解析唯一数据和重命名文件

Question

我试图创建一个 Perl 脚本来重命名文件（数百个不同名称的文件），但我没有成功。我首先需要找到唯一的文件编号，然后将其重命名为更易于阅读的名称。由于文件名不连续，因此很难。

文件名示例：重要性数在que序列之后

#                                   vv-- this number
lane8-s244-index--ATTACTCG-TATAGCCT-01_S244_L008_R1_001.fastq
lane8-s245-index--ATTACTCG-ATAGAGGC-02_S245_L008_R1_001.fastq
lane8-s246-index--TCCGGAGA-TATAGCCT-09_S246_L008_R1_001.fastq
lane8-s247-index--TCCGGAGA-ATAGAGGC-10_S247_L008_R1_001.fastq
lane8-s248-index--TCCGGAGA-CCTATCCT-11_S248_L008_R1_001.fastq
lane8-s249-index--TCCGGAGA-GGCTCTGA-12_S249_L008_R1_001.fastq
lane8-s250-index--TCCGGAGA-AGGCGAAG-13_S250_L008_R1_001.fastq
lane8-s251-index--TCCGGAGA-TAATCTTA-14_S251_L008_R1_001.fastq
lane7-s0007-index--ATTACTCG-TATAGCCT-193_S7_L007_R1_001.fastq
lane7-s0008-index--ATTACTCG-ATAGAGGC-105_S8_L007_R1_001.fastq
lane7-s0009-index--ATTACTCG-CCTATCCT-195_S9_L007_R1_001.fastq
lane7-s0010-index--ATTACTCG-GGCTCTGA-106_S10_L007_R1_001.fastq
lane7-s0011-index--ATTACTCG-AGGCGAAG-197_S11_L007_R1_001.fastq
lane7-s0096-index--AGCGATAG-CAGGACGT-287_S96_L007_R1_001.fastq

我创建了一个名为 RENAMING_parse_data.sh 的文件，它引用了 RENAMING_parse_data.pl

所以理论上，这个想法是解析数据以找到位于名称中间的样本#，并获取该唯一 ID 并重命名它。但我认为它甚至不会进入 IF 循环。 有什么想法吗？

这里是调用 perl scipt 的 .sh 文件

#!/bin/bash
#first part is the program
#second is the directory path
#third and fourth times are the names of the output files 
#./parse_data.pl /ACTF/Course/PATHTDIRECTORY Tabsummary.txt Strucsummary.txt 
#WHERE  ./parse_data.pl =name of the program 
#WHERE /ACTF/Course/PATHTODIRECTORY  = directory path were your field are saved AND is referred to as $dir_in = $ARGV[0] in the perl script;
#new files you recreating with the extracted data AND is refered to as $dir_in = $ARGV[1];

./RENAMING_parse_data.pl ./Test/ FishList.txt

这里是 PERL 脚本：

#!/usr/bin/perl
print (":)\n");
#Proesessing files in a directory
$dir_in = $ARGV[0];
$indv_list = $ARGV[1];

#open directory to acess those files, the folder where you have the files
opendir(DIR, $dir_in) || die ("Cannot open $dir_in");
@files = readdir(DIR);

#set all variables = 0 to void chaos
$j=0;

#open output header line for output file and print header line for tab delimited file
open(OUTFILETAB, ">", $indv_list);
print(OUTFILETAB "\t Fish ID", "\t");

#open each file 
foreach (@files){
       #re start all arrays to void chaos
        print("in loop [$j]");
        @acc_ID=(); 
       #find FISH name
        #EXAMPLE FISH NAMES: (lenth of fishname varies)
        #lane8-s251-index--TCCGGAGA-TAATCTTA-14_S251_L008_R1_001.fastq.gz
        #lane7-s0096-index--AGCGATAG-CAGGACGT-287_S96_L007_R1_001.final.fastq
         #NOTE: what is in btween () is the ID that is printed NOTE that value can change from 2 -3 depending on Sample #
        #Trials:
        #lane[0-9]{1}-[a-z]{1}[0-9]{4}-index--[A-Z]{8}[A-Z]{8}-([0-9]{3})[a-z]{1}[0-9]{2}_[A-Z]{1}[0-9]{3}_[a-z]{1}[0-9]{1}_[0-9]{3}.fastq
         #lane[0-9]{1}-[a-z]{1}[0-9]{4}-index--[A-Z]{8}[A-Z]{8}-([0-9]{3})*.fastq
          #lane*([0-9]{3})*.fastq
         #lane.*-([0-9]{2})_.*.fastq
         #lane.*-([0-9]{2})_*.fastq
         #lane[0-9]{1}-[a-z]{1}[0-9]{3}-index--[A-Z]{8}[A-Z]{8}-([0-9]{2})_[A-Z]{1}[0-9]{3}_L008_R1_001.fastq
        $string_FISH = @files;
        if ($string_FISH =~ /^lane[0-9]{1}-[a-z]{1}[0-9]{3}-index--[A-Z]{8}[A-Z]{8}-([0-9]{2})_[A-Z]{1}[0-9]{3}_L008_R1_001.fastq/){
            $FISH_ID =$1;
            @acc_ID[$j] = $FISH_ID;
            #print ("FISH. = |$FISH_ID[$j]| \n");
            rename($string_FISH, "FISH. = |$FISH_ID[$j]|");
            #print ($acc_ID[$j], "\n");
            print(OUTFILETAB "FISH. = |$FISH_ID[$j]| \n");      
        }
    $j= $j+1;
}

