I've got a question on failure function description from "Compilers: Principles, Techniques, and Tools" aka DragonBook
Firstly, the quote:
In order to process text strings rapidly and search those strings for a keyword,
it is useful to define, for keyword b1b2...bn, and position s in that keyword , a failure function, f (s) ...
The objective is that b1b2.. - bf(s) is the longest proper prefix of
b1...bs, that is also a suffix of b1...bs. The reason f (s) is important is that
if we are trying to match a text string for blb2..bn, and we have matched the
first s positions, but we then fail (i.e., the next position of the text string does
not hold bs+l), then f (s) is the longest prefix of b1..bn that could possibly
match the text string up to the point we are at. Of course, the next character of
the text string must be bf(s)+1 or else we still have problems and must consider
a yet shorter prefix, which will be bf(f(s)).
So, the questions:
1. If we've matched s positions with the text, why f (s) is the longest prefix of b1..bn that matches the string? I think s - is the longest prefix.
2. Next character of the text string must be bf(s)+1, why? We have a mismatch at this position, does it matter at all what the char is?
f(s) is the longest prefix at that position that might match the entire keyword. The idea is not to try to match the keyword with the text from the start, but to find a position where the keyword appears.
Consider a search for the word 'aaaba' in the text 'aaaabaa'. The match fails after the three first a's, but it's not necessary to retry from the second 'a', since we know that if the next letter is a 'b' (which it is), we may have a match there.
Related
I am tryng to get rid of shortcodes inside a Google Sheet column. I have many items such as [spacer type="1" height="20"][spacer] or [FinalTilesGallery id="37"] I just would like to cancel them. Is there any simple way to do it?
Thanks !
For in-place replacement, the quick option would be to use the Find and Replace dialog (Ctrl + H) with Search Using Regular Expressions turned on, which is more powerful than your standard Find and Replace.
Find: \[.*?\] - Match anything within an open-bracket up to the very next close-bracket. This should work assuming you have no nested brackets, e.g. [[no][no]].
If you do have nested brackets, you'll have to change this to \[[^\[\]]*\]. And continue to Replace All until all the codes are gone.
Replace: Nothing.
Replace All. If you don't want to affect other sheets that may be in your document, make sure you select the right range to work with, too.
This just erases everything within the brackets.
If you want to erase any redundant spaces left by this, simply Find and Replace again (with Regular Expressions) on + (space and plus), which will match 1 or more spaces and replace with (single space).
E.g.:
string [] [] string2 -> string string2 after the shortcode replacement.
After replacing spaces, it will become string string2.
Let's say your original strings are in the range A2:A. Place the following into B2 of an otherwise completely empty Column B (or the second cell of any other empty column):
=ArrayFormula(IF(A2:A="",,TRIM(REGEXREPLACE(A2:A,"\[[^\[\]]+\]",""))))
I can't see your data, so I don't know what kind of information is between these shortcodes. If you find that this leaves you with concatenated pieces of data where there should be spaces between them, replace the above with this version:
=ArrayFormula(IF(A2:A="",,TRIM(REGEXREPLACE(SUBSTITUTE(SUBSTITUTE(A2:A,"["," ["),"]","] "),"\[[^\[\]]+\]",""))))
I can't teach regular expression language here. But I will note that, since square brackets have specific meaning within regex, your literal square brackets must be indicated with the escape character: the backslash.
Here is the regex expression alone:
\[[^\[\]]+\]
The opening \[ and the closing \], then, reference your actual opening and closing bracket sets. If we remove those, we have this left:
[^\[\]]+
Again, you see the escaped opening and closing square brackets, which I'll replace with the word these:
[^these]+
What remains there are opening and closing brackets with regex meaning, i.e., "anything in this group." And the circumflex symbol ^ as the first character within this set of square brackets means "anything except." The + symbol means "in any string length of one or more characters."
So that whole regex expression then reads: "A literal open square bracket, followed by one or more characters that are anything except square brackets, ending with a literal closing square bracket."
And we are REGEXREPLACE-ing any instance of that with "" (i.e., nothing).
I have a plain text file with a one string per line. I'd like to identify any instances where a string contains a value outside of a restricted character set. In this particular instance, if the string contains any character outside of the set "[THADGRC.SMBN-WVKY]" I want to retain it and pass it along to a new file.
For example, let's say the original file "mystrings.txt" contained the following data:
THADGRC.SMBN-WVKY
YKVW-NBMS.CRGDHAT
THADGRC.SMBN-WVKYI
My intention is to retain only the third sequence, because it contains a character outside of the allowed set (I) in this case.
