是否有 Perl 快捷方式来计算字符串中的匹配项数？

假设我有:

my $string = "one.two.three.four";

我应该如何处理上下文来获得模式找到匹配的次数(3) ？这可以通过使用一行程序来完成吗？

我试过了:

my ($number) = scalar($string=~/\./gi);

我认为通过在 $number周围放置括号，我可以强制数组上下文，通过使用 scalar，我可以得到计数。然而，我得到的只有 1。

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小开

Try this:

my $string = "one.two.three.four";
my ($number) = scalar( @{[ $string=~/\./gi ]} );

It returns 3 for me. By creating a reference to an array the regular expression is evaluated in list context and the @{..} de-references the array reference.

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最佳答案

That puts the regex itself in scalar context, which isn't what you want. Instead, put the regex in list context (to get the number of matches) and put that into scalar context.

 my $number = () = $string =~ /\./gi;
小开

I think the clearest way to describe this would be to avoid the instant-cast to scalar. First assign to an array, and then use that array in scalar context. That's basically what the = () = idiom will do, but without the (rarely used) idiom:

my $string = "one.two.three.four";
my @count = $string =~ /\./g;
print scalar @count;
小开

Also, see Perlfaq4 :

There are a number of ways, with varying efficiency. If you want a count of a certain single character (X) within a string, you can use the tr/// function like so:

$string = "ThisXlineXhasXsomeXx'sXinXit";
$count = ($string =~ tr/X//);
print "There are $count X characters in the string";

This is fine if you are just looking for a single character. However, if you are trying to count multiple character substrings within a larger string, tr/// won't work. What you can do is wrap a while() loop around a global pattern match. For example, let's count negative integers:

$string = "-9 55 48 -2 23 -76 4 14 -44";
while ($string =~ /-\d+/g) { $count++ }
print "There are $count negative numbers in the string";

Another version uses a global match in list context, then assigns the result to a scalar, producing a count of the number of matches.

$count = () = $string =~ /-\d+/g;
小开

another way,

my $string = "one.two.three.four";
@s = split /\./,$string;
print scalar @s - 1;
小开

Is the following code a one-liner?

print $string =~ s/\./\./g;
小开

Friedo's method is: $a = () = $b =~ $c.

But it's possible to simplify this even further to just ($a) = $b =~ $c, like so :

my ($matchcount) = $text =~ s/$findregex/ /gi;

You could thank just wrap this up in a function, getMatchCount(), and not worry about it destroying the passed string.

On the other hand, you can add in a swap, which may be a bit more computation, but does not result in altering the string.

my ($matchcount) = $text =~ s/($findregex)/$1/gi;
小开 
my $count = 0;
my $pos = -1;
while (($pos = index($string, $match, $pos+1)) > -1) {
$count++;
}

checked with Benchmark, it's pretty fast

小开

I noticed that if you have an OR condition in your regular expression (eg /(K..K)|(V.AK)/gi ) then the array produced may have undefined elements which are included in the count at the end.

For example:

my $seq = "TSYCSKSNKRCRRKYGDDDDWWRSQYTTYCSCYTGKSGKTKGGDSCDAYYEAYGKSGKTKGGRNNR";
my $regex = '(K..K)|(V.AK)';
my $count = () = $seq =~ /$regex/gi;
print "$count\n";

Gives a value of count of 6.

I found the solution in this post How do I remove all undefs from array?

my $seq = "TSYCSKSNKRCRRKYGDDDDWWRSQYTTYCSCYTGKSGKTKGGDSCDAYYEAYGKSGKTKGGRNNR";
my $regex = '(K..K)|(V.AK)';
my @count = $seq =~ /$regex/gi;
@count = grep defined, @count;
my $count = scalar @count;
print "$count\n";

Which then gives the correct answer of three.


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