r - 从R中的字符串中提取单词

Question

我正在尝试提取字符串片段并从那些匹配的模式中创建新变量。我已经尝试了“strings”包中的许多函数，但似乎无法得到结果。下面的例子是由数据组成的。我想获取一个字符串并提取片段并将它们存储到新数据框的新列中。

例子

ex <- c("The Accountant (2016)Crime (vodmovies112.blogspot.com.es)","Miss Peregrine's Home for Peculiar Children (2016)FantasySci-Fi (vodmovies112.blogspot.com.es),"Fantastic Beasts And Where To Find Them (2016) TSAdventure (openload.co)","Ben-Hur (2016) HDActionAdventure (vodmovies112.blogspot.com.es)","The Remains (2016) 1080p BlurayHorror (openload.co)" ,"Suicide Squad (2016) HDAction (openload.co)")

     >ex
[1] "The Accountant (2016)Crime (vodmovies112.blogspot.com.es)"
[2] "Miss Peregrine's Home for Peculiar Children (2016)FantasySci-Fi (vodmovies112.blogspot.com.es)"
[3] "Fantastic Beasts And Where To Find Them (2016) TSAdventure (openload.co)"
[4] "Ben-Hur (2016) HDActionAdventure (vodmovies112.blogspot.com.es)"
[5] "The Remains (2016) 1080p BlurayHorror (openload.co)"
[6] "Suicide Squad (2016) HDAction (openload.co)"

genres <- c("Action","Adventure","Animation","Biography",
        "Comedy","Crime","Documentary","Drama","Family",
        "Fantasy","Film-Noir","History","Horror","Music",
        "Musical","Mystery","Romance","Sci-Fi","Sport","Thriller",
        "War","Western")

genres <- paste0("^",genres,"|")
genres[22] <- "^Western"
> genres
[1] "^Action|"      "^Adventure|"   "^Animation|"   "^Biography|"
[5] "^Comedy|"      "^Crime|"       "^Documentary|" "^Drama|"
[9] "^Family|"      "^Fantasy|"     "^Film-Noir|"   "^History|"
[13] "^Horror|"      "^Music|"       "^Musical|"     "^Mystery|"
[17] "^Romance|"     "^Sci-Fi|"      "^Sport|"       "^Thriller|"
[21] "^War|"         "^Western"

试图完成

> df
           title year                       domain genre
1 The Accountant 2016 vodmovies112.blogspot.com.es Crime

Answer 1

这里有一个可能性:

temp <- strsplit(ex, "\\(|\\)")
df <- setNames(as.data.frame(lapply(1:4,function(i) sapply(temp,"[",i)), stringsAsFactors = FALSE), c("title", "year", "genre", "domain"))
df <- df[ , c("title", "year", "domain", "genre")]
correct <- sapply(seq_along(df$genre), function(y) which(lengths(sapply(seq_along(genres), function(x) grep(genres[x], df$genre[y])))>0))
correct <- lapply(correct, function(x) paste0(genres[x], collapse = " "))
df$genre <- unlist(correct)

df
#                                         title year                       domain            genre
# 1                              The Accountant  2016 vodmovies112.blogspot.com.es            Crime
# 2 Miss Peregrine's Home for Peculiar Children  2016 vodmovies112.blogspot.com.es   Fantasy Sci-Fi
# 3     Fantastic Beasts And Where To Find Them  2016                  openload.co        Adventure
# 4                                     Ben-Hur  2016 vodmovies112.blogspot.com.es Action Adventure
# 5                                 The Remains  2016                  openload.co           Horror
# 6                               Suicide Squad  2016                  openload.co           Action

基本上，我们将向量 ex 分成 4 个部分，用括号分隔。然后我们创建包含 4 列的 data.frame df。最难的部分是正确提取类型(因为每部电影可能有不止一种类型)。我结合使用了 sapply、lapply 和 grep 来完成它。完成后，我们会“更正”专栏类型。

这是您的数据:

ex <- c("The Accountant (2016)Crime (vodmovies112.blogspot.com.es)", 
"Miss Peregrine's Home for Peculiar Children (2016)FantasySci-Fi (vodmovies112.blogspot.com.es)", 
"Fantastic Beasts And Where To Find Them (2016) TSAdventure (openload.co)", 
"Ben-Hur (2016) HDActionAdventure (vodmovies112.blogspot.com.es)", 
"The Remains (2016) 1080p BlurayHorror (openload.co)", "Suicide Squad (2016) HDAction (openload.co)"
)

genres <- c("Action", "Adventure", "Animation", "Biography", "Comedy", 
"Crime", "Documentary", "Drama", "Family", "Fantasy", "Film-Noir", 
"History", "Horror", "Music", "Musical", "Mystery", "Romance", 
"Sci-Fi", "Sport", "Thriller", "War", "Western")