I'm using iPhone SDK and need of assistance.
I have the following array of search terms:
f, f, last, m
And a string (the top numbers are just indexes for convenience):
0 10 21 30 45
firstname middlename lastname firstnameagain firstnomatch
Expected result ranges (location, length): (0, 1) (30, 1) (21, 4) (10, 1)
I would like to have a regular expression that matches all the search terms in the string but only if they are a prefix of a word. When the search term appears N times, it will match only the first N prefixes (In the example, "f" was entered twice thus the possible match at (45, 1) wasn't returned).
I've tried to write many possible regular expressions and all have failed to match with proper results. I've came to the conclusion that the "\b" meta-character cannot be used because the string can contain non word characters adjacent to letters ("firstname#", "?lastName", ...)
I'm not sure this is a job for a regex alone.
This could be accomplished with or without a regex, by exploding the string by spaces and then checking each string against a list of remaining prefixes.
You keep a list of prefixes as such:
[f, f, last, m]
And when you match against firstname, you remove the prefix that was found. In this case, you remove f:
[f, last, m]
When you match against firstnameagain, you remove f again, and no longer try to match f:
[last, m]
To perform the search, you could iterate through the array of search prefixes and use the NSString rangeOfString function to see if the search prefix is at the front. If you still want to use a regex, you can match against:
#"^(f|f|last|m)"
As you match prefixes, remove them from the array, then remake the regex.
Related
please i want to validate the inputs from a user, the format for the inputs would be: 3 uppercase characters, 3 integer numbers, an optional space, a -, an optional space, either a 'LAB or ((EN or ENLH) with 1 interger number ranging from a [1-9]).
The regex i wrote is
/\D{3}\d{3}\s?-\s?(LAB|(EN(LH)?\d{1}))/
am finding it difficult to stop inputs after the LAB so that when EEE333 - LAB1 is inputed it becomes invalid.
If you are asking how to prevent LAB1 at the end, use an end of line anchor $ in your regex test:
/\D{3}\d{3}\s?-\s?(LAB|(EN(LH)?\d{1}))$/
If you are trying to require exactly one digit at the end of the acceptable strings, move the single digit match outside of the optional groups:
/\D{3}\d{3}\s?-\s?(LAB|(EN(LH)?))\d{1}$/
I have wrote for you the following regular expression:
[A-Z]{3}[0-9]{3}\s?-\s?(?:LAB|(?:EN|LH))[1-9]{1}
The regex works a follows:
[A-Z]{3}
MATCH EXACTLY THREE UPPERCASE CHARACTERS RANGING FROM A TO Z
[0-9]{3}
MATCH EXACTLY THREE NUMBERS RANGING FROM 0 TO 9
\s?\-\s?
MATCH a space (optional) or a '-' (required) or a space (optional)
(?:LAB|(?:EN|LH))
MATCH 'LAB' OR ('EN' OR 'LH')?: omits capturing LAB OR EN OR LH
[1-9]{1}
MATCH EXACTLY ONE NUMBERS RANGING FROM 1 TO 9
You could place your regex between word boundaries \b.
You start your regex with \D which is any character that is not a digit. That would for example also match $%^. You could use [A-Z].
You use \d{1} which is a shorhand for [0-9], but you want to match a digit between 1 and 9 [1-9]. You could also omit the {1}.
Maybe this updated will work for you?
\b[A-Z]{3}\d{3} ?- ?(?:LAB|(?:EN(?:LH)?[1-9]))\b
Explanation
A word boundary \b
Match 3 uppercase characters [A-Z]{3}
Match 3 digits \d{3}
Match an optional whitespace, a hyphen and another optional whitespace ?- ?
A non capturing group which for example matches LAB or EN EN1 or ENLH or ENLH9 (?:EN(?:LH)?[1-9]))
A word boundary \b
So given a string like this "\"turkey AND ham\" NOT \"roast beef\"" I need to get an array with the inner strings like so: ["turkey AND ham", "roast beef"] and eliminate OR's, AND's and NOT's that may or may not be there.
With the help of Rubular I came up with this regex /\\["']([^"']*)\\["']/
which returns the following 2 groups:
Match 1
1. turkey AND ham
Match 2
1. roast beef
however when I use it with .scan keep getting and empty array.
I looked at this and this other SO posts, and a few others, but can not figure out where I am going wrong
Here is the result from my rails console:
=> q = "\"turkey and ham\" OR \"roast beef\""
=> q.scan(/\\["']([^"']*)\\["']/)
=> []
Expectation:
["turkey AND ham", "roast beef"]
I shall also mention I suck at regex.