It doesn't matter how many times, or in what order, an allowed character is present - all I care about is if a character exists in that string outside of the allowed set.
Originally I tried:
cat mystrings.txt | grep -v [THADGRC\.SMBN-WVKY] > badstrings.txt
but of course the third string contains those allowed character in addition to the non-allowed characters, thus this search ended up producing no "offending" strings.
Last thing: I'm not sure what characters outside of the allowed set might exist in this text file. It would be great to know ahead of time to just search for anything with an "I", but I don't actually know this ahead of time.
So the question: is there a way to use grep (or another tool, say awk?) to pass in a restricted list of characters, and flag any instances where a string contains any number of characters outside of that set?
Thanks for your consideration
I think that your problem is N-W. This doesn't match "N", "-" and "W", it matches a range from "N" to "W". You should move "-" to the end of the character class, or escape it. I suggest changing to:
grep '[^THADGRC.SMBNWVKY-]' mystrings.txt
Also, note that "." doesn't have to be escaped when it's inside a character class.
Your attempt says "remove any lines which contain one of these characters at least once". But you want "print any lines which contain at least one character not in this set."
(Also, quote your regular expressions , and lose the useless cat.)
grep '[^-THADGRC.SMBNWVKY]' mystrings.txt > badstrings.txt
I moved the dash to the beginning of the character class on the assumption that you want a literal dash, not the regex range N-W (i.e. N, O, P, Q, R, S, T, U, V, W).
I need to check wether matching parenthesis is present in a string that might have emoticons (like :) or :(). For example, "(:)())()", "(abcd)()ghijk)((mnop)qert)"
I have used the patterns "^[:\\(|:\\)]" to check for emoticons and "\\([^()]*\\)" to check for matching parenthesis present, but they are not detected. How can I do this?
The really simple solution to this problem is to count the parentheses, trying to solve it with regular expressions is hard though extended regular expressions can handle it. Here is a sketch of the simple algorithm:
Set openParenthesisCount to 0
Iterate over the string:
If current character is ( increment openParenthesisCount
If current character is ) decrement openParenthesisCount, if count goes negative then fail (too many closing)
If current character is : lookahead and skip next character if it is a parenthesis (skip smilies)
If openParenthesisCount is zero => succeed
HTH
As far as I can tell, you want to match a string if and only if it contains matching parentheses, after ignoring every occurrence of ":)" and ":(" in the string, if any.
So, try this:
^((?!:).)*\(.*(?<!:)\).*
It will match the following strings:
()
(abd)
(())(
(:))
(:(:))
(:)())()
(abcd)()ghijk)((mnop)qert)
(abc):
(:abc)
But will NOT match the following:
)(
(:)
(:(
:(:)
:()
(:)
:()(:)
(
)
(abc
abc)
(abc:)
:(abc)
I have some strings with a sentence and i need to subdivise it into a substring of maximum 40 characters.
But i don't want to split the sentence in the middle of a word.
I tried with .gsub function but it's return 40 characters maximum and avoid to cut the string in the middle of a word. But it's return only the first occurence.
sentence[0..40].gsub(/\s\w+$/,'')
I tried with split but i can select only the fist 40 characters and split in the middle of a word...
sentence.split(...){40}
My string is "Sure, we will show ourselves only when we know the east door has been opened.".
The string output i want is
["Sure, we will show ourselves only when we","know the east door has
been opened."]
Do you have a solution ? Thanks
Your first attempt:
sentence[0..40].gsub(/\s\w+$/,'')
almost works, but it has one fatal flaw. You are splitting on the number of characters before cutting off the last word. This means you have no way of knowing whether the bit being trimmed off was a whole word, or a partial word.
Because of this, your code will always cut off the last word.
I would solve the problem as follows:
sentence[/\A.{0,39}[a-z]\b/mi]
\A is an anchor to fix the regex to the start of the string.
.{0,39}[a-z] matches on 1 to 40 characters, where the last character must be a letter. This is to prevent the last selected character from being punctuation or space. (Is that desired behaviour? Your question didn't really specify. Feel free to tweak/remove that [a-z] part, e.g. [a-z.] to match a full stop, if desired.)
\b is a word boundary look-around. It is a zero-width matcher, on beginning/end of words.
/mi modifiers will include case insensitive (i.e. A-Z) and multi-line matches.
