字符串拆分后访问第n个元素

1) 数据帧 转换为数据帧,然后很容易拾取列或列的子集:

DF <- read.table(text = string, sep = ",", as.is = TRUE)

DF[[1]]
## [1] "A" "B" "A"

DF[[3]]
## [1] "some text"       "some other text" "yet another one"

DF[-1]
##   V2              V3  V4
## 1  1       some text 200
## 2  2 some other text 300
## 3  3 yet another one 100

DF[2:3]
##   V2              V3
## 1  1       some text
## 2  2 some other text
## 3  3 yet another one

2) data.table::transpose data.table包具有转换列表的功能,以便 stringt stringt[[3]] 是第三个场的向量,例如,与(1)类似。更紧凑的是数据 tstrsplit 以下@Henrik提及或同一包裹的 fread

library(data.table)

stringt <- transpose(strsplit(string, ","))

# or
stringt <- tstrsplit(string, ",")

stringt[[1]]
## [1] "A" "B" "A"

stringt[[3]]
## [1] "some text"       "some other text" "yet another one"

stringt[-1]
## [[1]]
## [1] "1" "2" "3"
##
## [[2]]
## [1] "some text"       "some other text" "yet another one"
##
## [[3]]
## [1] "200" "300" "100"

stringt[2:3]
## [[1]]
## [1] "1" "2" "3"
##
## [[2]]
## [1] "some text"       "some other text" "yet another one"

purrr也有一个 transpose

library(purrr)
transpose(strsplit(string, ","))

生成列表列表,而不是字符向量列表。

一个选择是使用 word stringr 具有 sep 论点

library(stringr)
word(string, 1, sep = ",")
#[1] "A" "B" "A"

word(string, 3, sep = ",")
#[1] "some text"       "some other text" "yet another one"

最糟糕的是,我发现了另一个在baser中使用正则表达式的选项。

#Get 1st element
sub("(?:[^,],){0}([^,]*).*", "\\1",string)
#[1] "A" "B" "A"

#Get 3rd element
sub("(?:[^,],){2}([^,]*).*", "\\1",string)
#[1] "some text"       "some other text" "yet another one"

这里有两组比赛。第一个匹配任何不是逗号的字符,后面跟一个逗号 n ?: )而第二组被捕获并返回。还要注意括号中的数字( {} )必须比我们想要的词少一个。所以 {0} {2} 返回第三个单词。

基准

string <- c("A,1,some text,200","B,2,some other text,300","A,3,yet another one,100")
string <- rep(string, 1e5)

library(microbenchmark)
microbenchmark(
  tmfmnk_sapply = sapply(strsplit(string, ","), function(x) x[1]),
  tmfmnk_tstrsplit = tstrsplit(string, ",")[[1]],
  avid_useR_sapply = sapply(strsplit(string, ","), '[', 1),
  avid_useR_str_split = str_split(string, ",", simplify = TRUE)[,1],
  Ronak_Shah_word = word(string, 1, sep = ","),
  Ronak_Shah_sub = sub("(?:[^,],){0}([^,]*).*", "\\1",string),
  G_Grothendieck ={DF <- read.table(text = string, sep = ",",as.is = TRUE);DF[[1]]},
  times = 5
)
#Unit: milliseconds
#               expr     min      lq    mean  median      uq     max neval
#      tmfmnk_sapply 1629.69 1641.61 2128.14 1834.99 1893.43 3640.96     5
#   tmfmnk_tstrsplit 1269.94 1283.79 1286.29 1286.68 1290.76 1300.30     5
#   avid_useR_sapply 1445.40 1447.64 1555.76 1498.14 1609.52 1778.13     5
#avid_useR_str_split  324.68  332.28  332.30  333.97  334.01  336.54     5
#    Ronak_Shah_word 6571.29 6810.92 6956.20 6930.86 7217.26 7250.69     5
#     Ronak_Shah_sub  349.76  354.77  356.91  358.91  359.17  361.94     5
#     G_Grothendieck  354.93  358.24  364.43  362.24  367.79  378.94     5