理想的最终结果

所以最后我希望它取文件名，找到唯一标识符并重命名它

来自：

lane8-s244-index--ATTACTCG-TATAGCCT-01_S244_L008_R1_001.fastq
lane7-s0007-index--ATTACTCG-TATAGCCT-193_S7_L007_R1_001.fastq

到：

Fish.01.fastq
Fish.193.fastq

非常感谢任何关于解决此问题或需要完全更改的想法或建议。

Answer 1

在 Perl 解决方案的核心，您可以使用

s/^.*-(\d+)_[^-]+(?=\.fastq\z)/Fish.$1/sa

例如，

$ ls -1 *.fastq
lane8-s244-index--ATTACTCG-TATAGCCT-01_S244_L008_R1_001.fastq
lane8-s245-index--ATTACTCG-ATAGAGGC-02_S245_L008_R1_001.fastq
lane8-s246-index--TCCGGAGA-TATAGCCT-09_S246_L008_R1_001.fastq
lane8-s247-index--TCCGGAGA-ATAGAGGC-10_S247_L008_R1_001.fastq
lane8-s248-index--TCCGGAGA-CCTATCCT-11_S248_L008_R1_001.fastq
lane8-s249-index--TCCGGAGA-GGCTCTGA-12_S249_L008_R1_001.fastq

$ rename 's/^.*-(\d+)_[^-]+(?=\.fastq\z)/Fish.$1/sa' *.fastq

$ ls -1 *.fastq
Fish.01.fastq
Fish.02.fastq
Fish.09.fastq
Fish.10.fastq
Fish.11.fastq
Fish.12.fastq

（有两个类似的工具名为rename。这个也称为prename。）

---

自己实现非常简单：

#!/usr/bin/perl

use strict;
use warnings;

my $errors = 0;
for (@ARGV) {
   my $old = $_;

   s/^.*-(\d+)_[^-]+(?=\.fastq\z)/Fish.$1/sa;

   my $new = $_;

   next if $new eq $old;

   if ( -e $new ) {
      warn( "Can't rename \"$old\" to \"$new\": Already exists\n" );
      ++$errors;
   }
   elsif ( !rename( $old, $new ) ) {
      warn( "Can't rename \"$old\" to \"$new\": $!\n" );
      ++$errors;
   }
}

exit( !!$errors );

提供要重命名的文件作为参数（例如，使用 shell 中的 *.fastq）。

$ ls -1 *.fastq
lane8-s244-index--ATTACTCG-TATAGCCT-01_S244_L008_R1_001.fastq
lane8-s245-index--ATTACTCG-ATAGAGGC-02_S245_L008_R1_001.fastq
lane8-s246-index--TCCGGAGA-TATAGCCT-09_S246_L008_R1_001.fastq
lane8-s247-index--TCCGGAGA-ATAGAGGC-10_S247_L008_R1_001.fastq
lane8-s248-index--TCCGGAGA-CCTATCCT-11_S248_L008_R1_001.fastq
lane8-s249-index--TCCGGAGA-GGCTCTGA-12_S249_L008_R1_001.fastq

$ ./a *.fastq

$ ls -1 *.fastq
Fish.01.fastq
Fish.02.fastq
Fish.09.fastq
Fish.10.fastq
Fish.11.fastq
Fish.12.fastq

存在性检查 (-e) 是为了防止意外地将一堆文件重命名为相同的名称，从而导致除了一个文件之外的所有文件都丢失。

---

以上是我经常使用的单线模式的清理版本。

dir /b ... | perl -nle"$o=$_; s/.../.../; $n=$_; rename$o,$n if!-e$n"

改编为sh：

\ls ... | perl -nle'$o=$_; s/.../.../; $n=$_; rename$o,$n if!-e$n'