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，则会获取与Year-1对应的行。
所需的输出
# 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，因此我需要首先按subclass进行分组。现在，数据以这种方式排序：
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)"
))
我尝试使用app之类的许多方法来获取cumsum(1:length(AppYear))，但是找不到成功的答案。
答案 0 :(得分：5)
解决有关具有多组subclass的df的后续问题。
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
N.B。 。我正在使用OP在问题的 Edit 部分下提供的数据。
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"))
答案 1 :(得分：2)
这是var query = context.Solutions
.Select(s => new SolutionDto {
Id = s.Id,
Name = s.Name,
Description = s.Description,
Category = new CategoryDto {
Id = s.Category.Id,
Name = s.Category.Name
},
Views = s.SolutionViews.Count()
});
版本。粗糙的部分是自连接的，以允许查找表执行延迟年份。
data.table这里与@ M-M的答案大致相同。请注意，library(data.table)
dt <- as.data.table(df1)
setorder(dt, AppYear, Patent)
dt[, app := rleid(AppYear), by = .(subclass)]
dt[unique(dt[, .(lagging_year = shift(AppYear)
,lagging_app = shift(app) + 1), by = subclass])
, on = .(subclass
,app = lagging_app )
, lag_year := lagging_year]
dt
Patent subclass AppYear app lag_year
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
#total of 30 rows.
一直在计数，因此它不是100％等效的。第一个.GRP为435 / 91.21，此处的subclass为5，而不是其他解决方案的1。
app这是表演：
library(data.table)
dt <- as.data.table(df1)
setorder(dt, AppYear, Patent)
dt[, `:=` (app = .GRP, app_cnt = .N), by = .( AppYear, subclass)]
dt[, lag_year := rep(shift(unique(AppYear)), unique(app_cnt)), by = .(subclass)]
dt[, app_cnt := NULL]
dt
数据和功能：
Unit: milliseconds
expr min lq mean median uq max neval
M_M_dplyr 6.5839 6.85235 7.442658 6.94440 7.26040 23.0357 100
cole_dt_join 6.0260 6.27025 6.616121 6.44040 6.84965 8.3686 100
cole_dt_rep 3.0404 3.15575 3.435112 3.26355 3.76085 4.7526 100
cole_dt_rep_rleid 3.4116 3.59275 3.911844 3.68695 4.01000 10.3520 100