Question

在此处查看问题： match-rows-based-on-first-field-and-combine-second-field

如果满足以下条件，您将如何解决这个问题：

单个CSV文件
有多个重复记录属于不同类别
您需要匹配第一个字段并合并/追加类别字段

示例文件：

External ID     Item Name       Item Description        Release Date    Expiry Date     Weight  Template ID     Enabled EntityId        Classifications Address N/A     City    State   Zipcode Country of domain purchase is made from Title   Cover Image     Link    Author
411280  Shade me        Shade me        04-May-2017     01-Jan-9999             0       Y       -1      Teen                                                    Shade me        MC.GIF        http://catalog.org/cgi-bin/koha/opac-detail.pl?biblionumber=411280 Brown, Jennifer
411280  Shade me        Shade me        04-May-2017     01-Jan-9999             0       Y       -1      Books                                                   Shade me        MC.GIF        http://catalog.org/cgi-bin/koha/opac-detail.pl?biblionumber=411280 Brown, Jennifer
413036  Now that's what I call music!   Now that's what I call music!   04-May-2017     01-Jan-9999             0       Y       -1      Teen                                                    Now that's what I call music!  MC.GIF http://catalog.org/cgi-bin/koha/opac-detail.pl?biblionumber=413036

挑战是匹配重复的ID并合并类别。

Shade me        MC.GIF        http://catalog.org/cgi-bin/koha/opac-detail.pl?biblionumber=411280 Brown, Jennifer
    411280  Shade me        Shade me        04-May-2017     01-Jan-9999             0       Y       -1      Teen;Books

更新

while ( <FILE> ) { 
        next if 1..1;
        chomp $_; 

        my ( $id, $name, $desc, $reldate, $expdate, $weight, $temp, $enabled, $ent, $class, $addr, $na, $city, $state, $zip, $country, $title, $img, $link, $auth ) = split /\t/ , $_; 

        if ( ! $merge{$id} ) { 
                $merge{$id} = "$id, $name, $desc, $reldate, $expdate, $weight, $temp, $enabled, $ent, $class, $addr, $na, $city, $state, $zip, $country, $title, $img, $link, $auth";
        } else {
                $merge{$class} .= ";$class" if ( $merge{$id} ne $class )
        }
}

p %merge;

给我提出问题的一句话是：

$merge{$class} .= ";$class" if ( $merge{$id} ne $class )

你可以看到我需要做什么 - 合并类字段。不工作

Answer 1

我会将文件加载到某个数据结构中并记住每个唯一的列值，然后根据需要打印它们。例如。如在此示例中一样（使用|，因为它的分隔符更好地显示为\t）：

#!/usr/bin/env perl 

use 5.024;
use warnings;
use Data::Dumper;

my $records;
my $numcols;
while(<DATA>) {
    chomp;
    my(@cols) = split /\|/, $_, -1;
    $numcols = @cols if( $. == 1 );
    die "Wrong number of columns (@{[scalar @cols]} instead of $numcols) in line $." unless (@cols == $numcols);
    $records->{$cols[0]}->[0] = $. unless $records->{$cols[0]}; #remember the line# of the 1st apperance
    for(my $c = 1; $c < $numcols; $c++) {   #skip the id (col[0])
        $records->{$cols[0]}->[$c]->{$cols[$c]}++;
    }
}
# if want, check the data-structure
#say Dumper($records);

for my $id (sort {$records->{$a}->[0] <=> $records->{$b}->[0]} keys %$records) {
    say join("|",
            $id,
            map { join(';', sort grep {/\S/} keys $records->{$id}->[$_]->%*) } 1 .. $#{$records->{$id}}  #skip col[0]
        );
}

__DATA__
ID|Name1|Name2|Name3
id1|c11|c12|c13
id1|c11|c12|c13
id2|c21|c22|c23
id1|c31|c12|c13
id3|c41||c43
id1|c51|c12|c13
id1|c31||c13
id1|c11||c13
id1|c31|c12|c13
id2|c21|c22|c83
id4|c91|c92|

打印

ID|Name1|Name2|Name3
id1|c11;c31;c51|c12|c13
id2|c21|c22|c23;c83
id3|c41||c43
id4|c91|c92|

将一些shell用于漂亮的列perl script.pl | sed 's/||/| |/g' | column -s'|' -t

ID   Name1        Name2  Name3
id1  c11;c31;c51  c12    c13
id2  c21          c22    c23;c83
id3  c41                 c43
id4  c91          c92

Answer 2

小文件的简单方法可以如上所述：

#!/usr/bin/env perl 

use common::sense;    
use DDP;

my %merge;

while ( <DATA> )
{

 next if 1..1;

 chomp $_;

 my ( $id, $text, $category ) = split /,/ , $_;

 if ( ! $merge{$id} )
 {
     $merge{$id} = "$id,$text,$category";
 }
 else
 {
       my ( undef, undef , $c ) = split /,/ , $merge{$id};

       if ( $c !~ /\b$category\b/ )
       {
          $merge{$id} .= ";$category";
       }
 }


}

p %merge;

__DATA__
Id    Title               Category
12345,My favorite martian,aliens
13444,Texas Meat,BBQ
12345,My favorite martian,aliens

<强>输出：

{
    12345   "12345,My favorite martian,aliens;space",
    13444   "13444,Texas Meat,BBQ"
}

考虑到你不想要重复的类别会有所帮助。

Perl：在第一个字段上匹配重复项合并最后一个字段

2 个答案: