在R'；s strplit（）函数：添加索引值有什么区别？

我正在学习R，我从一个练习中得到了以下代码，目标是编写代码，计算

rquote

中第一个u之前的R的数量

rquote <- "r's internals are irrefutably intriguing"
chars <- strsplit(rquote, split = "")[[1]]

# Initialize rcount
rcount <- 0

# Finish the for loop
for (char in chars) {
  if (char == "r"){
      rcount = rcount + 1
  }
  if (char == "u"){
      break
  }
  
}

# Print out rcount
print(rcount)

---

当您使用或不使用

[[1]]

时，

字符

是不同的

使用

[[1]]

时：

chars1 <- strsplit(rquote, split = "")[[1]]
chars1
# [1] "r" "'" "s" " " "i" "n" "t" "e" "r" "n" "a" "l" "s" " " "a" "r" "e" " " "i" "r" "r" "e"
#[23] "f" "u" "t" "a" "b" "l" "y" " " "i" "n" "t" "r" "i" "g" "u" "i" "n" "g"

length(chars1)
#[1] 40

class(chars1)
#[1] "character"

chars2 <- strsplit(rquote, split = "")
chars2
#[[1]] <- Don't forget this [[1]], it is important
# [1] "r" "'" "s" " " "i" "n" "t" "e" "r" "n" "a" "l" "s" " " "a" "r" "e" " " "i" "r" "r" "e"
#[23] "f" "u" "t" "a" "b" "l" "y" " " "i" "n" "t" "r" "i" "g" "u" "i" "n" "g"

length(chars2)
#[1] 1

class(chars2)
#[1] "list"

strsplit

返回一个列表，因此添加

[[1]]

将列表添加到字符向量。当您迭代任何对象时，您迭代的对象长度在

chars1

情况下为40，在

chars2

情况下为1

chars2 <- strsplit(rquote, split = "")
chars2
#[[1]] <- Don't forget this [[1]], it is important
# [1] "r" "'" "s" " " "i" "n" "t" "e" "r" "n" "a" "l" "s" " " "a" "r" "e" " " "i" "r" "r" "e"
#[23] "f" "u" "t" "a" "b" "l" "y" " " "i" "n" "t" "r" "i" "g" "u" "i" "n" "g"

length(chars2)
#[1] 1

class(chars2)
#[1] "list"