Perl：如何向多维数组中的每一行添加额外的列

如何解决Perl：如何向多维数组中的每一行添加额外的列

我有两个多维数组：array1(@wipr) 和 array2(@widesc)

#!/usr/bin/perl
use strict;
use warnings;
use Data::Dumper;
  
# Initializing and defining the array
my @wipr = ( ['1111','2221'],['1112','2222'],['1113','2223'],['1115','2225'],'2224'] );
my @widesc = ( ['1111','Bug1'],'Feature1'],'Feature2'],['1114','Feature2'] );
# Printing items from the array
#print Dumper(\@wipr);
#print Dumper(\@widesc);

# Using For loop
for(my $m = 0; $m <= $#wipr; $m++)
{   
   for(my $n = 0; $n <= $#widesc ; $n++)
   {  
      #print $wipr[$m][0],"\n";
      #print $widesc[$n][0];
      if ( "$wipr[$m][0]" == "$widesc[$n][0]" )
      {
          print "Matches: $wipr[$m][0] == $widesc[$n][0]";
          my $desc="$widesc[$n][1]";
          print "desc value is : $desc";
          #push(@{$wipr[$m][1][$m]},"$desc");
          #print "$wipr[$m][1][$m]";
          $wipr[$m][1][$m] = "'$desc'";
          #push (@wipr,$wipr[$m][1][$m]);
      }
      #else
      #{
        #  print "Doesn't Matches: $wipr[$m][0] == $widesc[$n][0]";
      #}
   }  
   print "\n";  
} 
print Dumper(\@wipr);

如果array1(@wipr)中第一列的值与array2(@widesc)中第一列的值匹配，我们需要将array2的第二个元素添加到array1中。 如果值不匹配，它应该只输入一个字符串“未找到”

所需的输出应该是：

 my @result= ( ['1111','2221','2222','2223','2225','Not Found'],'2224','Feature2'] );

我在 array1 中添加第三列时遇到了困难。你能帮忙吗。

错误：不能使用字符串（“2221”）作为数组引用，而在 arrays1.pl line33 中使用“strict refs”。

解决方法

天真的方法：

my @result;
for my $wipr (@wiprs)
   my $wi1 = $wipr->[0];

   for my $widesc (@widescs)
      my $wi2 = $widesc->[0];
      if ($wi1 == $wi2) {
         push @result,[ $wi1,$wipr->@[1..$#$wipr],$widesc->@[1..$#$widesc] ];
         last;
      }
   }
}

这很慢。如果有N个WI，则内循环执行N*N/2次。这意味着花费的时间与 N² 的平方成正比。

当某件事占用资源（例如时间）与 N² 的平方成正比时，我们说它是 O(N²)。

---

上述速度如此缓慢的原因是匹配记录的过程。如果数组被排序怎么办？

@wiprs   = sort { $a->[0] <-> $b->[0] } @wiprs;
@widescs = sort { $a->[0] <-> $b->[0] } @widescs;

my $pr_i = 0;
my $desc_i = 0;

my @results;
while ($pr_i < @wiprs && $desc_i < @widescs) {
   my $wipr   = $wiprs[$pr_i];
   my $widesc = $widescs[$desc_i];

   my $wi1 = $wipr->[0];
   my $wi2 = $wipdesc->[0];

   my $cmp = $wi1 <=> $wi2;
   if ($cmp < 0) {
      ++$pr_i;
   }
   elsif ($cmp > 0) {
      ++$desc_i;
   }
   else {
      push @result,$widesc->@[1..$#$widesc] ];
      ++$pr_i;
      ++$desc_i;
   }
}

这是 O(N log N)。这是很多更好。

如果我们假设有两个数组都具有相同的 WI，则上述简化为：

die "Bad data\n" if @wiprs != @widescs;

@wiprs   = sort { $a->[0] <-> $b->[0] } @wiprs;
@widescs = sort { $a->[0] <-> $b->[0] } @widescs;

my $pr_i = 0;
my $desc_i = 0;

my @results;
while ($pr_i < @wiprs && $desc_i < @widescs) {
   my $wipr   = $wiprs[$pr_i];
   my $widesc = $widescs[$desc_i];

   my $wi1 = $wipr->[0];
   my $wi2 = $wipdesc->[0];

   die "Bad data\n" if $wi1 != $wi2;

   push @result,$widesc->@[1..$#$widesc] ];
   ++$pr_i;
   ++$desc_i;
}

当然，如果数组已经排序，则简化为以下 O(N) 解决方案：

die "Bad data\n" if @wiprs != @widescs;

my $pr_i = 0;
my $desc_i = 0;

my @results;
while ($pr_i < @wiprs && $desc_i < @widescs) {
   my $wipr   = $wiprs[$pr_i];
   my $widesc = $widescs[$desc_i];

   my $wi1 = $wipr->[0];
   my $wi2 = $wipdesc->[0];

   die "Bad data\n" if $wi1 != $wi2;

   push @result,$widesc->@[1..$#$widesc] ];
   ++$pr_i;
   ++$desc_i;
}

---

最后，我们通过首先构建索引得到了一个没有上述假设的 O(N) 解决方案。

my %widescs_lookup;
for my $widesc (@widescs)
   my $wi = $widesc->[0];
   $widescs_lookup{$wi} = $widesc;
}

my @results;
for my $wipr (@wiprs)
   my $wi = $wipr->[0];

   my $wi_desc = $widescs_lookup{$wi}
      or next;

   push @result,$widesc->@[1..$#$widesc] ];
}