One very minor note is that because this regex is matching 1 to 40 characters (rather than zero), it is possible to get a null result. (Although this is seemingly very unlikely, since you'd need a 1-word, 41+ letter string!!) To account for this edge case, call .to_s on the result if needed.
Update: Thank you for the improved edit to your question, providing a concrete example of an input/result. This makes it much clearer what you are asking for, as the original post was somewhat ambiguous.
You could solve this with something like the following:
sentence.scan(/.{0,39}[a-z.!?,;](?:\b|$)/mi)
String#scan returns an array of strings that match the pattern - so you can then re-join these strings to reconstruct the original.
Again, I have added a few more characters (!?,;) to the list of "final characters in the substring". Feel free to tweak this as desired.
(?:\b|$) means "either a word boundary, or the end of the line". This fixes the issue of the result not including the final . in the substrings. Note that I have used a non-capture group (?:) to prevent the result of scan from changing.
I am writing a parse for a script language.
I need to recognize strings, integers and floats.
I successfully recognize strings with the rule:
[a-zA-Z0-9_]+ {return STRING;}
But I have problem recognizing Integers and Floats. These are the (wrong) rules I wrote:
["+"|"-"][1-9]{DIGIT}* { return INTEGER;}
["+"|"-"]["0." | [1-9]{DIGIT}*"."]{DIGIT}+ {return FLOAT;}
How can I fix them?
Furthermore, since a "abc123" is a valid string, how can I make sure that it is recognized as a string and not as the concatenation of a string ("abc") and an Integer ("123") ?
First problem: There's a difference between (...) and [...]. Your regular expressions don't do what you think they do because you're using the wrong punctuation.
Beyond that:
No numeric rule recognizes 0.
Both numeric rules require an explicit sign.
Your STRING rule recognizes integers.
So, to start:
[...] encloses a set of individual characters or character ranges. It matches a single character which is a member of the set.
(...) encloses a regular expression. The parentheses are used for grouping, as in mathematics.
"..." encloses a sequence of individual characters, and matches exactly those characters.
With that in mind, let's look at
["+"|"-"][1-9]{DIGIT}*
The first bracket expression ["+"|"-"] is a set of individual characters or ranges. In this case, the set contains: ", +, " (again, which has no effect because a set contains zero or one instances of each member), |, and the range "-", which is a range whose endpoints are the same character, and consequently only includes that character, ", which is already in the set. In short, that was equivalent to ["+|]. It will match one of those three characters. It requires one of those three characters, in fact.
The second bracket expression [1-9] matches one character in the range 1-9, so it probably does what you expected. Again, it matches exactly one character.
Finally, {DIGIT} matches the expansion of the name DIGIT. I'll assume that you have the definition:
DIGIT [0-9]
somewhere in your definitions section. (In passing, I note that you could have just used the character class [:digit:], which would have been unambiguous, and you would not have needed to define it.) It's followed by a *, which means that it will match zero or more repetitions of the {DIGIT} definition.
Now, an example of a string which matches that pattern:
|42
And some examples of strings which don't match that pattern:
-7 # The pattern must start with |, + or "
42 # Again, the pattern must start with |, + or "
+0 # The character following the + must be in the range [0-9]
Similarly, your float pattern, once the [...] expressions are simplified, becomes (writing out the individual pieces one per line, to make it more obvious):
["+|] # i.e. the set " + |
["0.|[1-9] # i.e. the set " 0 | [ 1 2 3 4 5 6 7 8 9
{DIGIT}* # Any number of digits
"." # A single period
] # A single ]
{DIGIT}+ # one or more digits
So here's a possible match:
"..]3
I'll skip over writing out the solution because I think you'll benefit more from doing it yourself.
Now, the other issues:
Some rule should match 0. If you don't want to allow leading zeros, you'll need to just a it as a separate rule.
Use the optional operator (?) to indicate that the preceding object is optional. eg. "foo"? matches either the three characters f, o, o (in order) or matches the empty string. You can use that to make the sign optional.
The problem is not the matching of abc123, as in your question. (F)lex always gives you the longest possible match, and the only rule which could match the starting character a is the string rule, so it will allow the string rule to continue as long as it can. It will always match all of abc123. However, it will also match 123, which you would probably prefer to be matched by your numeric rule. Here, the other (f)lex matching criterion comes into play: when there are two or more rules which could match exactly the same string, and none of the rules can match a longer string, (f)lex chooses the first rule in the file. So if you want to give numbers priority over strings, you have to put the number rule earlier in your (f)lex file than the string rule.
I hope that gives you some ideas about how to fix things.