I want to find whether a given regular expression is a subset of larger regular expression.
For example given a larger regular expression ((a*)(b(a*))), I want to find if a regular expression like (aab.*) or (a.*) matches to it. I am developing a program where I need to find all sub string of given length that can be formed from given regular expression.
$count=0;
$len=0;
sub match{
my $c=$_[1];
my $str=$_[0];
my $reg=$_[2];
#if($str.".*"!~/^$reg$/){
# return;
#}
if($c==$len){
if($str=~/^reg$/){
$count++;
}
return;
}
my $t=$str.'a';
&match($t,$c+1,$reg);
my $t=$str.'b';
&match($str.'b',$c+1,$reg);
}
for(<>){
#arr=split(/\s/,$_);
$len=$arr[1];
&match('a',1,$arr[0]);
&match('b',1,$arr[0]);
print $count;
}
So I thought that I would start strings of given length using recursion and when the string size reaches desired length, I would compare it to original exp. This works fine for small sub strings but runs into stack overflow for larger sub strings. So I was thinking that while generating part of string itself I would check the expression to given reg exp. But that didn't work. For above given reg exp ((a*)(b(a*))) if we compare it to partial string (aa) it will fail as the reg exp doesn’t match. So in order for it to work, I need to compare two regular expression by adding .* behind every partial sub stirng. I tried to find answer on web but was unsuccessful.
I tried the following code but naturally it failed. Can any one suggest some other approach.
if("a.*"=~/((a*)(b(a*)))/){
print match;
}
But here the first part is considered as an actual string. Can you help me how to convert code so I can compare (a.*) as a regular expression instead of string.
I think one approach is to find the length of matched string if it can be done. For instance if you match (aab) to (aac) than you can obtain length where the matched stopped.
Now compare the position where the match stopped, if its equal to length of your string than its equivalent to regex of str(.*). I read that it can be done in some other languages but I am not sure about perl.
Related
I use string.format(str, regex) of LUA to fetch some key word.
local RICH_TAGS = {
"texture",
"img",
}
--\[((img)|(texture))=
local START_OF_PATTER = "\\[("
for index = 1, #RICH_TAGS - 1 do
START_OF_PATTER = START_OF_PATTER .. "(" .. RICH_TAGS[index]..")|"
end
START_OF_PATTER = START_OF_PATTER .. "("..RICH_TAGS[#RICH_TAGS].."))"
function RichTextDecoder.decodeRich(str)
local result = {}
print(str, START_OF_PATTER)
dump({string.find(str, START_OF_PATTER)})
end
output
hello[img=123] \[((texture)|(img))
dump from: [string "utils/RichTextDecoder.lua"]:21: in function 'decodeRich'
"<var>" = {
}
The output means:
str = hello[img=123]
START_OF_PATTER = \[((texture)|(img))
This regex works well with some online regex tools. But it find nothing in LUA.
Is there any wrong using in my code?
You cannot use regular expressions in Lua. Use Lua's string patterns to match strings.
See How to write this regular expression in Lua?
Try dump({str:find("\\%[%("))})
Also note that this loop:
for index = 1, #RICH_TAGS - 1 do
START_OF_PATTER = START_OF_PATTER .. "(" .. RICH_TAGS[index]..")|"
end
will leave out the last element of RICH_TAGS, I assume that was not your intention.
Edit:
But what I want is to fetch several specific word. For example, the
pattern can fetch "[img=" "[texture=" "[font=" any one of them. With
the regex string I wrote in my question, regex can do the work. But
with Lua, the way to do the job is write code like string.find(str,
"[img=") and string.find(str, "[texture=") and string.find(str,
"[font="). I wonder there should be a way to do the job with a single
pattern string. I tryed pattern string like "%[%a*=", but obviously it
will fetch a lot more string I need.
You cannot match several specific words with a single pattern unless they are in that string in a specific order. The only thing you could do is to put all the characters that make up those words into a class, but then you risk to find any word you can build from those letters.
Usually you would match each word with a separate pattern or you match any word and check if the match is one of your words using a look up table for example.
So basically you do what a regex library would do in a few lines of Lua.
