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.
Related
I have a name spaced class..
"CommonCar::RedTrunk"
I need to convert it to an underscored string "common_car_red_trunk", but when I use
"CommonCar::RedTrunk".underscore, I get "common_car/red_trunk" instead.
Is there another method to accomplish what I need?
Solutions:
"CommonCar::RedTrunk".gsub(':', '').underscore
or:
"CommonCar::RedTrunk".sub('::', '').underscore
or:
"CommonCar::RedTrunk".tr(':', '').underscore
Alternate:
Or turn any of these around and do the underscore() first, followed by whatever method you want to use to replace "/" with "_".
Explanation:
While all of these methods look basically the same, there are subtle differences that can be very impactful.
In short:
gsub() – uses a regex to do pattern matching, therefore, it's finding any occurrence of ":" and replacing it with "".
sub() – uses a regex to do pattern matching, similarly to gsub(), with the exception that it's only finding the first occurrence (the "g" in gsub() meaning "global"). This is why when using that method, it was necessary to use "::", otherwise a single ":" would have been left. Keep in mind with this method, it will only work with a single-nested namespace. Meaning "CommonCar::RedTrunk::BigWheels" would have been transformed to "CommonCarRedTrunk::BigWheels".
tr() – uses the string parameters as arrays of single character replacments. In this case, because we're only replacing a single character, it'll work identically to gsub(). However, if you wanted to replace "on" with "EX", for example, gsub("on", "EX") would produce "CommEXCar::RedTrunk" while tr("on", "EX") would produce "CEmmEXCar::RedTruXk".
Docs:
https://apidock.com/ruby/String/gsub
https://apidock.com/ruby/String/sub
https://apidock.com/ruby/String/tr
This is a pure-Ruby solution.
r = /(?<=[a-z])(?=[A-Z])|::/
"CommonCar::RedTrunk".gsub(r, '_').downcase
#=> "common_car_red_trunk"
See (the first form of) String#gsub and String#downcase.
The regular expression can be made self-documenting by writing it in free-spacing mode:
r = /
(?<=[a-z]) # assert that the previous character is lower-case
(?=[A-Z]) # assert that the following character is upper-case
| # or
:: # match '::'
/x # free-spacing regex definition mode
(?<=[a-z]) is a positive lookbehind; (?=[A-Z]) is a positive lookahead.
Note that /(?<=[a-z])(?=[A-Z])/ matches an empty ("zero-width") string. r matches, for example, the empty string between 'Common' and 'Car', because it is preceeded by a lower-case letter and followed by an upper-case letter.
I don't know Rails but I'm guessing you could write
"CommonCar::RedTrunk".delete(':').underscore
I have strings in the format:
'I had a great time with #[2468] and #[1357]! #[1111] #[2321]#[1212]'
I want to be able to extract the numbers between the # and # symbols, but I do not want the included square brackets. For example I would like to return:
user_ids = [2468, 1357]
hash_tag_ids = [1111, 2321, 1212]
Any ideas?
Because you want to match all occurrences of the pattern, the string.scan method is what you want. Scan automatically returns everything that matches the pattern, so you don't need to use "capture groups" (the parentheses you see in most regular expressions), but you do need to use "lookahead" and "lookbehind" to match some stuff without including it in your result.
The two lines you need are:
string.scan(/(?<=#\[)\d+(?=\])/).map(&:to_i) # => [2468, 1357]
string.scan(/(?<=#\[)\d+(?=\])/).map(&:to_i) # => [1111, 2321, 1212]
The (?<=...) creates a "positive lookbehind" which ensures that the preceding characters match ..., but those characters aren't included in the matched text. In other words, (?<=#\[) will match "#[", but "#[" will not be included in the results returned by string.scan.
Notice the opening square bracket, and the closing square bracket have a slash in front of them. This is because square brackets have special meaning in a regular expression (they create a "character class"), but since we want to match a literal square bracket, we must "escape" them with a slash.
\d+ means to match 1 or more digits.
(?=...) creates a "positive lookahead" which ensures that the following characters match ..., but those characters aren't included in the matched text. Same as the lookbehind above, but checks the following characters instead of the preceding characters. In this case, (?=\]) matches "]" without including the "]" in the results returned by string.scan.
string.scan will return an array of strings. The .map(&:to_i) part will run string.to_i on each string to return an actual integer value.
string.scan(/(?<=#\[)[^\]]*(?=\])/) # => ["2468", "1357"]
string.scan(/(?<=#\[)[^\]]*(?=\])/) # => ["1111", "2321", "1212"]
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.
I have to validate username in my app so that it cannot contain two consecutive period symbols. I tried the following.
username.match(/(..)/)
but found out that this matches "a." and "a..". I expected to get nil as the output of the match operation for input "a.". Is my approach right ?
You need to put a \ in front of the periods, because period is a reserved character for "any character except newline".
So try:
username.match(/(\.\.)/)
Short answer
You can use something like this (see on rubular.com):
username.match(/\.{2}/)
The . is escaped by preceding with a backslash, {2} is exact repetition specifier, and the brackets are removed since capturing is not required in this case.
On metacharacters and escaping
The dot, as a pattern metacharacter, matches (almost) any character. To match a literal period, you have at least two options:
Escape the dot as \.
Match it as a character class singleton [.]
Other metacharacters that may need escaping are | (alternation), +/*/?/{/} (repetition), [/] (character class), ^/$ (anchors), (/) (grouping), and of course \ itself.
References
regular-expressions.info/Literals and metacharacters, Dot: ., Character Class: […], Anchors: ^$, Repetition: *+?{…}, Alternation: |, Optional: ?, Grouping: (…)
On finite repetition
To match two literal periods, you can use e.g. \.\. or [.][.], i.e. a simple concatenation. You can also use the repetition construct, e.g. \.{2} or [.]{2}.
The finite repetition specifier also allows you write something like x{3,5} to match at least 3 but at most 5 x.
Note that repetition has a higher precedence that concatenation, so ha{3} doesn't match "hahaha"; it matches "haaa" instead. You can use grouping like (ha){3} to match "hahaha".
On grouping
Grouping (…) captures the string it matches, which can be useful when you want to capture a match made by a subpattern, or if you want to use it as a backreference in the other parts of the pattern.
If you don't need this functionality, then a non-capturing option is (?:…). Thus something like (?:ha){3} still matches "hahaha" like before, but without creating a capturing group.
If you don't actually need the grouping aspect, then you can just leave out the brackets altogether.
Is there anything better than string.scan(/(\w|-)+/).size (the - is so, e.g., "one-way street" counts as 2 words instead of 3)?
string.split.size
Edited to explain multiple spaces
From the Ruby String Documentation page
split(pattern=$;, [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