将 data.frame 的 colnames 增加 1

Question

有一个 data.frame 和 colnames

nam <- c("a", paste0("a_", seq(12)))
"a" "a_1" "a_2" "a_3" "a_4" "a_5" "a_6" "a_7" "a_8" "a_9" "a_10" "a_11" "a_12"

我怎样才能将带数字的名字的数字加 1？

预期的结果是

"a" "a_2" "a_3" "a_4" "a_5" "a_6" "a_7" "a_8" "a_9" "a_10" "a_11" "a_12" "a_13"

到目前为止我的解决方案看起来很复杂...有没有比

更简单的方法

increment_names <- function(nam){
  where <- regexpr("\d", nam)
  ind <- which(where > 0)
  increment <- as.numeric(substring(nam[ind], where[ind])) + 1
  substring(nam[ind], where[ind]) <- as.character(increment)
  nam
}

> increment_names(nam)
 [1] "a" "a_2" "a_3" "a_4" "a_5" "a_6" "a_7" "a_8" "a_9" "a_10" "a_11" "a_12" "a_13"

Answer 1

只要你的模式是 "nonnumbers_numbers":

nums <- as.numeric(gsub("[^0-9]", "", nam))
nam[!is.na(nums)] <- paste0(gsub("[0-9]", "", nam), nums + 1)[!is.na(nums)]

生产：

 [1] "a"    "a_2"  "a_3"  "a_4"  "a_5"  "a_6"  "a_7"  "a_8"  "a_9"  "a_10" "a_11" "a_12"
 [13] "a_13"

Answer 2

使用 gsubfn 包你可以做一些简单的事情

library(gsubfn) 
gsubfn("\d+", function(x) as.numeric(x) + 1, nam)
## [1] "a"    "a_2"  "a_3"  "a_4"  "a_5"  "a_6"  "a_7"  "a_8"  "a_9"  "a_10" "a_11" "a_12" "a_13"

这适用于任何模式，您无需假定上述“nonnumbers_numbers”模式, 例如

(nam <- c("a", paste0(seq(12), "_a")))
## [1] "a"    "1_a"  "2_a"  "3_a"  "4_a"  "5_a"  "6_a"  "7_a"  "8_a"  "9_a"  "10_a" "11_a" "12_a"
gsubfn("\d+", function(x) as.numeric(x) + 1, nam)
## [1] "a"    "2_a"  "3_a"  "4_a"  "5_a"  "6_a"  "7_a"  "8_a"  "9_a"  "10_a" "11_a" "12_a" "13_a"

Answer 3

基础regmatches解法：

r <- regexpr("\d+", nam)
regmatches(nam, r) <- as.numeric(regmatches(nam, r)) + 1
nam
# [1] "a"    "a_2"  "a_3"  "a_4"  "a_5"  "a_6"  "a_7"  "a_8"  ...

Answer 4

你可以试试 "ore" 包，你的替换可以是函数，像这样：

nam <- c("a", paste0("a_", seq(12)))
nam
library(ore)
ore.subst("-?\d+", function(x) as.numeric(x) + 1, nam, all = TRUE)
#  [1] "a"    "a_2"  "a_3"  "a_4"  "a_5"  "a_6"  "a_7"  "a_8"  "a_9" 
# [10] "a_10" "a_11" "a_12" "a_13"

这与 "gsubfn" 包的功能相似，但（至少在这种情况下）效率更高。以下是一些基准：

library(stringi)
set.seed(1)
nam <- stri_rand_strings(10000, 5, pattern = "[A-J0-9]")

f_ORE <- function(invec = nam) {
  ore.subst("-?\d+", function(x) as.numeric(x) + 1, invec, all = TRUE)
} 

f_GSUBFN <- function(invec = nam) {
  gsubfn("\d+", function(x) as.numeric(x) + 1, invec)
}

f_BASE <- function(invec = nam) {
  r <- regexpr("\d+", invec)
  regmatches(invec, r) <- as.numeric(regmatches(invec, r))+1
  invec
}

system.time(f_GSUBFN())
#    user  system elapsed 
#    5.48    0.01    5.50 

library(microbenchmark)
microbenchmark(f_BASE(), f_ORE())
# Unit: milliseconds
#      expr       min        lq      mean    median        uq      max neval
#  f_BASE() 141.79743 149.58914 161.49041 152.81038 162.10550 357.6483   100
#   f_ORE()  57.35309  59.58433  65.84678  60.92218  68.40062 116.7714   100

请注意，虽然 "ore" 方法和 "gsubfn" 方法区域相同，但它们似乎与基本 R 方法略有不同。

考虑：

> identical(f_ORE(), f_GSUBFN())
[1] TRUE

## Edge case...
> nam[988]
[1] "0G019"
> f_ORE()[988]     ## 019 becomes 20 (without the leading zero)
[1] "1G20"
> f_GSUBFN()[988]  ## Same
[1] "1G20"
> f_BASE()[988]    ## This seems off...
[1] "1G019"

将 data.frame 的 colnames 增加 1

increment colnames of a data.frame by 1

regex

r

string

dataframe

increment