perl 多行变一行，保持顺序不变

tewithte

比如我有一堆数据
>1
1234
>2
2235
>3
5895
>4
5896
>5
7595
想通过perl变成
>1   1234
>2   2235
>3   5895
>4   5896
>5   7595，

#! perl -w

print "Please enter the folder name:\n";
chomp ($folder=<STDIN>);
print "The output folder name:\n";
chomp ($output=<STDIN>);

mkdir $output;

opendir(R,"$folder") || die "folder does not exist";
@num=readdir(R);
closedir(R);
shift(@num);
shift(@num);
$n=@num;
open (W,">","document.txt");
print W "Total files: $n \n\n";

foreach (@num){
        print W "$_ is read\n";
        $new = "com_"."$_";
        open (IN,"./$folder/$_")|| die "Could not open file $_";
        @all = <IN>;
        close (IN);

        foreach (@all) {
                chomp $_;
                s/^>//;
        }
       
        %notsorted = @all;
       
        open (OUT,">","$output/$new")|| die "Could not open new file: $new.\n ";
        foreach $seqname (keys %notsorted){
                $notsorted{$seqname} =~ tr/atcg/ATCG/;
                printf OUT ("%-30s%s",$seqname,$notsorted{$seqname});
                print OUT "\n";
                }
               
        close OUT;
}

close W;

但是输出后，文件的名字是乱序的，请问怎么能按顺序输出？

jason680

回复 1# tewithte

$ perl -lane '{if(m/^>/){print $s if($s);$s=$_}else{$s.="\t$_"}}END{print $s}' FILE
>1        1234
>2        2235
>3        5895
>4        5896
>5        7595

   

kk861123

1. #!/usr/bin/perl
   
2. use strict;
   
3. use warnings;
   
4. use Data::Dumper;
   
5. 
   
6. local $/ = '>';
   
7. while (<DATA>) {
   
8.     s/\n/ /;
   
9.     print;
   
10. }
    
11. 
    
12. __DATA__
    
13. >1
    
14. 1234
    
15. >2
    
16. 2235
    
17. >3
    
18. 5895
    
19. >4
    
20. 5896
    
21. >5
    
22. 7595

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104359176

1. use v6;
   
2. 
   
3. 'filename.txt'.IO.slurp.replace("\n", "\t").split(">\t");

复制代码

sunzhiguolu

1. perl -pe 's/\n/ / if(/\A>/)' f

复制代码

moperyblue

1. sed '/>/{N;s/\n/ /}' file
   
2. awk '/>/{getline x;print $0,x}' file

复制代码