最简单的方法来匹配字符串数组在Perl中进行搜索？ [英] Simplest way to match array of strings to search in perl?
问题描述
我想要做的就是检查字符串数组对我的搜索字符串，并得到相应的键，所以我可以保存它。有没有用Perl这样做的神奇的方式，还是我注定要使用一个循环？如果是这样，什么是最有效的方式做到这一点？
我是比较新的Perl的（我只写了其他2个脚本），所以我不知道很多神奇的是，只是Perl是魔术= D
引用数组：（1 ='佳能'，2 ='惠普'，3 ='索尼'）
搜索字符串：索尼的Cyber-shot DSC-S600
最终结果是：3
更新：
根据讨论中地图基于解决方案（参见选项＃1 ）可能是最简洁的解决方案，但前提是你不认为地图循环（答案短的版本是：这是一个循环就实现/性能，它不是从语言观理论上讲循环）。 假设，通过建立一个常规的前$ P $你不关心你是否获得3或索尼作为答案，你可以不用在简单的情况下，循环pssion用或逻辑（ | ）从数组，像这样的：
我@strings =（佳能，惠普，索尼）;
我的$ search_in =索尼的Cyber-shot DSC-S600
我的$ combined_search =加入（|，@字符串）;
我@which_found =（$ search_in =〜/（$ combined_search）/）;
打印$ which_found [0] \\ n;
从我的测试运行
结果：索尼
常规前pression会（一旦变量 $ combined_search 是由Perl的插值）采取的形式 /（佳能|惠普|索尼）/ 这是你想要的。
这将无法正常工作，就是如果有任何字符串包含regex的特殊字符（如 | 或）） - 在这种情况下，你需要躲避他们。
注意：我个人认为这个有点作弊，因为为了落实加入（），Perl本身必须某处做一个循环内interpeter。所以，这个答案可能无法满足你的愿望，保持循环少，取决于你是否想要避免出于性能考虑，一个循环，对有清洁剂或短code。
P.S。要获得3，而不是索尼，你将不得不使用一个循环 - 无论是在一个明显的方式，在骨子里的循环做1场;或使用节省你自己写的循环，但将有呼叫下一个循环。库
我公司将提供3替代解决方案。
＃1选项： - 我的最爱。使用地图，我个人仍然认为一个循环：
我@strings =（佳能，惠普，索尼）;
我的$ search_in =索尼的Cyber-shot DSC-S600
我的$ combined_search =加入（|，@字符串）;
我@which_found =（$ search_in =〜/（$ combined_search）/）;
打印$ which_found [0] \\ n;
死未找到除非@which_found;
我的$ strings_index = 0;
我的％strings_indexes = {图$ _ =＆GT; $ strings_index ++} @strings;
我的$指数= 1 + $ strings_indexes {$ which_found [0]};
＃需加1，因为在Perl数组是零索引开始，你想3
＃2选项：使用的背后隐藏着一个很好的CPAN库方法的循环：
使用List :: MoreUtils QW（firstidx）;
我@strings =（佳能，惠普，索尼）;
我的$ search_in =索尼的Cyber-shot DSC-S600
我的$ combined_search =加入（|，@字符串）;
我@which_found =（$ search_in =〜/（$ combined_search）/）;
死未找到！除非@which_found;
打印$ which_found [0] \\ n;
我的$ index_of_found = 1 + firstidx {$ _当量$ which_found [0]} @strings;
＃需加1，因为在Perl数组是零索引开始，你想3
＃3选项：这里是明显的循环方式：
我的$ found_index = -1;
我@strings =（佳能，惠普，索尼）;
我的$ search_in =索尼的Cyber-shot DSC-S600
我的foreach $指数（0 .. $＃字符串）{
接下来，如果$ search_in〜/ $字符串[$指数] /！;
$ found_index = $指数;
持续; ＃退出循环的早期，这就是为什么我没有用地图在这里
}
＃$检查对found_index -1;如果你想3，而不是2加1。
What I want to do is check an array of strings against my search string and get the corresponding key so I can store it. Is there a magical way of doing this with Perl, or am I doomed to using a loop? If so, what is the most efficient way to do this?
I'm relatively new to Perl (I've only written 2 other scripts), so I don't know a lot of the magic yet, just that Perl is magic =D
Reference Array: (1 = 'Canon', 2 = 'HP', 3 = 'Sony')
Search String: Sony's Cyber-shot DSC-S600
End Result: 3
UPDATE:
Based on the results of discussion in this question, depending on your intent/criteria of what constitutes "not using a loop", the map based solution below (see "Option #1) may be the most concise solution, provided that you don't consider map a loop (the short version of the answers is: it's a loop as far as implementation/performance, it's not a loop from language theoretical point of view).
Assuming you don't care whether you get "3" or "Sony" as the answer, you can do it without a loop in a simple case, by building a regular expression with "or" logic (|) from the array, like this:
my @strings = ("Canon", "HP", "Sony");
my $search_in = "Sony's Cyber-shot DSC-S600";
my $combined_search = join("|",@strings);
my @which_found = ($search_in =~ /($combined_search)/);
print "$which_found[0]\n";
Result from my test run: Sony
The regular expression will (once the variable $combined_search is interpolated by Perl) take the form /(Canon|HP|Sony)/ which is what you want.
This will NOT work as-is if any of the strings contain regex special characters (such as | or ) ) - in that case you need to escape them
NOTE: I personally consider this somewhat cheating, because in order to implement join(), Perl itself must do a loop somewhere inside the interpeter. So this answer may not satisfy your desire to remain loop-less, depending on whether you wanted to avoid a loop for performance considerations, of to have cleaner or shorter code.
P.S. To get "3" instead of "Sony", you will have to use a loop - either in an obvious way, by doing 1 match in a loop underneath it all; or by using a library that saves you from writing the loop yourself but will have a loop underneath the call.
I will provide 3 alternative solutions.
#1 option: - my favorite. Uses "map", which I personally still consider a loop:
my @strings = ("Canon", "HP", "Sony");
my $search_in = "Sony's Cyber-shot DSC-S600";
my $combined_search = join("|",@strings);
my @which_found = ($search_in =~ /($combined_search)/);
print "$which_found[0]\n";
die "Not found" unless @which_found;
my $strings_index = 0;
my %strings_indexes = map {$_ => $strings_index++} @strings;
my $index = 1 + $strings_indexes{ $which_found[0] };
# Need to add 1 since arrays in Perl are zero-index-started and you want "3"
#2 option: Uses a loop hidden behind a nice CPAN library method:
use List::MoreUtils qw(firstidx);
my @strings = ("Canon", "HP", "Sony");
my $search_in = "Sony's Cyber-shot DSC-S600";
my $combined_search = join("|",@strings);
my @which_found = ($search_in =~ /($combined_search)/);
die "Not Found!"; unless @which_found;
print "$which_found[0]\n";
my $index_of_found = 1 + firstidx { $_ eq $which_found[0] } @strings;
# Need to add 1 since arrays in Perl are zero-index-started and you want "3"
#3 option: Here's the obvious loop way:
my $found_index = -1;
my @strings = ("Canon", "HP", "Sony");
my $search_in = "Sony's Cyber-shot DSC-S600";
foreach my $index (0..$#strings) {
next if $search_in !~ /$strings[$index]/;
$found_index = $index;
last; # quit the loop early, which is why I didn't use "map" here
}
# Check $found_index against -1; and if you want "3" instead of "2" add 1.
这篇关于最简单的方法来匹配字符串数组在Perl中进行搜索？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
