Perl

Question

正在尝试执行如下操作。

#expand [a] to is, and isn't
#expand [b] to test, and demo
my $string = 'This [a] a [b]';

所以基本上我会得到

my @strings = ('This is a test', 'This isn\'t a test', 'This is a demo', 'This isn\'t a demo');

我在使用递归之前已经这样做了，但是那是在一个非常大的数据集和很多规则上，我觉得使用 map 或 grep 或 List::MoreUtils 可能有更简单的方法我可以别想了

Answer 1

给出表单的输入

my %dict = (
   a => ["is", "isn't"],
   b => ["test", "demo"],
);

my $template = 'This [a] a [b]';

算法::循环版本：

use Algorithm::Loops qw( NestedLoops );

my @loops;
for ($template) {
   if (/\G \[ /xgc) {
      /\G ( [^\]]* ) \] /xgc
         or die("Missing \"]\"\n");

      my $var = ;
      length($var)
         or die("Empty \"[]\"\n");

      $dict{$var}      
         or die("Unknown var \"$var\"\n");

      push @loops, $dict{$var};
      redo;
   }

   if (/\G ( [^\[]+ ) /xgc) {
      push @loops, [  ];
      redo;
   }

   /\G \z /xgc
      or die("Internal error");
}

my $iter = NestedLoops(\@loops);
while (my @parts = $iter->()) {
   print(join('', @parts), "\n");
}

输出：

This is a test
This is a demo
This isn't a test
This isn't a demo

glob基于版本：

$_ = '{'.join(',', map quotemeta($_), @$_).'}'
   for values(%dict);

my $glob;
for ($template) {
   if (/\G \[ /xgc) {
      /\G ( [^\]]* ) \] /xgc
         or die("Missing \"]\"\n");

      my $var = ;
      length($var)
         or die("Empty \"[]\"\n");

      $dict{$var}
         or die("Unknown var \"$var\"\n");

      $glob .= $dict{$var};
      redo;
   }

   if (/\G ( [^\[]+ ) /xgc) {
      $glob .= ;
      redo;
   }

   /\G \z /xgc
      or die("Internal error");
}

while (defined( my $string = glob($glob) )) {
   print($string, "\n");
}

没有错误检查和特定字典，这可以缩小很多：

$ perl -E'say for glob shift=~s/\[((a)|b)]|(.)/?"\Q":?"{is,isn'\''t}":"{test,demo}"/serg' \
   'This [a] a [b]'
This is a test
This is a demo
This isn't a test
This isn't a demo

Perl - 将字符串扩展到列表中。

Perl - Expand a string into a list.

parsing

templates

cartesian-product