I have some doubt about the outcome of a binding operator expression in perl. I mean expression like
string =~ /pattern/
I have done some simple test
$ss="a1b2c3";
say $ss=~/a/; # 1
say $ss=~/[a-z]/g; # abc
#aa=$ss=~/[a-z]/g;say #aa; # abc
$aa=#aa;say $aa; # 3
$aa=$ss=~/[a-z]/g;say $aa; # 1
note the comment part above is the running result.
So here comes the question, what on earth is returned by $ss=~/[a-z]/g, it seems that it returned an array according to code line 3,4,5. But what about the last line, why it gives 1 instead of 3 which is the length of array?
The return of the match operator depends on the context: in list context it returns all captured matches, in scalar context the true/false. The say imposes list context, but in the first example nothing is captured in the regex so you only get "success."
Next, the behavior of /g modifier also differs across contexts. In list context, with it the string keeps being scanned with the given pattern until all matches are found, and a list with them is returned. These are your second and third examples.
But in scalar context its behavior is a bit specific: with it the search will continue from the position of the last match, the next time round. One typical use is in the loop condition
while (/(\w+)/g) { ... }
This is a bit of a tokenizer: after the body of the loop runs the next word is found, etc.
Then the last example doesn't really make sense; you are getting the "normal" scalar-context matching success/fail, and /g doesn't do anything -- until you match on $ss the next time
perl -wE'
$s=shift||q(abc);
for (1..2) { $m = $s=~/(.)/g; say "$m: $1"; }
'
prints lines 1:a and then 1:b.
Outside of iterative structures (like while condition) the /g in scalar context is usually an error, pointless at best or a quiet bug.
See "Global matching" under "Using regular expressions" in perlretut for /g.
See regex operators in perlop in general, and about /g as well. A useful tool to explore /g workings is pos.
I have a list of records that are character vectors. Here's an example:
'1mil_0,1_1_1_lb200_ks_drivers_sorted.csv'
'1mil_0_1_lb100_ks_drivers_sorted.csv'
'1mil_1_1_lb2_100_100_ks_drivers_sorted.csv'
'1mil_1_1_lb100_ks_drivers_sorted.csv'
From these names I would like to extract whatever's between the two substrings 1mil_ and _ks_drivers_sorted.csv.
So in this case the output would be:
0,1_1_1_lb200
0_1_lb100
1_1_lb2_100_100
1_1_lb100
I'm using MATLAB so I thought to use regexp to do this, but I can't understand what kind of regular expression would be correct.
Or are there some other ways to do this without using regexp?
Let the data be:
x = {'1mil_0,1_1_1_lb200_ks_drivers_sorted.csv'
'1mil_0_1_lb100_ks_drivers_sorted.csv'
'1mil_1_1_lb2_100_100_ks_drivers_sorted.csv'
'1mil_1_1_lb100_ks_drivers_sorted.csv'};
You can use lookbehind and lookahead to find the two limiting substrings, and match everything in between:
result = cellfun(#(c) regexp(c, '(?<=1mil_).*(?=_ks_drivers_sorted\.csv)', 'match'), x);
Or, since the regular expression only produces one match, the following simpler alternative can be used (thanks #excaza for noticing):
result = regexp(x, '(?<=1mil_).*(?=_ks_drivers_sorted\.csv)', 'match', 'once');
In your example, either of the above gives
result =
4×1 cell array
'0,1_1_1_lb200'
'0_1_lb100'
'1_1_lb2_100_100'
'1_1_lb100'
For me the easy way to do this is just use espace or nothing to replace what you don't need in your string, and the rest is what you need.
If is a list, you can use a loop to do this.
Exemple to replace "1mil_" with "" and "_ks_drivers_sorted.csv" with ""
newChr = strrep(chr,'1mil_','')
newChr = strrep(chr,'_ks_drivers_sorted.csv','')
I need to search a cell array and return a single boolean value indicating whether any cell matches a regular expression.
For example, suppose I want to find out if the cell array strs contains foo or -foo (case-insensitive). The regular expression I need to pass to regexpi is ^-?foo$.
Sample inputs:
strs={'a','b'} % result is 0
strs={'a','foo'} % result is 1
strs={'a','-FOO'} % result is 1
strs={'a','food'} % result is 0
I came up with the following solution based on How can I implement wildcard at ismember function of matlab? and Searching cell array with regex, but it seems like I should be able to simplify it:
~isempty(find(~cellfun('isempty', regexpi(strs, '^-?foo$'))))
The problem I have is that it looks rather cryptic for such a simple operation. Is there a simpler, more human-readable expression I can use to achieve the same result?
