Question

我有一个名为例如的文件列表：

"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，我想得到以下列表：

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

有人能帮帮我吗？

提前谢谢

最佳

Answer 1

假设您确定您将始终拥有＆＃34; test1＆＃34;，＆＃34; test2＆＃34;和＆＃34; test3＆＃34;，您可以使用正则表达式grep执行此任务。

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)
  })
})

根据模式列表的子集文件

1 个答案: