Perl - 捕獲一個數組中出現多個元素的句子

Question

我有一個文本文件和一個包含單詞列表的數組。我需要找到一種方式，在這種情況下，我可以過濾掉出現超過1個句子的句子。我只是無法制定如何編寫代碼。這裏有一個例子：

輸入：

my @strings = (
    "i'm going to find the occurrence of two words if possible", 
    "i'm going to find the occurrence of two words if possible", 
    "to find a solution to this problem", 
    "i will try my best for a way to this problem" 
); 

my @words = ("find", "two", "way"); 

輸出：

i'm going to find the occurrence of two words if possible 
i'm going to find the occurrence of two words if possible 

而且我也知道這是一個簡單的問題，但我的心似乎已經打了路障。

Answer 1

如果你想使用關鍵字的兩個或多個實例的字符串：

my @keywords = ("find", "two", "way"); 
my %keywords = map { $_ => 1 } @keywords; 

for my $string (@strings) { 
    my @words = $string =~ /\w+/g; 
    my $count = grep { $keywords{$_} } @words; # Count words that are keywords. 
    if ($count >= 2) { 
     ... 
    } 
} 

短路交替（即好極長的字符串）：

my @keywords = ("find", "two", "way"); 
my %keywords = map { $_ => 1 } @keywords; 

for my $string (@strings) { 
    my $count = 0; 
    while ($string =~ /\w+/g) { 
     if ($keywords{$_} && ++$count == 2) { 
     ... 
     last; 
     } 
    } 
} 

---

如果你想具有兩個或多個關鍵字實例的字符串：

my @keywords = ("find", "two", "way"); 

for my $string (@strings) { 
    my @words = $string =~ /\w+/g; 
    my %seen; ++$seen{$_} for @words; 
    my $count = grep { $seen{$_} } @keywords; # Count keywords that were seen. 
    if ($count >= 2) { 
     ... 
    } 
} 

候補：

my @keywords = ("find", "two", "way"); 

for my $string (@strings) { 
    my @words = $string =~ /\w+/g; 
    my %seen = map { $_ => -1 } @keywords; 
    my $count = grep { ++$seen{$_} == 0 } @words; 
    if ($count >= 2) { 
     ... 
    } 
} 

的短路交替（即適用於極長的琴絃）：

my @keywords = ("find", "two", "way"); 

for my $string (@strings) { 
    my $count = 0; 
    my %seen = map { $_ => -1 } @keywords; 
    while ($string =~ /\w+/g) { 
     if (++$seen{$_} == 0 && ++$count == 2) { 
     ... 
     last; 
     } 
    } 
}