我有一个大型数据集，显示从“健康”事件到随后的“病态”事件的孩子的跟进
我正在尝试使用dplyr来计算“健康”事件和第一次“生病”事件之间的时间
模拟数据集
id <- c(1,1,1,1,1,1)
event <- c("healthy","","","sick","sick","")
date_follow_up <- c("4/1/15", "4/2/15", "4/3/15", "4/4/15", "4/5/15", "4/6/15")
df1 <- data_frame(id, event, date_follow_up)
模拟输出数据集
id <- c(1,1,1,1,1,1)
event <- c("healthy","","","sick","sick","")
date_follow_up <- c("4/1/15", "4/2/15", "4/3/15", "4/4/15", "4/5/15", "4/6/15")
diff_time <- c(3,"","","","","")
df1 <- data_frame(id, event, date_follow_up, diff_time)
我只能使用dplyr按“id”和“date_follow_up”对数据进行排序，然后按“id”分组：
df2 <- df1 %>% arrange(id, date_follow_up) %>% group_by(id)
请计算日期差异并将其添加到行旁边，并为每个人添加“健康”事件，这需要帮助：）
答案 0 :(得分：3)
使用@ akrun的示例数据，这是使用来自 data.table 的滚动连接的一种方式：
require(data.table)
dt = as.data.table(mydf)[, date_follow_up := as.Date(date_follow_up, format="%m/%d/%y")][]
dt1 = dt[event == "healthy"]
dt2 = dt[event == "sick"]
idx = dt2[dt1, roll = -Inf, which = TRUE, on = c("id", "date_follow_up")]
这个想法是：对于每个健康的日期（在dt1中），获取第一个生病日期（dt2）>=的健康日期的索引。
然后直接减去两个日期以获得最终结果。
dt[event == "healthy",
diff := as.integer(dt2$date_follow_up[idx] - dt1$date_follow_up)]
答案 1 :(得分：2)
我对您的数据进行了一些修改以彻底检查此案例。我的建议与alistaire建议的类似。我的建议可以在mydf中为id 2生成NA，而alistaire建议会创建Inf。首先，我将您的日期（字符）转换为日期对象。然后，我将数据按id分组，并通过减去healthy的第一天计算时差（即date_follow_up[event == "healthy"][1] ）从sick的第一天开始（即date_follow_up[event == "sick"][1]）。最后，我用NA替换了不相关行的时差。
id event date_follow_up
1 1 healthy 4/1/15
2 1 4/2/15
3 1 4/3/15
4 1 sick 4/4/15
5 1 sick 4/5/15
6 2 4/1/15
7 2 healthy 4/2/15
8 2 4/3/15
9 2 4/4/15
10 2 4/5/15
11 3 4/1/15
12 3 healthy 4/2/15
13 3 sick 4/3/15
14 3 4/4/15
15 3 4/5/15
library(dplyr)
mutate(mydf, date_follow_up = as.Date(date_follow_up, format = "%m/%d/%y")) %>%
group_by(id) %>%
mutate(foo = date_follow_up[event == "sick"][1] - date_follow_up[event == "healthy"][1],
foo = replace(foo, which(event != "healthy"), NA))
Source: local data frame [15 x 4]
Groups: id [3]
id event date_follow_up foo
<int> <chr> <date> <S3: difftime>
1 1 healthy 2015-04-01 3 days
2 1 2015-04-02 NA days
3 1 2015-04-03 NA days
4 1 sick 2015-04-04 NA days
5 1 sick 2015-04-05 NA days
6 2 2015-04-01 NA days
7 2 healthy 2015-04-02 NA days
8 2 2015-04-03 NA days
9 2 2015-04-04 NA days
10 2 2015-04-05 NA days
11 3 2015-04-01 NA days
12 3 healthy 2015-04-02 1 days
13 3 sick 2015-04-03 NA days
14 3 2015-04-04 NA days
15 3 2015-04-05 NA days
数据
mydf <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
3L, 3L, 3L, 3L, 3L), event = c("healthy", "", "", "sick", "sick",
"", "healthy", "", "", "", "", "healthy", "sick", "", ""), date_follow_up = c("4/1/15",
"4/2/15", "4/3/15", "4/4/15", "4/5/15", "4/1/15", "4/2/15", "4/3/15",
"4/4/15", "4/5/15", "4/1/15", "4/2/15", "4/3/15", "4/4/15", "4/5/15"
)), .Names = c("id", "event", "date_follow_up"), row.names = c(NA,
-15L), class = "data.frame")
答案 2 :(得分：2)
我们也可以使用data.table。转换＆＃39; data.frame＆＃39;到＆＃39; data.table＆＃39; （setDT(mydf)），使用Date将＆＃39; date_follow_up的类更改为as.Date，按＆＃39; id＆＃39;分组通过获取逻辑向量（event == "healthy"）的累积和来创建分组变量，我们得到了＆＃39; date_follow_up＆＃39;对于第一个＆＃34;病人＆＃34; ＆＃39;事件＆＃39;第一个＆＃39; date_follow_up＆＃39; （这将是＆＃34;健康＆＃34;）if有any＆＃34;生病＆＃34; ＆＃39;事件＆＃39;在该特定群组或else返回＆＃34; NA＆＃34;。
