如何从 R 中的字符串中提取数字？

Question

names(score)
 [1] "(Intercept)"              "aado2_calc(20,180]"       "aado2_calc(360,460]"     
 [4] "aado2_calc(460,629]"      "albumin[1,1.8]"           "albumin(1.8,2.2]"        
 [7] "albumin(2.2,2.8]"         "aniongap(15,18]"          "aniongap(18,20]"         
[10] "aniongap(20,22]"          "aniongap(22,25]"          "aniongap(25,49]"

我想提取括号内的两个数字（括号外的数字不需要）有“（”或“[”。第一个数字将分配给一个对象"low"和仅次于 "high"。

Answer 1

您可以使用 readr 包和函数 parse_number 以方便使用。要获得更多功能，您需要使用 r 中的基本正则表达式函数或 stringi

之类的包

Answer 2

scorenames <- c(
  "(Intercept)"              ,"aado2_calc(20,180]"       ,"aado2_calc(360,460]"     
 ,"aado2_calc(460,629]"      ,"albumin[1,1.8]"           ,"albumin(1.8,2.2]"        
 ,"albumin(2.2,2.8]"         ,"aniongap(15,18]"          ,"aniongap(18,20]"         
 ,"aniongap(20,22]"          ,"aniongap(22,25]"          ,"aniongap(25,49]"
)

第一步可能是提取 "parens" 分隔符内的所有内容（包括 ()、[] 和逗号 ,）。

mat <- regmatches(scorenames,
                  gregexpr("(?<=[\[\(,])[0-9.]+(?=[\]\),])", scorenames, perl = TRUE))
str(mat)
# List of 12
#  $ : chr(0) 
#  $ : chr [1:2] "20" "180"
#  $ : chr [1:2] "360" "460"
#  $ : chr [1:2] "460" "629"
#  $ : chr [1:2] "1" "1.8"
#  $ : chr [1:2] "1.8" "2.2"
#  $ : chr [1:2] "2.2" "2.8"
#  $ : chr [1:2] "15" "18"
#  $ : chr [1:2] "18" "20"
#  $ : chr [1:2] "20" "22"
#  $ : chr [1:2] "22" "25"
#  $ : chr [1:2] "25" "49"

从这里，我们可以看出 (1) 第一个有问题（毫不奇怪，你需要在这里弄清楚你想要什么），以及 (2) 其余的看起来是正确的。

这是处理此列表的一种粗略方法。这是非常信任和天真的......你应该添加检查以确保列表的长度为 2，所有内容都正确转换（可能在 tryCatch 中），等等

newnames <- lapply(mat, function(m) {
  if (! length(m)) return(list(low = NA, high = NA))
  setNames(as.list(as.numeric(m)), nm = c("low", "high"))
})
str(newnames)
# List of 12
#  $ :List of 2
#   ..$ low : logi NA
#   ..$ high: logi NA
#  $ :List of 2
#   ..$ low : num 20
#   ..$ high: num 180
#  $ :List of 2
#   ..$ low : num 360
#   ..$ high: num 460
# ...snip...

您可以通过以下方式将其变成 data.frame：

head(do.call(rbind.data.frame, newnames))
#     low  high
# 1    NA    NA
# 2  20.0 180.0
# 3 360.0 460.0
# 4 460.0 629.0
# 5   1.0   1.8
# 6   1.8   2.2

Answer 3

就像@jake-kaupp 说的-使用stringi :) 如您所见，stringi 解决方案更短、更容易理解并且更快-最多30 次！

简答：

arr <- stri_extract_all_regex(x, "(?<=[\[\(,])[0-9.]+(?=[\]\),])", simplify = NA)
data.frame(low = as.numeric(arr[,1]), high = as.numeric(arr[,2]))

长答案：

require(stringi)
require(microbenchmark)

grepFun <- function(x){
  mat <- regmatches(x,
                gregexpr("(?<=[\[\(,])[0-9.]+(?=[\]\),])", x, perl = TRUE))
  newnames <- lapply(mat, function(m) {
    if (! length(m)) return(list(low = NA, high = NA))
      setNames(as.list(as.numeric(m)), nm = c("low", "high"))
  })
  do.call(rbind.data.frame, newnames)
}

striFun <- function(x){
  arr <- stri_extract_all_regex(x, "(?<=[\[\(,])[0-9.]+(?=[\]\),])", simplify = NA)
  data.frame(low = as.numeric(arr[,1]), high = as.numeric(arr[,2]))
}

# both functions work the same
grepFun(scorenames)
     low  high
1     NA    NA
2   20.0 180.0
3  360.0 460.0
4  460.0 629.0
...
12  25.0  49.0
striFun(scorenames)
     low  high
1     NA    NA
2   20.0 180.0
3  360.0 460.0
4  460.0 629.0
...
12  25.0  49.0

# generating more complicated vector 
n <- 10000
x <- stri_paste(stri_rand_strings(n, length = 1:10), sample(c("(","["),n,TRUE),  
  sample(1000,n,TRUE), ",", sample(1000,n,TRUE),    sample(c(")","]"), n, TRUE))
head(x) # check first elements
[1] "O[68,434]"      "Ql[783,151)"    "Zk0(773,60)"    "ETfV(446,518]"  "Xixbr(576,855)" "G6QnHu(92,955)"

#short test using new data
grepFun(x[1:6])
  low high
1  68  434
2 783  151
3 773   60
4 446  518
5 576  855
6  92  955
striFun(x[1:6])
  low high
1  68  434
2 783  151
3 773   60
4 446  518
5 576  855
6  92  955

#and some benchmark to prove performance
microbenchmark(grepFun(x), striFun(x))
Unit: milliseconds
       expr       min        lq      mean    median        uq       max neval
 grepFun(x) 330.27733 366.09306 416.56330 406.08914 465.29829 568.15250   100
 striFun(x)  11.57449  11.97825  13.38157  12.46927  13.67699  25.97455   100

如何从 R 中的字符串中提取数字？

how can I extract numbers from a string in R?

string

numbers

r

stringr