不同年份的滚动计数

Question

这可能很容易，但是我还没有弄清楚。

这是我的数据集的一部分：

structure(list(Patent = c("4683202", "4683195", "4800159", "4965188", 
"4994368", "5328824", "4879214", "4921794", "4983728", "4994372"
), subclass = c("435/91.2", "435/91.2", "435/91.2", "435/91.2", 
"435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2"
), AppYear = c(1985L, 1986L, 1986L, 1987L, 1987L, 1987L, 1988L, 
1988L, 1990L, 1990L), app = 1:10, class = "data.frame", row.names = c(NA, 
-10L), .Names = c("Patent", "subclass", "AppYear", "app", "lag(AppYear)"
))


> data
# A tibble: 10 x 3
  Patent  subclass AppYear
   <chr>   <chr>      <int>
 1 4683202 435/91.2    1985
 2 4683195 435/91.2    1986
 3 4800159 435/91.2    1986
 4 4965188 435/91.2    1987
 5 4994368 435/91.2    1987
 6 5328824 435/91.2    1987
 7 4879214 435/91.2    1988
 8 4921794 435/91.2    1988
 9 4983728 435/91.2    1990
10 4994372 435/91.2    1990

首先，我需要获得不同年份“ app”的滚动计数。其次，我需要创建不同年份“ lag（AppYear）”的滞后，如果前一年相同，则将获取与year-1对应的行。

期望的输出

# A tibble: 10 x 5
   Patent  subclass AppYear   app `lag(AppYear)`
   <chr>   <chr>      <int> <int>          <int>
 1 4683202 435/91.2    1985     1             NA
 2 4683195 435/91.2    1986     2           1985
 3 4800159 435/91.2    1986     2           1985
 4 4965188 435/91.2    1987     3           1986
 5 4994368 435/91.2    1987     3           1986
 6 5328824 435/91.2    1987     3           1986
 7 4879214 435/91.2    1988     4           1987
 8 4921794 435/91.2    1988     4           1987
 9 4983728 435/91.2    1990     5           1988
10 4994372 435/91.2    1990     5           1988

编辑整个数据集包含许多子类，因此我需要先按进行分组subclass。现在以这种方式对数据进行排序：

data <- data %>% 
  select(Patent, subclass, AppYear) %>% 
  arrange(AppYear,Patent) %>% 
  group_by(subclass) %>% 
  mutate(app = 1:n(), lag(AppYear))

。

structure(list(Patent = c("4683202", "4683195", "4800159", "4965188", 
"4994368", "5328824", "4879214", "4921794", "4983728", "4994372", 
"5066584", "5075216", "5091310", "5093245", "5132215", "5185243", 
"5409818", "5409818", "6107023", "4994370", "5001050", "5023171", 
"5035996", "5035996", "5043272", "5045450", "5055393", "5085983", 
"5106729", "5106729"), subclass = c("435/91.2", "435/91.2", "435/91.2", 
"435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2", 
"435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2", 
"435/91.2", "435/91.21", "435/91.2", "435/91.2", "435/91.2", 
"435/91.2", "435/91.2", "435/91.2", "435/91.21", "435/91.2", 
"435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.21"), 
    AppYear = c(1985L, 1986L, 1986L, 1987L, 1987L, 1987L, 1988L, 
    1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 
    1988L, 1988L, 1988L, 1989L, 1989L, 1989L, 1989L, 1989L, 1989L, 
    1989L, 1989L, 1989L, 1989L, 1989L), app = c(1L, 2L, 3L, 4L, 
    5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 1L, 
    17L, 18L, 19L, 20L, 21L, 22L, 2L, 23L, 24L, 25L, 26L, 27L, 
    3L), `lag(AppYear)` = c(NA, 1985L, 1986L, 1986L, 1987L, 1987L, 
    1987L, 1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 1988L, 
    1988L, NA, 1988L, 1988L, 1988L, 1989L, 1989L, 1989L, 1988L, 
    1989L, 1989L, 1989L, 1989L, 1989L, 1989L)), class = "data.frame", row.names = c(NA, 
-30L), .Names = c("Patent", "subclass", "AppYear", "app", "lag(AppYear)"
))

我尝试app使用许多方法，例如cumsum(1:length(AppYear))但找不到成功的答案。

Answer 1

更新：

解决有关包含多个组的df的后续问题subclass。

library(dplyr)

df1 %>% 
  select(Patent, subclass, AppYear) %>% 
  arrange(AppYear, Patent) %>%
  group_by(subclass) %>% 
  group_map(~mutate(.,app=group_indices(.,AppYear),
                    lag_year = rep(lag(unique(.$AppYear)), count_(., "AppYear")$n)), 
            keep = T) %>% 
  bind_rows() %>% 
  arrange(AppYear, Patent) 

#> # A tibble: 30 x 5
#>    Patent  subclass AppYear   app lag_year
#>    <chr>   <chr>      <int> <int>    <int>
#>  1 4683202 435/91.2    1985     1       NA
#>  2 4683195 435/91.2    1986     2     1985
#>  3 4800159 435/91.2    1986     2     1985
#>  4 4965188 435/91.2    1987     3     1986
#>  5 4994368 435/91.2    1987     3     1986
#>  6 5328824 435/91.2    1987     3     1986
#>  7 4879214 435/91.2    1988     4     1987
#>  8 4921794 435/91.2    1988     4     1987
#>  9 4983728 435/91.2    1988     4     1987
#> 10 4994372 435/91.2    1988     4     1987
#> # ... with 20 more rows

注意：我正在使用OP在问题的“编辑”部分下提供的数据。

原始答案：

library(dplyr)

df1 %>% 
  arrange(AppYear, Patent) %>%
  mutate(app = group_indices(.,AppYear), 
        lag_year = rep(lag(unique(.$AppYear)), count_(., "AppYear")$n))

#> # A tibble: 10 x 5
#>    Patent  subclass AppYear   app lag_year
#>    <chr>   <chr>      <int> <int>    <int>
#>  1 4683202 435/91.2    1985     1       NA
#>  2 4683195 435/91.2    1986     2     1985
#>  3 4800159 435/91.2    1986     2     1985
#>  4 4965188 435/91.2    1987     3     1986
#>  5 4994368 435/91.2    1987     3     1986
#>  6 5328824 435/91.2    1987     3     1986
#>  7 4879214 435/91.2    1988     4     1987
#>  8 4921794 435/91.2    1988     4     1987
#>  9 4983728 435/91.2    1990     5     1988
#> 10 4994372 435/91.2    1990     5     1988

数据：

df1 <- structure(list(Patent=c("4683202", "4683195", "4800159", "4965188", 
                      "4994368", "5328824", "4879214", "4921794", "4983728", "4994372"), 
                 subclass=c("435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2",
                      "435/91.2", "435/91.2", "435/91.2", "435/91.2", "435/91.2"), 
                 AppYear=c(1985L, 1986L, 1986L, 1987L, 1987L, 1987L, 1988L, 
                      1988L, 1990L, 1990L)), 
                 row.names=c(NA, -10L), 
                 class=c("tbl_df", "tbl", "data.frame"))