根据模式列表的子集文件

Question

我有一个名为的文件列表，例如：

"Experiment_myex1_test1"     
"Experiment_myex1_test2"      
"Experiment_myex1_test3"       
"Experiment_myex2_test1"      
"Experiment_myex2_test2"    
"Experiment_myex2_test3"      
"Experiment_myex3_test1"      
"Experiment_myex3_test2"    
"Experiment_myex3_test3"    

等等(总共1000个文件)和一个如下所示的文件.txt：

df
List_of_ex      
myex1     
myex1    
myex1 
myex2     
myex2      
myex2
myex4    
myex4     
myex4     

(由于*test1、*test2和*test3，它们一式三份出现)。

我想从list.files()获取与df[,1]中的模式匹配的文件列表。换句话说，我希望得到以下列表：

"Experiment_myex1_test1"     
"Experiment_myex1_test2"      
"Experiment_myex1_test3"       
"Experiment_myex2_test1"      
"Experiment_myex2_test2"    
"Experiment_myex2_test3"      

Answer 1

假设您确信您将始终拥有"test1“、"test2”和"test3"，则可以使用带有grep的正则表达式来完成此任务。

lapply(df[, 1], function(pattern_ex) {
  lapply(c("test1", "test2", "test3"), function(pattern_test){
    grep(pattern = paste0(pattern_ex, ".*", pattern_test, "$"), 
         x = list.files(),
         value = T)
  })
})