r - 使用多个变量创建一个变量

Question

我正在尝试清理数据集并在名称下创建 3 个变量:Adventure、Action 和 Comedy。原始数据集有 3000 个观察值(导入文件名:dat)。我只展示了一些观察结果

id    Runtime        Genres                                       
37      75       animation, adventure, family, fantasy, musical   
1       162      action, adventure, fantasy, sci_fi       
95      126      action, fantasy   
100     101      comedy, drama, fantasy   
82      136      action, adventure, sci-fi    
99      117      animation, adventure, comedy, family, sport   
91      95       animation, comedy, crime, family

在 R 中导入数据集后，使用以下 R 代码将所有流派分成 5 个:

dat1 <- dat %>% separate (Genres, c("Genres1","Genres2" ,"Genres3" ,"Genres4" ,"Genres5" ), sep=",", extra = "drop", fill = "right")


id    Runtime    Genres1    Genres2    Genres3  Genres4  Genres5                                       
37      75       animation  adventure  family   fantasy  musical   
1       162      action     adventure  fantasy  sci_fi       
95      126      action     fantasy   
100     101      comedy     drama      fantasy   
82      136      action     adventure  sci-fi    
99      117      animation  adventure  comedy   family   sport   
91      95       animation  comedy     crime    family

如何将 Action 、冒险和喜剧的所有类型分别归为 1 个类别？

我尝试使用以下代码:

使用冒险创建了一个空列

dat1 ["adventure"] <- NA

dat1$adventure <- ifelse(dat1$Genres1=="adventure",1,(ifelse(dat1$Genres2=="adventure",1,0))) 

建议将代码缩短为

  dat1$adventure <- ifelse((dat1$Genres1=="adventure" | dat1$Genres2=="adventure" | dat1$Genres3=="adventure" | dat1$Genres4=="adventure" ),1, 0)


id    Runtime    Genres1    Genres2    Genres3  Genres4  Genres5  Adventure                                     
37      75       animation  adventure  family   fantasy  musical  0
1       162      action     adventure  fantasy  sci_fi            0
95      126      action     fantasy                               0
100     101      comedy     drama      fantasy                    0
82      136      action     adventure  sci-fi                     0
99      117      animation  adventure  comedy   family   sport    0   
91      95       animation  comedy     crime    family            0

代码能够提取 Genres1 的冒险，但为 Genres2 返回零。

我已经重新编辑了这个问题。我尝试了一些建议但不确定如何去做，因为有 3000 次观察。

运行后建议

流派列表，向量的形成并将其分配给 dat2

dat2 <- c( "adventure", "comedy", "action", "drama", "animation", "fantasy", "mystery", "family", "sci-fi", "thriller", "romance", "horror", "musical","history", "war", "documentary", "biography")

表(因子(dat2))表(因子(dat2))

 action   adventure   animation   biography      comedy documentary          drama 
      1           1           1           1           1           1           1 
 family     fantasy     history      horror     musical     mystery     romance 
      1           1           1           1           1           1           1 
 sci-fi    thriller         war 
      1           1           1                                                                   

创建函数

 fun1 <- function("adventure", "comedy", "action", "drama", "animation",
"fantasy", "mystery", "family", "sci-fi", "thriller", "romance", "horror", 
"musical","history", "war", "documentary", "biography")) {
 vector_of_cur_genres <- seperate(i, sep = ", ")
 result <- table(factor(vector_of_cur_genres, dat2))
 return(result)
 }  

  # Results         

 fun1 <- function("adventure", "comedy", "action", "drama", 
 "animation", "fantasy", "mystery", "family", "sci-fi", "thriller",  
 "romance", "horror", "musical","history", "war", "documentary", 
 "biography")) {
  Error: unexpected string constant in "fun1 <- function("adventure""
  >   vector_of_cur_genres <- separate(i, sep = ", ")
  Error: Please supply column name
  >   result <- table(factor(vector_of_cur_genres, dat2))
  Error in factor(vector_of_cur_genres, dat2) : 
  object 'vector_of_cur_genres' not found
  >   return(result)
  Error: no function to return from, jumping to top level
   > }
   Error: unexpected '}' in "}"

  mat <- mapply(fun1,dat2$Genres)
       Error in match.fun(FUN) : object 'fun1' not found                                                                                                                                                                                                        

Answer 1

您可以混合使用表和因子来获得您想要的结果。首先，您要确保所有流派每次都拼写完全相同 ("Adventure"!= "adventure")。然后，您应该创建一个包含所有可能流派的向量 c("Adventure", "Comedy", "Drama", ...")。

然后，对于每一行，您调用 table(factor(genres, list_of_possible_genres))，它将返回一个计数表。然后你可以用这样的东西构造一个矩阵

mat <- mapply(
    function(i) {
        table(factor(separate(i, ...),list_of_possible_genres))
    },df$Genres)
#you want to use the original Data.Frame after import

new.df <- cbind(df,mat) #they should both have the same number of rows here

使单独调用中的 ... 与原始函数中的相同。如果您对各个功能或步骤的作用有任何疑问，我可以在评论中解释。

我在 mapply 调用 function (i) ... 中定义了一个函数，这类似于在 Python 中定义 lambda。该函数接受一串流派，并返回一个命名向量，其中包含每种可能流派出现的次数。

编辑:

fun1 <- function(string_of_genres)) {
    vector_of_cur_genres <- seperate(i, sep = ", ")
    result <- table(factor(vector_of_cur_genres, list_of_possible_genres))
    return(result)
}
mat <- mapply(fun1,df$Genres)