perl编程哈希回忆

Question

我在努力编码一个程序，并且无法检测到代码中的错误。

我有两个文件是这样的：

文件1

Z 712 8571 + A 
X 712 714 + A 
Y 8569 8571 + A 
Z 24137 24264 + B 
X 24137 24139 + B 
Z 24322 24391 + B 
Z 24490 26064 + B 
Y 26062 26064 + B 
Z 26704 26740 + C 
X 26704 26706 + C 
Z 26814 27170 + C 
Z 27257 27978 + C 
Y 27976 27978 + C 
Z 30488 32170 + D 
X 30488 30490 + D 
Y 32168 32170 + D 
Z 32689 32811 + E 
X 32689 32691 + E 
Z 33038 33259 + E 
Z 33309 35147 + E 
Y 35145 35147 + E 

和

文件2

1 A 
2 B 
3 C 
4 D 
5 E 

这是我的代码，我不知道什么是错的这和它的麻烦。

use strict; 

if (@ARGV != 2) { 
    print "Invalid arguments"; 
    print "Usage: perl code.pl [file1][file2]"; 
    exit(0); 
} 

my $FILE1 = $ARGV[0]; 
    my %data1 =(); 
    my $xyz = ""; 
    my $z_id = 0; 
    my $start = 0; 
    my $end = 0; 
    my $positive = ""; 
    my $letter = ""; 

my $FILE2 = $ARGV[1]; 
    my %data2 =(); 
    my $alphabet_id = 0; 
    my $alphabet = ""; 

open (FILE1DATA, $FILE1); 
open (FILE2DATA, $FILE2); 

while (my $fileline1 = <FILE1DATA>) { 
    chomp $fileline1; 

    my @line1 = split /\t/, $fileline1; 
    $xyz = $line1[0]; 
    if ($xyz eq "Z") {$z_id++;} 
    $start = $line1[1]; 
    $end = $line1[2]; 
    $positive = $line1[3]; 
    $letter = $line1[4]; 

    $data1{$letter}{ZID} = $z_id; 
    $data1{$letter}{XYZ} = $xyz; 
    $data1{$letter}{START} = $start; 
    $data1{$letter}{ENDD} = $end; 
    $data1{$letter}{POSTIVE} = $positive; 
    $data1{$letter}{LETTER} = $letter; 

    while (my $fileline2 = <FILE2DATA>) { 
     chomp $fileline2; 

     my @line2 = split /\t/, $fileline2; 
     $alphabet_id = $line2[0]; 
     $alphabet = $line2[1]; 
     $data2{$alphabet}{ID} = $alphabet_id; 
     $data2{$alphabet}{ALPHA} = $alphabet; 
     foreach (%data2) { 
      foreach ($data1{$letter}{LETTER}) { 
       if ($data1{$letter}{LETTER} eq $data2{$alphabet}{ALPHA}){ 
        $data1{$letter}{XYZ} = $data2{$alphabet}{ID}; 
       } 
      } 
     } 
    } 
    print $data1{$letter}{ZID},"\t",$data1{$letter}{XYZ},"\t",$data1{$letter}{START},"\t",$data1{$letter}{ENDD},"\t",$data1{$letter}{POSTIVE},"\n"; 
} 

close (FILE1DATA); 
close (FILE2DATA); 

exit; 

如果我执行这样的结果是这样的：

1 1 712 8571 + 
1 X 712 714 + 
1 Y 8569 8571 + 
2 Z 24137 24264 + 
2 X 24137 24139 + 
3 Z 24322 24391 + 
4 Z 24490 26064 + 
4 Y 26062 26064 + 
5 Z 26704 26740 + 
5 X 26704 26706 + 
6 Z 26814 27170 + 
7 Z 27257 27978 + 
7 Y 27976 27978 + 
8 Z 30488 32170 + 
8 X 30488 30490 + 
8 Y 32168 32170 + 
9 Z 32689 32811 + 
9 X 32689 32691 + 
10 Z 33038 33259 + 
11 Z 33309 35147 + 
11 Y 35145 35147 + 

，但假设是这样的：

1 1 712 8571 + 
1 X 712 714 + 
1 Y 8569 8571 + 
2 2 24137 24264 + 
2 X 24137 24139 + 
3 2 24322 24391 + 
4 2 24490 26064 + 
4 Y 26062 26064 + 
5 3 26704 26740 + 
5 X 26704 26706 + 
6 3 26814 27170 + 
7 3 27257 27978 + 
7 Y 27976 27978 + 
8 4 30488 32170 + 
8 X 30488 30490 + 
8 Y 32168 32170 + 
9 5 32689 32811 + 
9 X 32689 32691 + 
10 5 33038 33259 + 
11 5 33309 35147 + 
11 Y 35145 35147 + 

这意味着给一个ID，字母 “Z”当“Z”遇到时（跳过X和Y），并且当最后一个字母与file2中的字母相同时，必须将file2中的alphabet_id返回到file1中的“Z”。

Answer 1

这可能会做你想要什么：

use strict; 
use warnings; 

if (@ARGV != 2) { 
    print "Invalid arguments"; 
    print "Usage: perl code.pl [file1][file2]"; 
    exit(0); 
} 

my $FILE1 = $ARGV[0]; 
    my %data1 =(); 
    my $xyz = ""; 
    my $z_id = 0; 
    my $start = 0; 
    my $end = 0; 
    my $positive = ""; 
    my $letter = ""; 

my $FILE2 = $ARGV[1]; 
    my %data2 =(); 
    my $alphabet_id = 0; 
    my $alphabet = ""; 

open (FILE1DATA, $FILE1); 
open (FILE2DATA, $FILE2); 

while (my $fileline2 = <FILE2DATA>) { 
    chomp $fileline2; 

    my @line2 = split /\t/, $fileline2; 
    $alphabet_id = $line2[0]; 
    $alphabet = $line2[1]; 
    $data2{$alphabet}{ID} = $alphabet_id; 
    $data2{$alphabet}{ALPHA} = $alphabet; 
} 


while (my $fileline1 = <FILE1DATA>) { 
    chomp $fileline1; 

    my @line1 = split /\t/, $fileline1; 
    $xyz = $line1[0]; 
    $start = $line1[1]; 
    $end = $line1[2]; 
    $positive = $line1[3]; 
    $letter = $line1[4]; 

    if ($xyz eq 'Z') { 
    $xyz = $data2{$letter}{ID}; 
    $z_id++; 
    } 
    print "$z_id\t$xyz\t$start\t$end\t$positive\n"; 
} 


close (FILE1DATA); 
close (FILE2DATA); 

exit; 

Answer 2

下面是一些不同的方法，根据您的数据集创建您想要的结果。有评论一些行：

use strict; 
use warnings; 

@ARGV == 1 or die "Invalid usage. Usage: perl $0 [file]"; 

my ($z_id, $val) = 0; 

while (<>) { 
    # Capture then delete last column letter and calc Z replacement val 
    s/\s+(\w)$/$val = (ord $1) - 64; ''/e; 

    # Replace Z w/val and inc Z count 
    $z_id++ if s/^Z/$val/; 

    print "$z_id\t$_"; 
} 

的钻石操作（<>）读他的名字通过@ARGV发送的文件。请注意，“Z”是用1-5代替它的替换的一部分。当$1为“A”时，表达式(ord $1) - 64评估为1，当“B”等时为2，所以只需要主数据文件（您的文件1）。如果Z替换s/^Z/$val/成功，则$z_id递增。

希望这会有所帮助！