我还没有包括Christoph的解决方案,因为我不清楚它如何适用于变量 N

我们可以将OP的代码简化为:

sapply(strsplit(string, ","), '[', 1)
# [1] "A" "B" "A"

sapply(strsplit(string, ","), '[', 3)
# [1] "some text"       "some other text" "yet another one"

还有 stringr::str_split simplify = TRUE ,我们可以直接为列编制索引,因为输出将是一个矩阵:

library(stringr)
str_split(string, ",", simplify = TRUE)[,1]
# [1] "A" "B" "A"

str_split(string, ",", simplify = TRUE)[,3]
# [1] "some text"       "some other text" "yet another one"

sapply() :

sapply(strsplit(string, ","), function(x) x[1])

[1] "A" "B" "A"

sapply(strsplit(string, ","), function(x) x[3])

[1] "some text"       "some other text" "yet another one"

或者使用另一种可能性 tstrsplit data.table :

tstrsplit(string, ",")[[1]]

[1] "A" "B" "A"

不同解决方案的基准:

library(microbenchmark)
microbenchmark(
 tmfmnk_sapply = sapply(strsplit(string, ","), function(x) x[1]),
 tmfmnk_tstrsplit = tstrsplit(string, ",")[[1]],
 avid_useR_sapply = sapply(strsplit(string, ","), '[', 1),
 avid_useR_str_split = str_split(string, ",", simplify = TRUE)[,1],
 Ronak_Shah = word(string, 1, sep = ","),
 times = 5
)

                expr     min      lq     mean  median      uq     max neval cld
       tmfmnk_sapply  34.543  36.395  45.8782  47.150  48.540  62.763     5  a 
    tmfmnk_tstrsplit  33.072  33.554  39.1166  35.012  36.116  57.829     5  a 
    avid_useR_sapply  39.612  45.292  61.1936  46.730  47.398 126.936     5  a 
 avid_useR_str_split  27.313  34.095  49.3412  43.834  43.977  97.487     5  a 
          Ronak_Shah 146.875 147.277 199.4978 162.995 218.322 322.020     5   b

复制“字符串”上的基准:

string <- rep(string, 1e5)

microbenchmark(
 tmfmnk_sapply = sapply(strsplit(string, ","), function(x) x[1]),
 tmfmnk_tstrsplit = tstrsplit(string, ",")[[1]],
 avid_useR_sapply = sapply(strsplit(string, ","), '[', 1),
 avid_useR_str_split = str_split(string, ",", simplify = TRUE)[,1],
 Ronak_Shah = word(string, 1, sep = ","),
 Christoph = regmatches(string, regexpr("^[^,]", string)),
 times = 5
)

                expr       min        lq      mean    median        uq       max neval
       tmfmnk_sapply 1529.8955 1608.2909 1926.7776 1820.0443 2105.9736 2569.6836     5
    tmfmnk_tstrsplit 1277.8712 1281.0371 1482.4520 1314.0074 1599.7686 1939.5757     5
    avid_useR_sapply 1428.7175 1470.9002 1487.5425 1483.1127 1521.3735 1533.6087     5
 avid_useR_str_split  306.2633  316.7539  360.8785  334.8516  335.5375  510.9863     5
          Ronak_Shah 5541.6199 5657.3593 5955.9653 6068.1067 6166.7249 6346.0157     5
           Christoph  231.0496  244.1049  383.9702  246.0421  273.2877  925.3667     5

可以使用base R和regepr完成:

regmatches(string, regexpr("^[^,]", string))
[1] "A" "B" "A"
regmatches(string, regexpr("[^,]*$", string))
[1] "200" "300" "100"
regmatches(string, regexpr("[^,]*,[^,]*$", string))
[1] "some text,200"       "some other text,300" "yet another one,100"