NOTE: The answer refers to the original regexp in the question: '-?foo'
You can avoid the find:
any(~cellfun('isempty', regexpi(strs, '-?foo')))
Another possibility: concatenate first all cells into a single string:
~isempty(regexpi([strs{:}], '-?foo'))
Note that you can remove the "-" sign in any of the above:
any(~cellfun('isempty', regexpi(strs, 'foo')))
~isempty(regexpi([strs{:}], 'foo'))
And that allows using strfind (with lower) instead of regexpi:
~isempty(strfind(lower([strs{:}]),'foo'))
I have to parse various strings and determine a prefix, number, and suffix. The problem is the strings can come in a wide variety of formats. The best way for me to think about how to parse it is to find the longest number in the string, then take everything before that as a prefix and everything after that as a suffix.
Some examples:
0001 - No prefix, Number = 0001, No suffix
1-0001 - Prefix = 1-, Number = 0001, No suffix
AAA001 - Prefix = AAA, Number = 001, No suffix
AAA 001.01 - Prefix = AAA , Number = 001, Suffix = .01
1_00001-01 - Prefix = 1_, Number = 00001, Suffix = -01
123AAA 001_01 - Prefix = 123AAA , Number = 001, Suffix = _01
The strings can come with any mixture of prefixes and suffixes, but the key point is the Number portion is always the longest sequential list of digits.
I've tried a variety of RegEx's that work with most but not all of these examples. I might be missing something, or perhaps a RegEx isn't the right way to go in this case?
(The RegEx should be .NET compatible)
UPDATE: For those that are interested, here's the C# code I came up with:
var regex = new System.Text.RegularExpressions.Regex(#"(\d+)");
if (regex.IsMatch(m_Key)) {
string value = "";
int length;
var matches = regex.Matches(m_Key);
foreach (var match in matches) {
if (match.Length >= length) {
value = match.Value;
length = match.Length;
}
}
var split = m_Key.Split(new String[] {value}, System.StringSplitOptions.RemoveEmptyEntries);
m_KeyCounter = value;
if (split.Length >= 1) m_KeyPrefix = split(0);
if (split.Length >= 2) m_KeySuffix = split(1);
}
You're right, this problem can't be solved purely by regular expressions. You can use regexes to "tokenize" (lexically analyze) the input but after that you'll need further processing (parsing).
So in this case I would tokenize the input with (for example) a simple regular expression search (\d+) and then process the tokens (parse). That would involve seeing if the current token is longer than the tokens seen before it.
To gain more understanding of the class of problems regular expressions "solve" and when parsing is needed, you might want to check out general compiler theory, specifically when regexes are used in the construction of a compiler (e.g. http://en.wikipedia.org/wiki/Book:Compiler_construction).
You're input isn't regular so, a regex won't do. I would iterate over the all groups of digits via (\d+) and find the longest and then build a new regex in the form of (.*)<number>(.*) to find your prefix/suffix.
Or if you're comfortable with string operations you can probably just find the start and end of the target group and use substr to find the pre/suf fix.
I don't think you can do this with one regex. I would find all digit sequences within the string (probably with a regex) and then I would select the longest with .NET code, and call Split().
This depends entirely on your Regexp engine. Check your Regexp environment for capturing, there might be something in it like the automatic variables in Perl.
OK, let's talk about your question:
Keep in mind, that both, NFA and DFA, of almost every Regexp engine are greedy, this means, that a (\d+) will always find the longest match, when it "stumbles" over it.
Now, what I can get from your example, is you always need middle portion of a number, try this:
/^(.*\D)?(\d+)(\D.*)?$/ig
The now look at variables $1, $2, $3. Not all of them will exist: if there are all three of them, $2 will hold your number in question, the other vars, parts of the prefix. when one of the prefixes is missing, only variable $1 and $2 will be set, you have to see for yourself, which one is the integer. If both prefix and suffix are missing, $1 will hold the number.
The idea is to make the engine "stumble" over the first few characters and start matching a long number in the middle.
Since the modifier /gis present, you can loop through all available combinations, that the machine finds, you can then simply take the one you like most or something.
This example is in PCRE, but I'm sure .NET has a compatible mode.