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dplyr:在标记时间段之前的年内过滤

dummy-variable time-series dplyr r

ulima2_ · 技术社区 · 7 年前

我想有一份年度国家特定假人的清单,我还想标记年份两年

library(tidyverse)

df <- tribble(
  ~year, ~country, ~occurrence, 
  #--|--|----
  2003, "USA", 1,
  2004, "USA", 0,
  2005, "USA", 0,
  2006, "USA", 0,
  2007, "USA", 0,
  2008, "USA", 0,
  2009, "USA", 0,
  2010, "USA", 0,
  2011, "USA", 1,
  2012, "USA", 0,
  2013, "USA", 0,
  2005, "FRA", 0,
  2006, "FRA", 0,
  2007, "FRA", 1,
  2008, "FRA", 1,
  2009, "FRA", 0,
  2010, "FRA", 0,
  2011, "FRA", 0,
  2012, "FRA", 0,
  2013, "FRA", 0,
  2014, "FRA", 0,
  2015, "FRA", 1
)

所以对于 "USA" 我还想放一个 1 occurence 2009年和2010年以及 FRA 2005年、2006年、2013年和2014年。

我想过做这样的事情:

df %>%
  group_by(country) %>%
  mutate(occurence = ifelse("not sure what to put here"),
                            1,
                            0))

2 回复 | 直到 7 年前

akrun 7 年前

按“国家”分组后,我们最多可以选择2个 lead 并获取 max pmax 获取“occurrence”中的预期输出

df %>%
  group_by(country) %>%
  mutate(occurrence = pmax(occurrence, lead(occurrence, default = 0),
                     lead(occurrence, default=0, n=2)))

或者这可以通过 data.table

library(data.table)
setDT(df)[,  occurrence := do.call(pmax, shift(occurrence, n = 0:2,
     type = "lead", fill = 0)), country]
df
#    year country occurrence
# 1: 2003     USA          1
# 2: 2004     USA          0
# 3: 2005     USA          0
# 4: 2006     USA          0
# 5: 2007     USA          0
# 6: 2008     USA          0
# 7: 2009     USA          1
# 8: 2010     USA          1
# 9: 2011     USA          1
#10: 2012     USA          0
#11: 2013     USA          0
#12: 2005     FRA          1
#13: 2006     FRA          1
#14: 2007     FRA          1
#15: 2008     FRA          1
#16: 2009     FRA          0
#17: 2010     FRA          0
#18: 2011     FRA          0
#19: 2012     FRA          0
#20: 2013     FRA          1
#21: 2014     FRA          1
#22: 2015     FRA          1

LmW. 7 年前

这里是另一个dplyr解决方案:

df %>% 
  group_by(country) %>% 
  mutate(
        occurrence=ifelse( lead(occurrence, 1) %in% 1 | 
                           lead(occurrence, 2) %in% 1, 
                           1, occurrence)
         )

# A tibble: 22 x 3
# Groups:   country [2]
    year country occurrence
   <dbl>   <chr>      <dbl>
 1  2003     USA          1
 2  2004     USA          0
 3  2005     USA          0
 4  2006     USA          0
 5  2007     USA          0
 6  2008     USA          0
 7  2009     USA          1
 8  2010     USA          1
 9  2011     USA          1
10  2012     USA          0
11  2013     USA          0
12  2005     FRA          1
13  2006     FRA          1
14  2007     FRA          1
15  2008     FRA          1
16  2009     FRA          0
17  2010     FRA          0
18  2011     FRA          0
19  2012     FRA          0
20  2013     FRA          1
21  2014     FRA          1
22  2015     FRA          1

lead(occurrence, 1) %in% 1 使用而不是 lead(occurrence, 1) == 1 NA .