获取R中组的第一个日期和上一个组的最后一个日期之间的日期差异答案

【问题标题】：Get date difference between first date of a group and last date of previous group in R获取R中组的第一个日期和上一个组的最后一个日期之间的日期差异
【发布时间】：2018-11-27 12:22:11
【问题描述】：

我有一个看起来像这样的数据集：

x = data.frame(id = c("A","A","A","A","B","B","B","B"), group = c(1,1,2,2,3,3,4,4),
               date1 = c("25/03/2017",  "26/03/2017","03/04/2017","04/04/2017",
                         "04/05/2017","26/08/2017","28/08/2017","30/08/2017"),    
               date2 = c("26/03/2017","29/03/2017","04/04/2017","04/05/2017",
                         "18/05/2017","28/08/2017","29/08/2017","31/08/2017")
                )
> x
  id group      date1      date2
1  A     1 25/03/2017 26/03/2017
2  A     1 26/03/2017 29/03/2017
3  A     2 03/04/2017 04/04/2017
4  A     2 04/04/2017 04/05/2017
5  B     3 04/05/2017 18/05/2017
6  B     3 26/08/2017 28/08/2017
7  B     4 28/08/2017 29/08/2017
8  B     4 30/08/2017 31/08/2017

我想做的是让每个人获取第二组 date1 中的第一个日期和前一组 date2 中的最后一个日期的日期差异。因此，例如，id = A 的人，我想获得“03/04/2017”和“29/03/2017”的天数差异。患者 B 也是如此。我每个人都有多个组。我想最终得到一个这样的数据集：

y = data.frame(id = c("A","A","B","B"), group = c(1,2,3,4),
               date1 = c("26/03/2017","03/04/2017","26/08/2017","28/08/2017"),    
               date2 = c("29/03/2017","04/04/2017","28/08/2017","29/08/2017"),
               datediff = c(NA,5,NA,0)
              ) 
> y
  id group      date1      date2 datediff
1  A     1 26/03/2017 29/03/2017       NA
2  A     2 03/04/2017 04/04/2017        5
3  B     3 26/08/2017 28/08/2017       NA
4  B     4 28/08/2017 29/08/2017        0

为此，我环顾四周，发现并回答了在同一组中减去第一个和最后一个观察结果，但对不同组的最后一个和第一个观察结果一无所知。任何帮助将非常感激。谢谢。

【问题讨论】：

不清楚您希望解决方案有多通用。您是否总是每个 id 有 2 个组，每个组有 2 行？
我并不总是每个 id 有 2 个组，也不总是有 2 个 id。这只是我的数据集的一个示例，用于说明。我提到我在 id 中有多个组，但也许我应该更清楚。我希望解决方案尽可能通用。

标签： r dplyr data-manipulation lubridate

【解决方案1】：

这是一种更通用的方法，每个 id 应该使用 3+ 组和/或每组 3+ 行：

library(dplyr)
library(lubridate)

# update dates (if needed)
x = x %>% mutate_at(vars(matches("date")), dmy)

# get appropriate rows based on first group 
x1 = x %>%
  group_by(id) %>%
  filter(group == min(group)) %>%
  filter(date1 == max(date1)) %>%
  ungroup()

# get appropriate rows based on last group 
x2 = x %>%
  group_by(id) %>%
  filter(group == max(group)) %>%
  filter(date2 == min(date2)) %>%
  ungroup()

# combine datasets and calculate date difference
x1 %>%
  bind_rows(x2) %>%
  arrange(id, group) %>%
  group_by(id) %>%
  mutate(datediff = as.numeric(date1 - lag(date2))) %>%
  ungroup()

# # A tibble: 4 x 5
#   id    group date1      date2      datediff
#   <fct> <dbl> <date>     <date>        <dbl>
# 1 A         1 2017-03-26 2017-03-29       NA
# 2 A         2 2017-04-03 2017-04-04        5
# 3 B         3 2017-08-26 2017-08-28       NA
# 4 B         4 2017-08-28 2017-08-29        0

【讨论】：

【解决方案2】：

使用lubridate::dmy 解析您的字符串日期。然后您可以使用dplyr 计算date1 与date2 的滞后值之间的差异。最后，过滤那些代表新组的行。

library(dplyr)
library(lubridate)
x = data.frame(id = c("A","A","A","A","B","B","B","B"), group = c(1,1,2,2,3,3,4,4),
               date1 = dmy(c("25/03/2017",  "26/03/2017","03/04/2017","04/04/2017",
                         "04/05/2017","26/08/2017","28/08/2017","30/08/2017")),    
               date2 = dmy(c("26/03/2017","29/03/2017","04/04/2017","04/05/2017",
                         "18/05/2017","28/08/2017","29/08/2017","31/08/2017"))
)



x %>%
  group_by(id) %>%
  filter(group != lag(group) | group != lead(group)) %>%
  mutate(diff = date1 - lag(date2)) %>%
  ungroup()



# A tibble: 4 x 5
  id    group date1      date2      diff     
  <fct> <dbl> <date>     <date>     <time>   
1 A         1 2017-03-26 2017-03-29 NA days  
2 A         2 2017-04-03 2017-04-04 " 5 days"
3 B         3 2017-08-26 2017-08-28 NA days  
4 B         4 2017-08-28 2017-08-29 " 0 days"

如果您想要数字输出，请使用mutate(diff = as.numeric(date1 - lag(date2)))。只要您的数据已排序 (x <- x[with(x, order(id, group)), ])，无论有多少人和组，它都应该可以正常工作。

【讨论】：