library(data.table)
setDT(mydf)[, date_follow_up := as.Date(date_follow_up, "%m/%d/%y")
][, foo := if(any(event == "sick"))
as.integer(date_follow_up[which(event=="sick")[1]] -
date_follow_up[1] )
else NA_integer_ ,
by = .(grp= cumsum(event == "healthy"), id)]
然后，我们可以改变＆＃34; foo＆＃34;到＆＃34; NA＆＃34;为所有＆＃34;事件＆＃34;那不健康＆＃34;。
mydf[event!= "healthy", foo := NA_integer_]
mydf
# id event date_follow_up foo
# 1: 1 healthy 2015-04-01 3
# 2: 1 2015-04-02 NA
# 3: 1 2015-04-03 NA
# 4: 1 sick 2015-04-04 NA
# 5: 1 sick 2015-04-05 NA
# 6: 2 2015-04-01 NA
# 7: 2 healthy 2015-04-02 NA
# 8: 2 2015-04-03 NA
# 9: 2 2015-04-04 NA
#10: 2 2015-04-05 NA
#11: 3 2015-04-01 NA
#12: 3 healthy 2015-04-02 1
#13: 3 sick 2015-04-03 NA
#14: 3 2015-04-04 NA
#15: 3 2015-04-05 NA
#16: 4 2015-04-01 NA
#17: 4 healthy 2015-04-02 3
#18: 4 2015-04-03 NA
#19: 4 2015-04-04 NA
#20: 4 sick 2015-04-05 NA
#21: 4 sick 2015-04-06 NA
#22: 4 2015-04-07 NA
#23: 4 healthy 2015-04-08 2
#24: 4 2015-04-09 NA
#25: 4 sick 2015-04-10 NA
注意：在这里，我准备了数据，其中可能有多个＆＃34;健康/生病＆＃34; ＆＃39;事件＆＃39;特定的＆＃34; id＆＃34;。
mydf <- structure(list(id = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3,
3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), event = c("healthy", "",
"", "sick", "sick", "", "healthy", "", "", "", "", "healthy",
"sick", "", "", "", "healthy", "", "", "sick", "sick", "", "healthy",
"", "sick"), date_follow_up = c("4/1/15", "4/2/15", "4/3/15",
"4/4/15", "4/5/15", "4/1/15", "4/2/15", "4/3/15", "4/4/15", "4/5/15",
"4/1/15", "4/2/15", "4/3/15", "4/4/15", "4/5/15", "4/1/15", "4/2/15",
"4/3/15", "4/4/15", "4/5/15", "4/6/15", "4/7/15", "4/8/15", "4/9/15",
"4/10/15")), .Names = c("id", "event", "date_follow_up"), row.names = c(NA,
25L), class = "data.frame")
答案 3 :(得分：0)
这是一种方法，但如果每个ID有多个“健康”事件，您可能需要对其进行调整以使其更加健壮：
# turn dates into subtractable Date class
df1 %>% mutate(date_follow_up = as.Date(date_follow_up, '%m/%d/%y')) %>%
group_by(id) %>%
# Add new column. If there is a "healthy" event,
mutate(diff_time = ifelse(event == 'healthy',
# subtract the date from the minimum "sick" date
min(date_follow_up[event == 'sick']) - date_follow_up,
# else if it isn't a "healthy" event, return NA.
NA))
## Source: local data frame [6 x 4]
##
## id event date_follow_up diff_time
## <dbl> <chr> <date> <dbl>
## 1 1 healthy 2015-04-01 3
## 2 1 2015-04-02 NA
## 3 1 2015-04-03 NA
## 4 1 sick 2015-04-04 NA
## 5 1 sick 2015-04-05 NA
## 6 1 2015-04-06 NA
答案 4 :(得分：0)
这是使用library(dplyr)
df1$date_follow_up <- as.Date(df1$date_follow_up, "%m/%d/%y")
df1 %>% group_by(id, event) %>%
filter(event %in% c("healthy", "sick")) %>%
slice(which.min(date_follow_up)) %>% group_by(id) %>%
mutate(diff_time = lead(date_follow_up) - date_follow_up) %>%
right_join(df1, by = c("id", "event" , "date_follow_up"))
# Output
Source: local data frame [6 x 4]
Groups: id [?]
id event date_follow_up diff_time
<dbl> <chr> <date> <S3: difftime>
1 1 healthy 2015-04-01 3 days
2 1 2015-04-02 NA days
3 1 2015-04-03 NA days
4 1 sick 2015-04-04 NA days
5 1 sick 2015-04-05 NA days
6 1 2015-04-06 NA days
的另一种方法（尽管与早期解决方案相比有点长）
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