22

我在一个列表中有一些数据，我需要寻找连续运行的整数（我的大脑认为rle但不知道如何在这里使用它）。

查看数据集并解释我所追求的更容易。

这是数据视图：

$greg
 [1]  7  8  9 10 11 20 21 22 23 24 30 31 32 33 49

$researcher
[1] 42 43 44 45 46 47 48

$sally
 [1] 25 26 27 28 29 37 38 39 40 41

$sam
 [1]  1  2  3  4  5  6 16 17 18 19 34 35 36

$teacher
[1] 12 13 14 15

期望的输出：

$greg
 [1]  7:11, 20:24, 30:33, 49

$researcher
 [1] 42:48

$sally
 [1] 25:29, 37:41

$sam
 [1]  1:6, 16:19 34:36

$teacher
 [1] 12:15

使用基本包如何用最高和最低之间的冒号和非连续部分之间的逗号替换连续跨度？请注意，数据从整数向量列表到字符向量列表。

MWE数据：

z <- structure(list(greg = c(7L, 8L, 9L, 10L, 11L, 20L, 21L, 22L, 
    23L, 24L, 30L, 31L, 32L, 33L, 49L), researcher = 42:48, sally = c(25L, 
    26L, 27L, 28L, 29L, 37L, 38L, 39L, 40L, 41L), sam = c(1L, 2L, 
    3L, 4L, 5L, 6L, 16L, 17L, 18L, 19L, 34L, 35L, 36L), teacher = 12:15), .Names = c("greg", 
    "researcher", "sally", "sam", "teacher"))
4

6 回答 6

12

我认为diff是解决方案。您可能需要一些额外的摆弄来处理单例，但是：

lapply(z, function(x) {
  diffs <- c(1, diff(x))
  start_indexes <- c(1, which(diffs > 1))
  end_indexes <- c(start_indexes - 1, length(x))
  coloned <- paste(x[start_indexes], x[end_indexes], sep=":")
  paste0(coloned, collapse=", ")
})

$greg
[1] "7:11, 20:24, 30:33, 49:49"

$researcher
[1] "42:48"

$sally
[1] "25:29, 37:41"

$sam
[1] "1:6, 16:19, 34:36"

$teacher
[1] "12:15"
于 2013-02-14T05:52:56.047 回答
7

使用IRanges：

require(IRanges)
lapply(z, function(x) {
    t <- as.data.frame(reduce(IRanges(x,x)))[,1:2]
    apply(t, 1, function(x) paste(unique(x), collapse=":"))
})

# $greg
# [1] "7:11"  "20:24" "30:33" "49"   
# 
# $researcher
# [1] "42:48"
# 
# $sally
# [1] "25:29" "37:41"
# 
# $sam
# [1] "1:6"   "16:19" "34:36"
# 
# $teacher
# [1] "12:15"
于 2013-02-14T06:01:10.200 回答
6

这是使用diff和tapply返回字符向量的尝试

runs <- lapply(z, function(x) {
  z <- which(diff(x)!=1); 
  results <- x[sort(unique(c(1,length(x), z,z+1)))]
  lr <- length(results)
  collapse <- rep(seq_len(ceiling(lr/2)),each = 2, length.out = lr)
  as.vector(tapply(results, collapse, paste, collapse = ':'))
  })

runs
$greg
[1] "7:11"  "20:24" "30:33" "49"   

$researcher
[1] "42:48"

$sally
[1] "25:29" "37:41"

$sam
[1] "1:6"   "16:19" "34:36"

$teacher
[1] "12:15"
于 2013-02-14T06:07:08.177 回答
5

我有一个与 Marius 非常相似的解决方案，他的作品和我的一样，但机制略有不同，所以我想我不妨把它贴出来：

findIntRuns <- function(run){
  rundiff <- c(1, diff(run))
  difflist <- split(run, cumsum(rundiff!=1))
  unname(sapply(difflist, function(x){
    if(length(x) == 1) as.character(x) else paste0(x[1], ":", x[length(x)])
  }))
}

lapply(z, findIntRuns)

产生：

$greg
[1] "7:11"  "20:24" "30:33" "49"   

$researcher
[1] "42:48"

$sally
[1] "25:29" "37:41"

$sam
[1] "1:6"   "16:19" "34:36"

$teacher
[1] "12:15"
于 2013-02-14T06:06:08.240 回答
5

另一个简短的解决方案lapplyand tapply：

lapply(z, function(x)
  unname(tapply(x, c(0, cumsum(diff(x) != 1)), FUN = function(y) 
    paste(unique(range(y)), collapse = ":")
  ))
)

结果：

$greg
[1] "7:11"  "20:24" "30:33" "49"   

$researcher
[1] "42:48"

$sally
[1] "25:29" "37:41"

$sam
[1] "1:6"   "16:19" "34:36"

$teacher
[1] "12:15"
于 2013-02-14T07:55:07.190 回答
3

派对迟到了，但这里有一个deparse基础的单线：

lapply(z,function(x) paste(sapply(split(x,cumsum(c(1,diff(x)-1))),deparse),collapse=", "))
$greg
[1] "7:11, 20:24, 30:33, 49L"

$researcher
[1] "42:48"

$sally
[1] "25:29, 37:41"

$sam
[1] "1:6, 16:19, 34:36"

$teacher
[1] "12:15"
于 2013-03-29T11:13:26.820 回答