我有数据列出大量客户的不同产品的开始和结束日期。不同产品的购买间隔可以重叠或有时间间隔：

library(lubridate)
library(Hmisc)
library(dplyr)

user_id <- c(rep(12, 8), rep(33, 5))

start_date <- dmy(Cs(31/10/2010,    18/12/2010, 31/10/2011, 18/12/2011, 27/03/2014, 18/12/2014, 27/03/2015, 18/12/2016, 01/07/1992, 20/08/1993, 28/10/1999, 31/01/2006, 26/08/2016))

end_date <- dmy(Cs(31/10/2011,  18/12/2011, 28/04/2014, 18/12/2014, 27/03/2015, 18/12/2016, 27/03/2016, 18/12/2017,
               01/07/2016,  16/08/2016, 15/11/2012, 28/02/2006, 26/01/2017))

data <- data.frame(user_id, start_date, end_date)

data
   user_id start_date   end_date
1       12 2010-10-31 2011-10-31
2       12 2010-12-18 2011-12-18
3       12 2011-10-31 2014-04-28
4       12 2011-12-18 2014-12-18
5       12 2014-03-27 2015-03-27
6       12 2014-12-18 2016-12-18
7       12 2015-03-27 2016-03-27
8       12 2016-12-18 2017-12-18
9       33 1992-07-01 2016-07-01
10      33 1993-08-20 2016-08-16
11      33 1999-10-28 2012-11-15
12      33 2006-01-31 2006-02-28
13      33 2016-08-26 2017-01-26

我想计算他/她持有任何产品的活跃天数或月份总数.

如果产品总是重叠，那不会有问题，因为那样我就可以简单地采取

data %>% 
group_by(user_id) %>% 
dplyr::summarize(time_diff = max(end_date) - min(start_date))

然而，正如您在用户 33 中看到的，产品并不总是重叠，它们的间隔必须单独添加到所有“重叠”间隔中。

有没有一种快速而优雅的方法来编码它，希望在dplyr?

使用有什么用IRanges and intersect?

library(IRanges)
data %>% 
  group_by(user_id) %>% 
  summarise(days_held=sum(width(reduce(IRanges(as.numeric(start_date), as.numeric(end_date)))))) 
# A tibble: 2 × 2
  user_id active_days
    <dbl>       <int>
1      12        2606
2      33        8967

这里是使用 Nathan Wert 的基准big_data。 ORange 方法似乎更快一些。

my_result <- function(x) {
x %>% 
    group_by(user_id) %>% 
    summarise(days_held=sum(width(reduce(IRanges(as.numeric(start_date), as.numeric(end_date)))))) 
}


library(microbenchmark)
microbenchmark(
  a <- my_result(big_data),
  b <- my_answer(big_data), times=2
)
Unit: seconds
                     expr      min       lq     mean   median       uq      max neval cld
 a <- my_result(big_data) 14.97008 14.97008 14.98896 14.98896 15.00783 15.00783     2  a 
 b <- my_answer(big_data) 17.59373 17.59373 17.76257 17.76257 17.93140 17.93140     2   b

all.equal(a, b)
[1] TRUE

Edit

为了可视化范围，您还可以绘制数据......

library(Gviz)
library(GenomicRanges)
a <- sapply(split(data, data$user_id), function(x) {
  AnnotationTrack(start = as.numeric(x$start_date), end = as.numeric(x$end_date),
                  chromosome = "chrNA", stacking = "full", name = as.character(unique(x$user_id)))
})
plotTracks(trackList = a)