dplyr-排列，分組，計算日期差

Question

我有一個龐大的數據集，顯示了從“健康”事件到后續“病假”事件的孩子的跟進情況

我正在嘗試使用dplyr計算“健康”事件與第一個“病假”事件之間的時間

模擬數據集

 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)

模擬輸出數據集

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”對數據進行排序，然后按“ id”對數據進行分組：

df2 <- df1 %>% arrange(id, date_follow_up) %>% group_by(id)

請在計算日期差異並將其添加到每個人的“健康”事件的行旁邊時需要幫助：)

Answer 1

使用@ akrun的數據。例如，下面是一個使用滾動單程從data.table 加入 ：

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")]

這個想法是：對於每個健康日期（以dt1 ），獲取第一個患病日期的索引（以dt2 ） >=健康日期。

然后，直接減去兩個日期即可得出最終結果。

dt[event == "healthy", 
     diff := as.integer(dt2$date_follow_up[idx] - dt1$date_follow_up)]

Answer 2

我還對您的數據進行了一些修改，以徹底檢查這種情況。 我的建議類似於利斯特主義者的建議。 我的建議可以為mydf id 2生成NA，而利斯特的建議可以創建Inf。 首先，我將您的日期（以字符形式）轉換為Date對象。然后，我將數據按id分組，並減去healthy的第一天來計算時間差（即date_follow_up[event == "healthy"][1] ）從sick的第一天開始（即date_follow_up[event == "sick"][1] ）。 最后，對於不相關的行，我用NA替換了時差。

   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

數據

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")

Answer 3

我們還可以使用data.table 。 將'data.frame'轉換為'data.table'（ setDT(mydf) ），使用as.Date將'date_follow_up'的類更改為Date ，按'id'分組，並通過獲取邏輯和的累加值創建分組變量向量（ event == "healthy" ）， if存在any “病態”事件，我們將獲得第一個“病態”“事件”與第一個“ date_follow_up”（即“健康”）的“ date_follow_up”差異在該特定組中， else返回“ NA”。

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，對於所有不“健康”的“事件”，我們可以將“ foo”更改為“ NA”。

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：在這里，我准備的數據可能對於一個特定的“ id”可能有多個“健康/病假”“事件”。

數據

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")

Answer 4

這是一種方法，但是如果每個ID有多個“健康”事件，則可能需要對其進行調整以變得更加健壯：

        # 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

Answer 5

這是使用dplyr的另一種方法（盡管與以前的解決方案相比要更長一些）

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