumsum(sales)) |>  # Cumulate sales across years
  ungroup()

sol
#> # A tibble: 6 × 4
#>   Neighbourhood  year sales cum_sales
#>   <chr>         <dbl> <dbl>     <dbl>
#> 1 ABC               1     1         1
#> 2 ABC               2     0         1
#> 3 ABC               3     0         1
#> 4 XYZ               1     1         1
#> 5 XYZ               2     1         2
#> 6 XYZ               3     0         2