When the regex used with scan contains a capture group (#davidhu2000's approach), one generally can use lookarounds1 instead. It's just a matter of personal preference. To allow for double-quoted strings that contain either single- or (escaped) double-quoted strings, you could use the following regex.
r = /
(?<=") # match a double quote in a positive lookbehind
[^"]+ # match one or more characters that are not double-quotes
(?=") # match a double quote in a positive lookahead
| # or
(?<=') # match a single quote in a positive lookbehind
[^']+ # match one or more characters that are not single-quotes
(?=') # match a single quote in a positive lookahead
/x # free-spacing regex definition mode
"\"turkey AND ham\" NOT 'roast beef'".scan(r)
#=> ["turkey AND ham", "roast beef"]
As '"turkey AND ham" NOT "roast beef"' #=> "\"turkey AND ham\" NOT \"roast beef\"" (i.e., how the single-quoted string is saved), we need not be concerned about that being an additional case to deal with.
1 For any in the audience who still consider regular expressions to be black magic, there are four kinds of lookarounds (positive and negative lookbehinds and lookaheads) as elaborated in the doc for Regexp. Sometimes they are regarded as "zero-width" matches as they are not part of the matched text.
You regex is trying to match \, which won't match anything in the string, since the \ existed to escape the double quote, and won't be part of the string.
So if you remove \\ in your regex
res = q.scan(/["']([^"']*)["']/)
This will return a 2d array
res = [["turkey and ham"], ["roast beef"]]
Each inner array is all the matching groups from the regex, so if you have two capture groups in your regex, you will see two items in the inner array.
If you want a simple array, you can run flatten method on the array.
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.
Is there anything better than string.scan(/(\w|-)+/).size (the - is so, e.g., "one-way street" counts as 2 words instead of 3)?
string.split.size
Edited to explain multiple spaces
From the Ruby String Documentation page
split(pattern=$;, [limit]) → anArray
Divides str into substrings based on a delimiter, returning an array
of these substrings.
If pattern is a String, then its contents are used as the delimiter
when splitting str. If pattern is a single space, str is split on
whitespace, with leading whitespace and runs of contiguous whitespace
characters ignored.
If pattern is a Regexp, str is divided where the pattern matches.
Whenever the pattern matches a zero-length string, str is split into
individual characters. If pattern contains groups, the respective
matches will be returned in the array as well.
If pattern is omitted, the value of $; is used. If $; is nil (which is
the default), str is split on whitespace as if ' ' were specified.
If the limit parameter is omitted, trailing null fields are
suppressed. If limit is a positive number, at most that number of
fields will be returned (if limit is 1, the entire string is returned
as the only entry in an array). If negative, there is no limit to the
number of fields returned, and trailing null fields are not
suppressed.
" now's the time".split #=> ["now's", "the", "time"]
While that is the current version of ruby as of this edit, I learned on 1.7 (IIRC), where that also worked. I just tested it on 1.8.3.
I know this is an old question, but this might be useful to someone else looking for something more sophisticated than string.split. I wrote the words_counted gem to solve this particular problem, since defining words is pretty tricky.
The gem lets you define your own custom criteria, or use the out of the box regexp, which is pretty handy for most use cases. You can pre-filter words with a variety of options, including a string, lambda, array, or another regexp.
counter = WordsCounted::Counter.new("Hello, Renée! 123")
counter.word_count #=> 2
counter.words #=> ["Hello", "Renée"]
# filter the word "hello"
counter = WordsCounted::Counter.new("Hello, Renée!", reject: "Hello")
counter.word_count #=> 1
counter.words #=> ["Renée"]
# Count numbers only
counter = WordsCounted::Counter.new("Hello, Renée! 123", rexexp: /[0-9]/)
counter.word_count #=> 1
counter.words #=> ["123"]
The gem provides a bunch more useful methods.
If the 'word' in this case can be described as an alphanumeric sequence which can include '-' then the following solution may be appropriate (assuming that everything that doesn't match the 'word' pattern is a separator):
>> 'one-way street'.split(/[^-a-zA-Z]/).size
=> 2
>> 'one-way street'.split(/[^-a-zA-Z]/).each { |m| puts m }
one-way
street
=> ["one-way", "street"]
However, there are some other symbols that can be included in the regex - for example, ' to support the words like "it's".
This is pretty simplistic but does the job if you are typing words with spaces in between. It ends up counting numbers as well but I'm sure you could edit the code to not count numbers.
puts "enter a sentence to find its word length: "
word = gets
word = word.chomp
splits = word.split(" ")
target = splits.length.to_s
puts "your sentence is " + target + " words long"
The best way to do is to use split method.
split divides a string into sub-strings based on a delimiter, returning an array of the sub-strings.
split takes two parameters, namely; pattern and limit.
pattern is the delimiter over which the string is to be split into an array.
limit specifies the number of elements in the resulting array.
For more details, refer to Ruby Documentation: Ruby String documentation
str = "This is a string"
str.split(' ').size
#output: 4
The above code splits the string wherever it finds a space and hence it give the number of words in the string which is indirectly the size of the array.
The above solution is wrong, consider the following:
"one-way street"
You will get
["one-way","", "street"]
Use
'one-way street'.gsub(/[^-a-zA-Z]/, ' ').split.size
This splits words only on ASCII whitespace chars:
p " some word\nother\tword|word".strip.split(/\s+/).size #=> 4
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.