将标点符号添加到 data.frame r 中的列表和 coreace

Question

我有一个单词和标点符号库。我正在尝试用它制作一个数据框，以便以后使用。原始数据集有 2,000,000 行带标点符号但它是一个列表。我无法从其余单词中解析出列表中的标点符号。我想要单词中每个标点符号之间的空格。我可以在 excel 中通过查找替换轻松地做到这一点。但是想在 R 中做。我有一个例子叫做 = df，我想要在 r 中的输出叫做 = output。我附上了下面的代码和我目前所拥有的。我尝试了 str_split 的 How 但它删除了 "How " 并且没有返回任何内容 "".

#--------Upload 1st dataset and edit-------#
library("stringr")
sent1<-c("How did Quebec? 1 2 3")
sent2<-c("Why does valve = .245? .66")
sent3<-c("How do I use a period (.) comma [,] and hyphen {-} to columns?")
df <- data.frame(text = c(sent1,sent2,sent3))
df <- as.matrix(df)
str_split(df, " ")#spaces

#-------------output-------------#
words1<-c("How", "did" ,"Quebec"," ? ","1", "2" ,"3")
words2<-c('Why', "does", "valve"," = ",".245","?" ,".66")
words3<-c("How" ,"do", "I", "use", "a", "period", '(',".",')', "comma" ,'[',",","]" ,"and" ,"hyphen" ,"{","-",'}' ,"to" ,"columns",'?')
output<-data.frame(words1,words2,words3)

Answer 1

这是完成工作的粗略概念：

首先拆分所有不是单词字符的字符（受another answer启发）。然后取最大长度，把其他的都填成一样的长度。

dfsplt <- strsplit( gsub("([^\w])","~\1~", df, perl = TRUE), "~")
dfsplt <- lapply(dfsplt, function(x) x[!x %in% c("", " ")])
n <- max(lengths(dfsplt))
sapply(dfsplt, function(x) {x <- rep(x, ceiling(n / length(x))); x[1:n]})
# or
sapply(dfsplt, function(x) x[(1:n - 1) %% length(x) + 1])

      [,1]     [,2]    [,3]     
 [1,] "How"    "Why"   "How"    
 [2,] "did"    "does"  "do"     
 [3,] "Quebec" "valve" "I"      
 [4,] "?"      "="     "use"    
 [5,] "1"      "."     "a"      
 [6,] "2"      "245"   "period" 
 [7,] "3"      "?"     "("      
 [8,] "How"    "."     "."      
 [9,] "did"    "66"    ")"      
[10,] "Quebec" "Why"   "comma"  
[11,] "?"      "does"  "["      
[12,] "1"      "valve" ","      
[13,] "2"      "="     "]"      
[14,] "3"      "."     "and"    
[15,] "How"    "245"   "hyphen" 
[16,] "did"    "?"     "{"      
[17,] "Quebec" "."     "-"      
[18,] "?"      "66"    "}"      
[19,] "1"      "Why"   "to"     
[20,] "2"      "does"  "columns"
[21,] "3"      "valve" "?"

Answer 2

这是一个选项，我们在标点字符之间创建一个 space，然后单独 scan

do.call(cbind, lapply(gsub("([[:punct:]])", " \1 ", 
       df$text), function(x) scan(text = x, what = "", quiet = TRUE)))
#      [,1]     [,2]    [,3]     
# [1,] "How"    "Why"   "How"    
# [2,] "did"    "does"  "do"     
# [3,] "Quebec" "valve" "I"      
# [4,] "?"      "="     "use"    
# [5,] "1"      "."     "a"      
# [6,] "2"      "245"   "period" 
# [7,] "3"      "?"     "("      
# [8,] "How"    "."     "."      
# [9,] "did"    "66"    ")"      
#[10,] "Quebec" "Why"   "comma"  
#[11,] "?"      "does"  "["      
#[12,] "1"      "valve" ","      
#[13,] "2"      "="     "]"      
#14,] "3"      "."     "and"    
#[15,] "How"    "245"   "hyphen" 
#[16,] "did"    "?"     "{"      
#[17,] "Quebec" "."     "-"      
#[18,] "?"      "66"    "}"      
#[19,] "1"      "Why"   "to"     
#[20,] "2"      "does"  "columns"
#[21,] "3"      "valve" "?"

将标点符号添加到 data.frame r 中的列表和 coreace

Add punctuation to a list in and coreace to data.frame r

string

r

word