根据条件识别特定行之后的行

Question

我有一个数据框/小标题，其中包含多个国家/地区的年度观察结果。在特定事件发生的年份中，变量 event 的值是 1。

我现在正在尝试指定一个新列event.10yrs，它在事件结束后的 9 年内获得值 1（如果事件持续数年，则为事件的最后一年）。在新事件发生且不是新事件的最后一年的年份中，新列event.10yrs 的值为 0。

低于一个国家的数据。列event.10yrs 是所需的输出。

 df <-structure(list(year = c(1970, 1971, 1972, 1973, 1974, 1975, 1976, 
1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 
1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 
1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 
2010, 2011, 2012, 2013, 2014, 2015), ccode = c(516, 516, 516, 
516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 
516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 
516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 516, 
516, 516, 516, 516), event = c(0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 
1, 1, 1, 0, 0, 0, 0, 0, 0, 0, NA, NA, NA, NA, NA), event.last.y = c(0, 
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, NA, 
NA, NA, NA, NA), event.10yrs = c(NA, 0, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 
0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, NA, NA, NA)), row.names = c(NA, 
-46L), vars = "ccode", drop = TRUE, class = c("grouped_df", "tbl_df", 
"tbl", "data.frame"), indices = list(0:45), group_sizes = 46L, biggest_group_size = 46L, labels = structure(list(
    ccode = 516), row.names = c(NA, -1L), vars = "ccode", drop = TRUE, class = "data.frame", .Names = "ccode"), .Names = c("year", 
"ccode", "event", "event.last.y", "event.10yrs"))

到目前为止我使用 dplyr 包的尝试：

df <- df %>%
  mutate(event.10yrs=case_when(event!=1 & year-9 < year[event.last.y==1] ~ 1,
                               TRUE ~ 0))

然而，这会产生以下警告：

Warning message:
In year < year[rs.war.last.y == 1] :
  longer object length is not a multiple of shorter object length

感谢任何提示。

Answer 1

也许只是一个嵌套的 ifelse（或 dplyr::if_else）

require(dplyr)
df %>% mutate(ev_10 = if_else(event == 0, 1, 
                           if_else(event.last.y ==1, 1, 0), 
                                                           0))

编辑

这篇文章在这里帮助了我：Find the index position of the first non-NA value in an R vector?

但我们不仅要替换第一次出现的 'x' ...
所以我用一个帮助列做了一些解决方法

index_1 <- unlist(lapply(which(df$event.last.y ==1 ), 
           function(x) seq(x, length.out=9))) 
# this makes a vector with all the index of the last 9 positions 
# after the last value == 1 
df$last_code <- df$event.last.y #just to duplicate your column
df$last_code[index_1] <- 1 #replacing the indices with '1'

现在我们可以像以前一样使用简单的嵌套条件语句了

df <- df %>% mutate(ev_10 = if_else(event == 0 & last_code==1, 1, 
#added the condition that last_code needs to be '1'
                          if_else(event.last.y ==1, 1, 0), 
                          0))

 head(df[c(2:13, 31:40),], 20) #printing only example rows here
# A tibble: 20 x 7
# Groups:   ccode [1]
    year ccode event event.last.y event.10yrs last_code ev_10
   <dbl> <dbl> <dbl>        <dbl>       <dbl>     <dbl> <dbl>
 1  1971   516  0            0           0         0     0   
 2  1972   516  1.00         1.00        1.00      1.00  1.00
 3  1973   516  0            0           1.00      1.00  1.00
 4  1974   516  0            0           1.00      1.00  1.00
 5  1975   516  0            0           1.00      1.00  1.00
 6  1976   516  0            0           1.00      1.00  1.00
 7  1977   516  0            0           1.00      1.00  1.00
 8  1978   516  0            0           1.00      1.00  1.00
 9  1979   516  0            0           1.00      1.00  1.00
10  1980   516  0            0           1.00      1.00  1.00
11  1981   516  0            0           1.00      0     0   
12  1982   516  0            0           0         0     0  
... 
13  2000   516  1.00         0           0         1.00  0   
14  2001   516  1.00         0           0         1.00  0   
15  2002   516  1.00         0           0         1.00  0   
16  2003   516  1.00         1.00        1.00      1.00  1.00
17  2004   516  0            0           1.00      1.00  1.00
18  2005   516  0            0           1.00      1.00  1.00
19  2006   516  0            0           1.00      1.00  1.00
20  2007   516  0            0           1.00      1.00  1.00