如何循环使用R中的列名

这是我的风格载体

另一个数据集具有相同的列名

这是数据集的列名

genres=c("Action","Adventure","Animation","Biography","Comedy","Crime",
 "Documentary","Drama","Family","Game.Show","Horror","Music","Musical",
 "Mystery","Romance","Sci.Fi","Short","Thriller","War","Western")

我想对所有类型运行此命令，用值替换每个类型

"Title"        "Genre"        "imdbRating"   "Release_Year" 
"Action"       "Adventure"    "Animation"    "Biography"    "Comedy"    
"Crime"        "Documentary"  "Drama"        "Family" 
"Fantasy"      "Game.Show"    "Horror"       "Music"
"Musical"      "Mystery"      "N.A"          "Romance"
"Sci.Fi"       "Short"        "Sport"        "Thriller"   
"War"          "Western"  

data\u predict$Genre[grepl（“*genres*”，data\u predict$Genre）]=1
原始数据集
数据预测Try

或

data\u predict[-1]体裁是“科幻”“动作、冒险、科幻”“动作、戏剧、战争”我想更新那些包含体裁的列。就像一行包含动作剧和战争一样，它的动作剧和战争应该成为1我正在使用此代码
data\u predict$Action[grepl（“动作”，data\u predict$体裁）]=1数据预测$冒险[grepl（“*冒险*”，数据预测$流派）]=1数据预测$动画[grepl（“*动画*”，数据预测$流派）]=1数据预测$传记[grepl（“*传记*”，数据预测$流派）]=1数据预测$喜剧[grepl（“*喜剧*”，数据预测$流派）]=1数据预测$犯罪[grepl（“*犯罪*，数据预测$流派）]=1数据预测$纪录片[grepl（“*纪录片*”，数据预测$流派）]=1数据预测$戏剧[grepl（“*戏剧*”，数据预测$流派）]=1
我想减少步骤。你能帮我吗？你可以在数据集中看到，我只想更新那些包含流派的列。例如，第二行包含动作、冒险和科幻，所以动作、冒险、科幻应该为这一行设置1。粘贴（流派，折叠=“|”）未工作图像没有帮助，因为我必须手动键入要测试的数据。请检查指南我是新的，因此很抱歉造成的麻烦。谢谢alot@PriyankPuri没问题。很高兴帮助你
     data_predict$genres[grepl("*genres*", data_predict$Genre)]=1

Orignal Data set


        data_predict<-structure(list(Genre = structure(c(3L, 1L, 2L), .Label = c("Action, Adventure, Sci-Fi", 
"Action, Drama, War", "Sci-Fi"), class = "factor"), Action = c(0, 
0, 0), Adventure = c(0, 0, 0), Animation = c(0, 0, 0), Biography = c(0, 
0, 0), Comedy = c(0, 0, 0), Crime = c(0, 0, 0), Documentary = c(0, 
0, 0), Drama = c(0, 0, 0), Family = c(0, 0, 0), Game.Show = c(0, 
0, 0), Horror = c(0, 0, 0), Music = c(0, 0, 0), Musical = c(0, 
0, 0), Mystery = c(0, 0, 0), Romance = c(0, 0, 0), Sci.Fi = c(0, 
0, 0), Short = c(0, 0, 0), Thriller = c(0, 0, 0), War = c(0, 
0, 0), Western = c(0, 0, 0)), .Names = c("Genre", "Action", "Adventure", 
"Animation", "Biography", "Comedy", "Crime", "Documentary", "Drama", 
"Family", "Game.Show", "Horror", "Music", "Musical", "Mystery", 
"Romance", "Sci.Fi", "Short", "Thriller", "War", "Western"), row.names = c(NA, 
3L), class = "data.frame") 

data_predicted<-structure(list(Genre = structure(c(3L, 1L, 2L), .Label = c("Action, Adventure, Sci-Fi", 
    "Action, Drama, War", "Sci-Fi"), class = "factor"), Action = c(0, 
    1, 1), Adventure = c(0, 1, 0), Animation = c(0, 0, 0), Biography = c(0, 
    0, 0), Comedy = c(0, 0, 0), Crime = c(0, 0, 0), Documentary = c(0, 
    0, 0), Drama = c(0, 0, 1), Family = c(0, 0, 0), Game.Show = c(0, 
    0, 0), Horror = c(0, 0, 0), Music = c(0, 0, 0), Musical = c(0, 
    0, 0), Mystery = c(0, 0, 0), Romance = c(0, 0, 0), Sci.Fi = c(0, 
    0, 0), Short = c(0, 0, 0), Thriller = c(0, 0, 0), War = c(0, 
    0, 1), Western = c(0, 0, 0)), .Names = c("Genre", "Action", "Adventure", 
    "Animation", "Biography", "Comedy", "Crime", "Documentary", "Drama", 
    "Family", "Game.Show", "Horror", "Music", "Musical", "Mystery", 
    "Romance", "Sci.Fi", "Short", "Thriller", "War", "Western"), row.names = c(NA, 
    3L), class = "data.frame")

library(qdapTools)
mtabulate(strsplit(as.character(data_predict$Genre), ', '))

 data_predict[-1] <- lapply(names(data_predict)[-1],
      function(x) as.numeric(grepl(x, data_predict$Genre)))