概述

Question

问题：

我有一个字符串向量foo：

>foo = c("1x2","3x4","5x6","7x8","9x10")

我将单个字符串拆分为“x”并将结果粘贴到goo：

>goo = strsplit(foo, "x")
>goo
[[1]]
[1] "1" "2"

[[2]]
[1] "3" "4"

[[3]]
[1] "5" "6"

[[4]]
[1] "7" "8"

[[5]]
[1] "9"  "10"

如何从此列表中提取第一个和第二个“列”？（我想要（1,3,5,7,9）和（2,4,6,8,10））

Answer 1

使用sapply使用“[[”：

连续'提取'

 sapply(goo, "[[" , 1)
[1] "1" "3" "5" "7" "9"

我一直认为这应该是结果，但我可能不理解这些问题。

Answer 2

> result <-  do.call(rbind, goo)
> result
     [,1] [,2]
[1,] "1"  "2" 
[2,] "3"  "4" 
[3,] "5"  "6" 
[4,] "7"  "8" 
[5,] "9"  "10"
> result[, 1] # column 1
[1] "1" "3" "5" "7" "9"
> result[, 2] # column 2
[1] "2"  "4"  "6"  "8"  "10"

Answer 3

最简单的方法是将结果包装在sapply语句

中

odd_results <- sapply(goo, function(x) x[1])

和

even_results <- sapply(goo, function(x) x[2])

Answer 4

概述

您需要按位置指定列表元素的提取。

# load data
foo <- c("1x2","3x4","5x6","7x8","9x10")

# split foo by 'x'
foo.list <- strsplit( x = foo, split = "x", fixed = TRUE )

# store the first and second elements of each list in a data frame
foo.df <-
data.frame(
    First_Element = unlist( lapply( X = foo.list, FUN = "[[", FUN.VALUE = 1 ) )
    , Second_Element = unlist( lapply( X = foo.list, FUN = "[[", FUN.VALUE = 2 ) )
  )

Answer 5

为了多样化，您可以使用

goo <- unlist(strsplit(foo, "x"))

a <- goo[seq(1, length(goo), 2)]
b <- goo[seq(2, length(goo), 2)]

哪个收益

[1] "1" "3" "5" "7" "9"

和

[1] "2"  "4"  "6"  "8"  "10"

分别

如何提取R的strsplit（）的结果

5 个答案:

概述