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]"

我想在括号内提取两个数字（不需要括号外的数字）并且有“（”或“[”。第一个数字将被分配给一个对象“低”而第二个数字将分配给“高” “。

Answer 1

您可以使用class MyClass(object): value = 5 object1 = MyClass() object2 = MyClass() print object1.value #This displays 5 print object2.value #This displays 5 print MyClass.value #This displays 5 #So far so good object1.value = 6 print object1.value #This displays 6 print object2.value #This displays 5 print MyClass.value #This displays 5 #Still good, but the confusing part is coming up... MyClass.value = 10 print object1.value #This displays 6. I was expecting it to be 10! print object2.value #This displays 10 print MyClass.value #This displays 10包和函数readr以方便使用。为了获得更多功能，您需要使用r中的基本正则表达式函数或类似parse_number

之类的包。

Answer 2

就像@jake-kaupp所说 - 使用stringi :)正如您所看到的，stringi解决方案更短，更容易理解，速度更快 - 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]))

答案很长：

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

Answer 3

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” - delimiters中的所有内容（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，我们可以看到（1）第一个是有问题的（毫不奇怪，你需要弄清楚你想要什么），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...你应该添加检查以确保列表长度为2，一切都正确转换（可能在tryCatch）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...

您可以将其转换为带有以下内容的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

如何从R中的字符串中提取数字？

3 个答案: