这可能很容易,但我还没有弄清楚。
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“应用程序”的滚动计数。其次,我需要创建不同年份“滞后(AppYear)”的滞后,如果前一年相同,则将获取第 1 年的行。
期望输出
# 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 整个数据集包括许多子类,因此我需要首先按
subclass
分组.数据现在以这种方式排序: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)"
))
我试图得到
app
使用多种方法,例如 cumsum(1:length(AppYear))
但找不到成功的答案。
最佳答案
更新:
解决关于具有多组 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
注意我正在使用问题的“编辑”部分下由 OP 提供的数据。原答案:
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"))
关于不同年份的滚动计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56840843/