R 中的连续引文编号:如果连续，则用连字符分隔数字 - 如果不是，则添加逗号

Question

我想为 R 中的图形生成连续的引文编号。如果数字是连续的，则应用连字符分隔。否则，数字用逗号分隔。例如，数字1, 2, 3, 5, 6, 8, 9, 10, 11 and 13应显示为1-3,5,6,8-11,13。

这个问题之前已经针对 c# 回答过，我已经编写了一个适用于 R 的函数，但这个函数可以改进。我发布这个问题作为其他可能有类似需求的人的参考。如果您发现R的类似问题(我没有(，请投票关闭，我将删除该问题。

下面的功能不是很优雅，但似乎可以完成这项工作。如何让功能更短更优雅？

x <- c(1,2,3,5,6,8,9,10,11,13)
library(zoo) ## the function requires zoo::na.approx function 
##' @title Generate hyphenated sequential citation from an integer vector
##' @param x integer vector giving citation or page numbers
##' @importFrom zoo na.approx
seq.citation <- function(x) {
## Result if lenght of the integer vector is 1. 
if(length(x) == 1) return(x) else {
## Sort
x <- sort(x)
## Difference
df <- diff(x)
## Index to determine start and end points
ind <- c("start", rep("no", length(df)-1), "end")
ind[which(df > 1)] <- "end"
## Temporary start point vector
sts <- which(ind == "end") + 1
ind[sts[sts < length(ind)]] <- "start"
## Replace the first index element
ind[1] <- "start"
## Replace the last index element, if preceding one is "end"
if(ind[length(ind)-1] == "end") ind[length(ind)] <- "start"
## Groups for comma separation using "start" as the determining value.
grp <- rep(NA, length(x))
grp[which(ind == "start")] <- 1:length(grp[which(ind == "start")])
grp <- zoo::na.approx(grp, method = "constant", rule = 2)
## Split sequences by group
seqs <- split(x, grp)
seqs <- lapply(seqs, function(k) {
if(length(k) == 1) k else {
if(length(k) == 2) paste(k[1], k[2], sep = ",") else {
paste(k[1], k[length(k)], sep = "-")
}}
})
## Result
return(do.call("paste", c(seqs, sep = ",")))
}
}
seq.citation(x)
# [1] "1-3,5,6,8-11,13"

Answer 1

您可以使用tapply通过基本 R 轻松完成此操作，

paste(tapply(x, cumsum(c(1, diff(x) != 1)), function(i) 
ifelse(length(i) > 2, paste0(head(i, 1), '-', tail(i, 1)), 
paste(i, collapse = ','))), collapse = ',')
[1] "1-3,5,6,8-11,13"

Answer 2

这适用于您的示例，并且应该相当通用。

# get run lengths of differences, with max value of 2
r <- rle(c(1, pmin(diff(x), 2)))
# paste selected x values with appropriate separator
res <- paste0(x[c(1, cumsum(r$lengths))], c("-", ",")[r$values], collapse="")
# drop final character, which is a separator
res <- substr(res, 1, nchar(res)-1)

这返回

res
[1] "1-3,5-6,8-11,13"

Answer 3

当然，还有来自"R.utils"包的seqToHumanReadable函数。

library(R.utils)
seqToHumanReadable(x)
# [1] "1-3, 5, 6, 8-11, 13"
seqToHumanReadable(x, tau = 1) ## If you want 5-6 and not 5, 6
# [1] "1-3, 5-6, 8-11, 13"

结果的外观也可以控制：

seqToHumanReadable(x, delimiter = "...", collapse = " | ")
# [1] "1...3 | 5 | 6 | 8...11 | 13"

Answer 4

与 Imo 的答案相比，您可以通过避免ifelse调用并替换一些paste0/paste调用来节省一些计算时间：

paste0(
tapply(x, cumsum(c(1, diff(x) != 1)), function(i){
len <- length(i)
if(len == 1)
i else sprintf(if(len == 2) "%d,%d" else "%d-%d", i[1], i[len])
}), collapse = ",")
#R> [1] "1-3,5,6,8-11,13"

这更快，如下所示：

# check computation time
bench::mark(
new = paste0(
tapply(x, cumsum(c(1, diff(x) != 1)), function(i){
len <- length(i)
if(len == 1)
i else sprintf(if(len == 2) "%d,%d" else "%d-%d", i[1], i[len])
}), collapse = ","),
Imo = paste(tapply(x, cumsum(c(1, diff(x) != 1)), function(i)
ifelse(length(i) > 2, paste0(head(i, 1), '-', tail(i, 1)),
paste(i, collapse = ','))), collapse = ','),
min_time = 1)
#R> # A tibble: 2 x 13
#R>   expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time 
#R>   <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> 
#R> 1 new          81.6µs   85.9µs    11228.    24.1KB     22.8  8848    18      788ms 
#R> 2 Imo         116.7µs    127µs     7613.    15.8KB     22.4  6123    18      804ms
# same with longer vector
set.seed(1)
x <- sort(sample.int(1e6, 1e5))
bench::mark(
new = paste0(
tapply(x, cumsum(c(1, diff(x) != 1)), function(i){
len <- length(i)
if(len == 1)
i else sprintf(if(len == 2) "%d,%d" else "%d-%d", i[1], i[len])
}), collapse = ","),
Imo = paste(tapply(x, cumsum(c(1, diff(x) != 1)), function(i)
ifelse(length(i) > 2, paste0(head(i, 1), '-', tail(i, 1)),
paste(i, collapse = ','))), collapse = ','),
min_time = 1)
#R> # A tibble: 2 x 13
#R>   expression     min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time 
#R>   <bch:expr> <bch:t> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> 
#R> 1 new          341ms    355ms      2.58    29.5MB     6.88     3     8      1.16s 
#R> 2 Imo          625ms    658ms      1.52    29.2MB    15.2      2    20      1.31s