如何在Perl中从字符串中删除一个特定字符?

I'm trying to remove a specific character from a string in Perl:

我试图从Perl中的字符串中删除一个特定的字符:

my $string="MATTHATBAT";
substr($string, 2, 1, '');

EDIT: This does work, sorry. Leaving this here in case someone needs to know how to do this.

编辑:对不起，这是可行的。把这个放在这里以防有人需要知道怎么做。

Also, is there a more efficient way of doing this?

还有更有效的方法吗?

The string should now be MATHATBAT.

绳子现在应该是MATHATBAT。

Am I missing something? I know that I can use regex s///, but I am iterating through the string, looking for a certain character (this char changes), then removing the character (but only that character at that offset). So eventually, I will be removing the second or third occurence of the character (i.e. MATTHABAT, MATTHATBA and even MATHABAT etc)

我遗漏了什么东西?我知道我可以使用regex s///，但是我正在遍历字符串，寻找一个特定的字符(这个字符改变了)，然后删除字符(但只删除那个偏移量中的字符)。所以最终，我将会删除角色的第二或第三个角色(即MATTHABAT, MATTHATBA，甚至MATHABAT等)

Can I maybe do this using search and replace? I am using a for loop to iterate through offsets.

我可以用搜索和替换吗?我使用for循环来迭代偏移量。

3 个解决方案

#1

Here is a benchmark comparing regexp vs substr:

这里有一个对比regexp和substr的基准:

#!/usr/bin/perl 
use 5.10.1;
use warnings;
use strict;
use Benchmark qw(:all);

my $count = -3;
my $r = cmpthese($count,
  {
    'substring' => sub {
        my $string = "MATTHATBAT";
        substr($string, 2, 1, ''); 
    },
    'regexp' => sub {
        my $string = "MATTHATBAT";
        $string =~ s/(.{2})./$1/;
    },
  }
);

Result:

结果:

               Rate    regexp substring
regexp     162340/s        --      -93%
substring 2206122/s     1259%        --

As you can see, substr is about 13.5 times as fast as regex.

可以看到，substr的速度是regex的13.5倍。

@Sinan Ünür 1259% is 13.5 times and not 12.5 times.

@Sinan Unur 1259%是13.5倍而不是12.5倍。

#2

Your example does work. The $string will contain MATHATBAT as you wanted to have, the problem is somewhere else, not in this part.

你的例子。$string将包含您想要的MATHATBAT，问题在其他地方，而不是这一部分。

#3

You can loop the reg.exp matches with //g

你可以循环这个reg。exp匹配/ / g

from perlrequick :

从perlrequick:

   $x = "cat dog house"; # 3 words
   while ($x =~ /(\w+)/g) {
       print "Word is $1, ends at position ", pos $x, "\n";
   }

think you can modify $x while iterating .. or you could store pos $x in an array and remove then afterwards

我想你可以在迭代时修改$x。或者可以将pos $x存储在一个数组中，然后删除

#1