Question

我们如何从文件列表中仅使用名称“log2.read.counts.2289_Tail”grep文件。

data =

"log2.read.counts.2289_12_Tumor_NF4_CTTGTAA_L002" 
"log2.read.counts.2289_1_Tail_cont_ATCACGA_L002"   
"log2.read.counts.2289_2_Tail_Lmyc_CGATGTA_L002"        
"log2.read.counts.2289_3_Tail_Nfib_TTAGGCA_L002"        
 "log2.read.counts.2289_4_Cell_LmycS3_TGACCAA_L002"

输出=

     "log2.read.counts.2289_1_Tail_cont_ATCACGA_L002"   
     "log2.read.counts.2289_2_Tail_Lmyc_CGATGTA_L002"   
     "log2.read.counts.2289_3_Tail_Nfib_TTAGGCA_L002"

Answer 1

如果搜索匹配，

grepl（）将返回TRUE。用它来过滤输入向量。如果你不熟悉正则表达式，花一些时间学习它们可能是明智之举。在这种情况下。它正在搜索你的字符串，中间有一个或多个数字。

input <- c("log2.read.counts.2289_12_Tumor_NF4_CTTGTAA_L002", 
           "log2.read.counts.2289_1_Tail_cont_ATCACGA_L002",   
           "log2.read.counts.2289_2_Tail_Lmyc_CGATGTA_L002",        
           "log2.read.counts.2289_3_Tail_Nfib_TTAGGCA_L002",        
           "log2.read.counts.2289_4_Cell_LmycS3_TGACCAA_L002" )
> input[grepl("log2\\.read\\.counts\\.2289_[0-9]+_Tail", input)]
[1] "log2.read.counts.2289_1_Tail_cont_ATCACGA_L002"
[2] "log2.read.counts.2289_2_Tail_Lmyc_CGATGTA_L002"
[3] "log2.read.counts.2289_3_Tail_Nfib_TTAGGCA_L002"

Answer 2

这是grep的一种方式：

fls <- c("log2.read.counts.2289_12_Tumor_NF4_CTTGTAA_L002", 
"log2.read.counts.2289_1_Tail_cont_ATCACGA_L002",   
"log2.read.counts.2289_2_Tail_Lmyc_CGATGTA_L002",        
"log2.read.counts.2289_3_Tail_Nfib_TTAGGCA_L002",        
"log2.read.counts.2289_4_Cell_LmycS3_TGACCAA_L002")

grep("^log2\\.read.counts\\.2289_\\d+_Tail", fls, value=TRUE)

## [1] "log2.read.counts.2289_1_Tail_cont_ATCACGA_L002"
## [2] "log2.read.counts.2289_2_Tail_Lmyc_CGATGTA_L002"
## [3] "log2.read.counts.2289_3_Tail_Nfib_TTAGGCA_L002"

Answer 3

假设您还没有文件列表： ls -1 | grep“log2.read.counts.2289_ [0-9] {1，} _ Tail”

[0-9] {1。}将匹配任何长度数字序列（在您的示例中，它们都是单个数字 - 1,2,3 - 但这也将匹配23和34）。

这些时期需要被转义，因为它们具有特殊含义（它自己的句号将匹配前面一个字符）。

Answer 4

grep'^ log2.read.counts.2289_ 尾巴'

grep在文件中搜索多个字符串

4 个答案: