in Pyparsing, what are the differences between MatchFirst, Or, and oneOf
when there are shared characters in the strings like
word, wording, words
Or(['word', 'wording', 'words'])
MatchFirst(['word', 'wording', 'words'])
oneOf(['word', 'wording', 'words'])
From the online docs (https://pythonhosted.org/pyparsing/)
MatchFirst - If two expressions match, the first one listed is the one that will match.
Or - If two expressions match, the expression that matches the longest string will be used.
oneOf - Helper to quickly define a set of alternative Literals, and makes sure to do longest-first testing when there is a conflict, regardless of the input order, but returns a MatchFirst for best performance.
MatchFirst tests the current parse location with each string in its constructor, stopping at the first one to match.
Or tests the current parse location against all of the strings given in its constructor, and will return the longest match.
oneOf generates a Regex or MatchFirst to match the longest match, by reordering the input list when there are alternatives with common start strings to test the longer string first.
oneOf operates on str understood as space separated strings and can be simplistically defined as
oneOf = lambda xs: Or(Literal(x) for x in xs.split(" "))
While Or operates on expressions - ParseElement instances.
So you can see either oneOf as specialization of Or or Or being a generalization of oneOf.
You can write oneOf('foo bar') as Literal('foo') ^ Literal('bar')
but you can't write every Or expression using oneOf.
MatchFirst is the same as Or except conflict resolution method - Or yields the longest match while MatchFirst returns the first match in definition order.
So
expr = Literal('bar') ^ Words(alphanums)
expr.parseString("barstool").asList() == ["barstool"]
but
expr = Literal('bar') | Words(alphanums)
expr.parseString("barstool").asList() == ["bar"]
Related
I am trying to match a pattern using rgep() function as below -
grep("XYZ31__Sheqwqet1__CSV.csv", "^(XYZ)+[0-9]{2}[a-zA-Z_]+(csv)+$")
However unfortunately above expression results in no match. Any pointer towards the right direction will be very helpful.
Thanks for your time
Before the csv there is also a . and some digits. In addition, the order of arguments is pattern, followed by the input x. (if we pass arguments via name, the order wouldn't matter though)
grep( "^(XYZ)+[0-9]{2}[[:alnum:]_.]+(csv)$", "XYZ31__Sheqwqet1__CSV.csv")
#[1] 1
Pattern match is
^- start of the string
(XYZ)+ - one or more occurence of those letters
[0-9]{2} - two digits
[[:alnum:]_.]+ - one or more alpha numeric characters including the additional two
(csv)$- csv at the end of the string
My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.
You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).
location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c
How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.
Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/
Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.
Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?
The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.
Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/
import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary
What is the equivalent in zsh for $BASH_REMATCH, and how is it used?
Alternatively, one could simply use
$match[1]
in place of
$BASH_REMATCH[1]
To make zsh behave the same as bash, use:
setopt BASH_REMATCH
Or within a function consider:
setopt local_options BASH_REMATCH
(this will only set the option within the scope of the function)
Then just use $BASH_REMATCH as you would in bash.
The manual says about BASH_REMATCH:
When set, matches performed with the =~ operator will set the BASH_REMATCH array variable, instead of the default MATCH and match variables. The first element of the BASH_REMATCH array will contain the entire matched text and subsequent elements will contain extracted substrings. This option makes more sense when KSH_ARRAYS is also set, so that the entire matched portion is stored at index 0 and the first substring is at index 1. Without this option, the MATCH variable contains the entire matched text and the match array variable contains substrings.
Then =~ will behave like in bash, but if you want the full behaviour as described in the manual:
string =~ regexp
true if string matches the regular expression regexp. If the option RE_MATCH_PCRE is set regexp is tested as a PCRE regular expression using the zsh/pcre module, else it is tested as a POSIX extended regular expression using the zsh/regex module. Upon successful match, some variables will be updated; no variables are changed if the matching fails.
If the option BASH_REMATCH is not set the scalar parameter MATCH is set to the substring that matched the pattern and the integer parameters MBEGIN and MEND to the index of the start and end, respectively, of the match in string, such that if string is contained in variable var the expression ‘${var[$MBEGIN,$MEND]}’ is identical to ‘$MATCH’. The setting of the option KSH_ARRAYS is respected. Likewise, the array match is set to the substrings that matched parenthesised subexpressions and the arrays mbegin and mend to the indices of the start and end positions, respectively, of the substrings within string. The arrays are not set if there were no parenthesised subexpresssions. For example, if the string ‘a short string’ is matched against the regular expression ‘s(...)t’, then (assuming the option KSH_ARRAYS is not set) MATCH, MBEGIN and MEND are ‘short’, 3 and 7, respectively, while match, mbegin and mend are single entry arrays containing the strings ‘hor’, ‘4’ and ‘6’, respectively.
If the option BASH_REMATCH is set the array BASH_REMATCH is set to the substring that matched the pattern followed by the substrings that matched parenthesised subexpressions within the pattern.
I am trying to build a regular expression in Qt for the following set of strings:
The set can contain all the set of strings of length 1 which does not include r and z.
The set also includes the set of strings of length greater than 1, which start with z, followed by any number of z's but must terminate with a single character that is not r and z
So far I have developed the following:
[a-qs-y]?|z+[a-qs-y]
But it does not work.
The question mark in your regular expression causes the first alternative to either match lowercase strings of length 1 excluding r and z or the empty string, and as the empty string can be matched within any string, the second alternative will never be matched against. The rest of your regular expression matches your specification, although you will probably want to make your regular expression only match entire strings by anchoring it:
QRegularExpression re("^[a-qs-y]$|^z+[a-qs-y]$");
QRegularExpressionMatch match = re.match("zzza");
if (match.hasMatch()) {
QString matched = match.captured(0);
// ...
}
Is it possible to set the grammar to match case insensitively.
so for example a rule:
checkName = 'CHECK' Word;
would match check name as well as CHECK name
Creator of PEGKit here.
The only way to do this currently is to use a Semantic Predicate in a round-about sort of way:
checkName = { MATCHES_IGNORE_CASE(LS(1), #"check") }? Word Word;
Some explanations:
Semantic Predicates are a feature lifted directly from ANTLR. The Semantic Predicate part is the { ... }?. These can be placed anywhere in your grammar rules. They should contain either a single expression or a series of statements ending in a return statement which evaluates to a boolean value. This one contains a single expression. If the expression evaluates to false, matching of the current rule (checkName in this case) will fail. A true value will allow matching to proceed.
MATCHES_IGNORE_CASE(str, regexPattern) is a convenience macro I've defined for your use in Predicates and Actions to do regex matches. It has a case-sensitive friend: MATCHES(str, regexPattern). The second argument is an NSString* regex pattern. Meaning should be obvious.
LS(num) is another convenience macro for your use in Predicates/Actions. It means fetch a Lookahead String and the argument specifies how far to lookahead. So LS(1) means lookahead by 1. In other words, "fetch the string value of the first upcoming token the parser is about to try to match".
Notice that I'm still matching Word twice at the end there. The first Word is necessary for matching 'check' (even though it was already tested in the predicate, it was not matched and consumed). The second Word is for your name or whatever.
Hope that helps.