Question

我有一个文件列表函数，如下所示：

files <- rev( list.files( list.dirs( "../runs" ), "*.gene.read.count", full.names=TRUE ) )

这会产生以下结果：

> files
 [1] "../runs/Sample_7316/7316_AACCGA_L003_R.gene.read.count"                                
 [2] "../runs/Sample_7315/7315_GAATCT_L003_R.gene.read.count"                                
 [3] "../runs/Sample_7314/7314_CCTTGC_L003_R.gene.read.count"                                
 [4] "../runs/Sample_7313/7313_AGGCCA_L003_R.gene.read.count"                                
 [5] "../runs/Sample_7312/7312_GCGAAG_L003_R.gene.read.count"                                
 [6] "../runs/Sample_7311/7311_TCTCAG_L003_R.gene.read.count"                                
 [7] "../runs/Sample_7310/7310_CTCTGG_L003_R.gene.read.count"                                
 [8] "../runs/Sample_7309/7309_ATGGCG_L008_R.gene.read.count"                                
 [9] "../runs/project-043.Allreports/project043.raw.gene.read.count.hetro.homo.csv"          
[10] "../runs/project-043.Allreports/project043.raw.gene.read.count.csv"                     
[11] "../runs/project-043.Allreports/project043.gene.read.count.hetro.homo.csv"              
[12] "../runs/project-043.Allreports/project043.gene.read.count.csv"                         
[13] "../runs/project-043.Allreports/analysis-project043.raw.gene.read.count.html"           
[14] "../runs/project-043.Allreports/analysis-project043.raw.gene.read.count.hetro.homo.html"
[15] "../runs/project-043.Allreports/analysis- project043.gene.read.count.html"              
[16] "../runs/project-043.Allreports/analysis-project043.gene.read.count.hetro.homo.html"

问题是我只需要从Sample_73

开始的dirs中的文件

我尝试过很多不同的东西，但对我来说没有任何作用： files <- rev( list.files( list.dirs( "../runs/Sample*" ), "*.gene.read.count", full.names=TRUE ) )

希望有办法只选择这些目录：

 [1] "../runs/Sample_7316/7316_AACCGA_L003_R.gene.read.count"                                
 [2] "../runs/Sample_7315/7315_GAATCT_L003_R.gene.read.count"                                
 [3] "../runs/Sample_7314/7314_CCTTGC_L003_R.gene.read.count"                                
 [4] "../runs/Sample_7313/7313_AGGCCA_L003_R.gene.read.count"                                
 [5] "../runs/Sample_7312/7312_GCGAAG_L003_R.gene.read.count"                                
 [6] "../runs/Sample_7311/7311_TCTCAG_L003_R.gene.read.count"                                
 [7] "../runs/Sample_7310/7310_CTCTGG_L003_R.gene.read.count"                                
 [8] "../runs/Sample_7309/7309_ATGGCG_L008_R.gene.read.count"

Answer 1

list.files与模式与完整文件名不匹配。与上面的注释类似，另一种方法是将grep应用于结果，或将其应用于您作为参数提供的路径列表，例如：

files <- rev( list.files( grep( "Sample_73", list.dirs( "../runs" ), value=TRUE), "*.gene.read.count", full.names=TRUE))

Answer 2

如celiomsj所述，pattern的{{1}}参数仅匹配文件名，而不是路径，所以最好用两行完成。以下答案使用list.files作为str_detect的可读替代方案。

grepl

list.files（）无法获取正确的列表

2 个答案: