Question

我有一个数据框，需要根据ID的每个子集在某些行中列出的日期对列进行有条件的重新编码。我试图找出如何最好地使用dplyr中的mutate函数来实现这一点。欢迎提出建议和替代解决方案，但我要避免使用for循环。

我知道如何编写一个非常冗长且效率低下的for循环来解决此问题，但想知道如何更有效地进行。

示例数据框：

df<-data.frame(ID = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2),
               date = as.Date(c("2016-02-01","2016-02-01","2016-02-01","2016-03-21", "2016-03-21", "2016-03-21", "2016-10-05", "2016-10-05", "2016-10-05", "2016-10-05", "2016-03-01","2016-03-01","2016-03-01","2016-04-21", "2016-04-21", "2016-04-21", "2016-11-05", "2016-11-05", "2016-11-05", "2016-11-05")),
               trial = c(NA, NA, NA, 1, 1, 1, NA, NA, NA, NA, NA, NA, NA, 1, 1, 1, NA, NA, NA, NA)

我的伪代码-前两个case_when语句中的第二个逻辑参数是我被困住的地方。

df%>%
  group_by(ID)%>%
  mutate(results = case_when(
     is.na(trial) & date < date where trial = 1 ~ 0,
     is.na(trial) & date > date where trial = 1 ~ 2,
     trial == trial
  ))

预期结果为：

data.frame(ID = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2),
               date = as.Date(c("2016-02-01","2016-02-01","2016-02-01","2016-03-21", "2016-03-21", "2016-03-21", "2016-10-05", "2016-10-05", "2016-10-05", "2016-10-05", "2016-03-01","2016-03-01","2016-03-01","2016-04-21", "2016-04-21", "2016-04-21", "2016-11-05", "2016-11-05", "2016-11-05", "2016-11-05")),
               trial = c(0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2)
)

Answer 1

一种选择是按“ ID”分组并通过在“试验”列的（rleid）上应用run-length-id来转换“试验”

library(dplyr)
library(data.table)
df %>%
   group_by(ID) %>% 
   mutate(trial = rleid(trial)-1)
# A tibble: 20 x 3
# Groups:   ID [2]
#      ID date       trial
#   <dbl> <date>     <dbl>
# 1     1 2016-02-01     0
# 2     1 2016-02-01     0
# 3     1 2016-02-01     0
# 4     1 2016-03-21     1
# 5     1 2016-03-21     1
# 6     1 2016-03-21     1
# 7     1 2016-10-05     2
# 8     1 2016-10-05     2
# 9     1 2016-10-05     2
#10     1 2016-10-05     2
#11     2 2016-03-01     0
#12     2 2016-03-01     0
#13     2 2016-03-01     0
#14     2 2016-04-21     1
#15     2 2016-04-21     1
#16     2 2016-04-21     1
#17     2 2016-11-05     2
#18     2 2016-11-05     2
#19     2 2016-11-05     2
#20     2 2016-11-05     2

或使用rle

df %>% 
  group_by(ID) %>%
  mutate(trial = with(rle(is.na(trial)), 
             rep(seq_along(values), lengths))-1)

Answer 2

将您的伪代码转换为代码，我们可以使用which.max(trial == 1)来获得第一次出现，其中每个组的trial = 1。这还假设每个trial在ID中至少有一个1项。

library(dplyr)

df %>%
  group_by(ID) %>%
  mutate(trial = case_when(is.na(trial) & date < date[which.max(trial == 1)] ~ 0, 
                             is.na(trial) & date > date[which.max(trial == 1)] ~ 2, 
                             TRUE ~ trial))


#      ID date       trial
#    <dbl> <date>     <dbl>
# 1     1 2016-02-01     0
# 2     1 2016-02-01     0
# 3     1 2016-02-01     0
# 4     1 2016-03-21     1
# 5     1 2016-03-21     1
# 6     1 2016-03-21     1
# 7     1 2016-10-05     2
# 8     1 2016-10-05     2
# 9     1 2016-10-05     2
#10     1 2016-10-05     2
#11     2 2016-03-01     0
#12     2 2016-03-01     0
#13     2 2016-03-01     0
#14     2 2016-04-21     1
#15     2 2016-04-21     1
#16     2 2016-04-21     1
#17     2 2016-11-05     2
#18     2 2016-11-05     2
#19     2 2016-11-05     2
#20     2 2016-11-05     2

多条件变异

2 个答案: