Question

拥有带有colnames的data.frame

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"

如何将数字的数字增加1？

预期结果将是

"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非常复杂......是否有比

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

基础regmatches解决方案：

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 2

只要你的模式是＆＃34; nonnumbers_numbers＆＃34;：

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

产地：

 [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 3

使用gsubfn包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，您不需要假设＆＃34; nonnumbers_numbers ＆＃34;上面提到的模式，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 4

你可以试试＆＃34; ore＆＃34;包，你的替换可以是函数，如：

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＆＃34; gsubfn＆＃34;的功能类似。包，但（至少在这种情况下）效率更高。以下是一些基准测试：

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＆＃34; ore＆＃34;方法和＆＃34; gsubfn＆＃34;进近区域相同，它们似乎与基础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

4 个答案: