正则表达式匹配每个字符至少一次

Question

我正在写一个解决谜题的应用程序。我正在搜索一些字符组合的文本，如果我有一组字符[abcd]，那么我需要在文本中查找仅包含字符abcd的子字符串，并且还必须包含至少一个字符 - 以便字符abcd匹配dcba或abbcdd，但不匹配acd，bbcd或abced。如果使用正则表达式[abcd] +，那么我将得到不包含每个字符的子字符串

Answer 1

为什么要在这里使用正则表达式？

def hasChars(search_string, chars): 
    return all(x in search_string for x in chars) 

>>> hasChars('aaabcd', 'abc') 
True 

Answer 2

如果字符串必须至少包含abcd的，但可以包含其他人，那么这将工作

(?=.*a)(?=.*b)(?=.*c)(?=.*d)

如果，以及它们只能包含abcd的，那么这可能是更好的

^(?=.*a)(?=.*b)(?=.*c)(?=.*d)[abcd]+$

更新

要回答你的问题，如果你正在寻找一个浮动的版本，这个你想要做什么：

(?=([abcd]{4,}))(?=[bcd]*a)(?=[acd]*b)(?=[abd]*c)(?=[abc]*d)\1

扩展：

 # At POSition 
(?=    # Lookahead 
    (     # Capture grp 1 
     [abcd]{4,}   # Get 4 or more (greedy) 'a' or 'b' or 'c' or 'd' characters 
    ) 
) 
(?=    # Lookahead, check for 'a' (still at POS) 
    [bcd]*a    # 0 or more [bcd]'s then 'a' 
) 
(?=    # Lookahead, check for 'b' (still at POS) 
    [acd]*b    # 0 or more [acd]'s then 'b' 
) 
(?=    # Lookahead, check for 'c' (still at POS) 
    [abd]*c    # 0 or more [abd]'s then 'c' 
) 
(?=    # Lookahead, check for 'd' (still at POS) 
    [abc]*d    # 0 or more [abc]'s then 'd' 
) 
\1     # Backref to capt grp 1, consume it 

    # Passed test, now at POSition + length of capture group 1 

更

你可能能够从搜索字符串中系统地构造正则表达式。我不太了解python，所以这里是一个如何在Perl中完成的示例。但请注意，字符串越长，查找匹配所需的时间就越长，但这应该相当快。

use strict; 
use warnings; 

my $samp = 'bddaaabcabbad characters abcd matches dcba or abbcdd, but not acd, bbcd or abced'; 

my $regex = '(?=([abcd]{4,}))(?=[bcd]*a)(?=[acd]*b)(?=[abd]*c)(?=[abc]*d)\1'; 

while ($samp =~/$regex/xg) 
{ 
    print "Found '$1'\n"; 
} 

# Regex construction 
# ------------------------------ 
my @AryofSearchStrs = (
'abcd', 
'%^&*', 
'hi()there', 
'==-yes', 
); 

for my $search_string (@AryofSearchStrs) 
{ 
    my $str = $search_string; 
    while($str =~ s/(.)(.*)\1/$1$2/g) {} 

    my @astr = split '', $str; 

    my $rxformed = '(?=([' . quotemeta($str) . ']{' . length($str) . ',}))'; 
    for (my $i = 0; $i < @astr; $i++) 
    { 
     $rxformed .= 
     '(?=[' 
     . join('', map { quotemeta($_) } @astr[0..($i-1), ($i+1)..$#astr]) 
     . ']*' 
     . quotemeta($astr[$i]) 
     . ')'; 
    } 
    $rxformed .= '\1'; 

    print "\n\n============\n"; 
    print "Search string = '$search_string'\n"; 
    print "Normalized = '$str'\n"; 
    print "Formed regex = \n$rxformed\n"; 
} 

输出

Found 'bddaaabcabbad' 
Found 'abcd' 
Found 'dcba' 
Found 'abbcdd' 


============ 
Search string = 'abcd' 
Normalized = 'abcd' 
Formed regex = 
(?=([abcd]{4,}))(?=[bcd]*a)(?=[acd]*b)(?=[abd]*c)(?=[abc]*d)\1 


============ 
Search string = '%^&*' 
Normalized = '%^&*' 
Formed regex = 
(?=([\%\^\&\*]{4,}))(?=[\^\&\*]*\%)(?=[\%\&\*]*\^)(?=[\%\^\*]*\&)(?=[\%\^\&]*\*)\1 


============ 
Search string = 'hi()there' 
Normalized = 'hi()ter' 
Formed regex = 
(?=([hi\(\ \)ter]{8,}))(?=[i\(\ \)ter]*h)(?=[h\(\ \)ter]*i)(?=[hi\ \)ter]*\()(?=[hi\(\)ter]*\)(?=[hi\(\ ter]*\))(?=[hi\(\ \)er]*t)(?=[hi\(\ \)tr]*e)(?=[hi\(\ \)te]*r)\1 


============ 
Search string = '==-yes' 
Normalized = '=-yes' 
Formed regex = 
(?=([\=\-yes]{5,}))(?=[\-yes]*\=)(?=[\=yes]*\-)(?=[\=\-es]*y)(?=[\=\-ys]*e)(?=[\=\-ye]*s)\1