分割和替换数据框中的字符变量R

Question

我有一个数据帧，其中包含多个不同长度的字符变量，我想将每个变量转换为列表，每个元素包含每个单词，并用空格分隔。

说我的数据是这样的：

char <- c("This is a string of text", "So is this") 
char2 <- c("Text is pretty sweet", "Bet you wish you had text like this") 

df <- data.frame(char, char2) 

# Convert factors to character 
df <- lapply(df, as.character) 

> df 
$char 
[1] "This is a string of text" "So is this"    

$char2 
[1] "Text is pretty sweet"    "Bet you wish you had text like this" 

现在我可以用strsplit（）由单个单词的分割每一列：

df <- transform(df, "char" = strsplit(df[, "char"], " ")) 
> df$char 
[[1]] 
[1] "This" "is"  "a"  "string" "of"  "text" 

[[2]] 
[1] "So" "is" "this" 

我想要做的就是创建一个循环或功能，这将允许我一次为这两列执行此操作，例如：

for (i in colnames(df) { 
    df <- transform(df, i = strsplit(df[, i], " ")) 
} 

但是，此p roduces错误：

Error in data.frame(list(char = c("This is a string of text", "So is this", : 
    arguments imply differing number of rows: 6, 8 

我也曾尝试：

splitter <- function(colname) { 
    df <- transform(df, colname = strsplit(df[, colname], " ")) 
} 

分路器（colnames（DF））

还告诉我：

Error in strsplit(df[, colname], " ") : non-character argument 

我很困惑，为什么对变换的调用适用于单个列，但不适用于在循环或函数中应用。任何帮助将非常感激！

Answer 1

我不transform

char <- c("This is a string of text", "So is this") 
char2 <- c("Text is pretty sweet", "Bet you wish you had text like this") 
df <- data.frame(char, char2) 
# Convert factors to character 
df <- lapply(df, as.character) 

所需的输出，我把

lapply(df, strsplit, split= " ") 

要获得

$char 
$char[[1]] 
[1] "This" "is"  "a"  "string" "of"  "text" 

$char[[2]] 
[1] "So" "is" "this" 


$char2 
$char2[[1]] 
[1] "Text" "is"  "pretty" "sweet" 

$char2[[2]] 
[1] "Bet" "you" "wish" "you" "had" "text" "like" "this" 

而且亚历克斯提到：从你的代码中的第lapply df <- lapply(df, as.character)能通过将df <- data.frame(char, char2)更改为来消除