How to find string which is started with ".."? - ruby

I was trying to find strings out which is followed by only "..",but couldn't get that :
["..ab","...cc","..ps","....kkls"].each do |x|
puts x if /../.match(x)
end
..ab
...cc
..ps
....kkls
=> ["..ab", "...cc", "..ps", "....kkls"]
["..ab","...cc","..ps","....kkls"].each do |x|
puts x if /(.)(.)/.match(x)
end
..ab
...cc
..ps
....kkls
=> ["..ab", "...cc", "..ps", "....kkls"]
Expected output:
["..ab","..ps"]

What you want is
/^\.\.(?!\.)/
The caret ^ at the beginning means match the beginning of the string; periods must be escaped by a backslash as \. because in regular expressions a plain period . matches any character; the (?!\.) is a negative look-ahead meaning the next character is not a period. So the expression means, "at the beginning of the string, match two periods, which must be followed by a character which is not a period."
>> /^\.\.(?!\.)/.match "..ab"
=> #<MatchData "..">
>> /^\.\.(?!\.)/.match "...cc"
=> nil

Try selecting on /^\.\.[^\.]/ (starts with two dots and then not a dot).
ss = ["..ab","...cc","..ps","....kkls"]
ss.select { |x| x =~ /^\.\.[^\.]/ } # => ["..ab", "..ps"]

Try using /^\.{2}\w/ as the regular expression.
A quick explanation:
^ means the start of the string. Without this, it can match dots that are found in the middle of the string.
\. translates to . -- if you use the dot on its own, it will match any non-newline character
{2} means that you're looking for two of the dots. (you could rewrite /\.{2}/ as /\.\./)
Finally, the \w matches any word character (letter, number, underscore).
A really good place to test Ruby regular expressions is http://rubular.com/ -- it lets you play with the regex and test it right in your browser.

You don't need regex for this at all, you can just extract the appropriate leading chunks using String#[] or String#slice and do simple string comparisons:
>> a = ["..ab", "...cc", "..ps", "....kkls", ".", "..", "..."]
>> a.select { |s| s[0, 2] == '..' && s[0, 3] != '...' }
=> ["..ab", "..ps", ".."]

Maybe this:
["..ab","...cc","..ps","....kkls"].each {|x| puts x if /^\.{2}\w/.match(x) }
Or if you want to make sure the . doesn't match:
["..ab","...cc","..ps","....kkls"].each {|x| puts x if /^\.{2}[^\.]/.match(x) }

Related

Simple regex - ignoring certain characters

I'm trying to use the match method with an argument of a regex to select a valid phone number, by definition, any string with nine digits.
For example:
9347584987 is valid,
(456)322-3456 is valid,
(324)5688890 is valid.
But
(340)HelloWorld is NOT valid and
456748 is NOT valid.
So far, I'm able to use \d{9} to select the example string of 9 digit characters in a row, but I'm not sure how to specifically ignore any character, such as '-' or '(' or ')' in the middle of the sequence.
What kind of Regex could I use here?
Given:
nums=['9347584987','(456)322-3456','(324)5688890','(340)HelloWorld', '456748 is NOT valid']
You can split on a NON digit and rejoin to remove non digits:
> nums.map {|s| s.split(/\D/).join}
["9347584987", "4563223456", "3245688890", "340", "456748"]
Then filter on the length:
> nums.map {|s| s.split(/\D/).join}.select {|s| s.length==10}
["9347584987", "4563223456", "3245688890"]
Or, you can grab a group of numbers that look 'phony numbery' by using a regex to grab digits and common delimiters:
> nums.map {|s| s[/[\d\-()]+/]}
["9347584987", "(456)322-3456", "(324)5688890", "(340)", "456748"]
And then process that list as above.
That would delineate:
> '123 is NOT a valid area code for 456-7890'[/[\d\-()]+/]
=> "123" # no match
vs
> '123 is NOT a valid area code for 456-7890'.split(/\D/).join
=> "1234567890" # match
I suggest using one regular expression for each valid pattern rather than constructing a single regex. It would be easier to test and debug, and easier to maintain the code. If, for example, "123-456-7890" or 123-456-7890 x231" were in future deemed valid numbers, one need only add a single, simple regex for each to the array VALID_PATTERS below.
VALID_PATTERS = [/\A\d{10}\z/, /\A\(\d{3}\)\d{3}-\d{4}\z/, /\A\(\d{3}\)\d{7}\z/]
def valid?(str)
VALID_PATTERS.any? { |r| str.match?(r) }
end
ph_nbrs = %w| 9347584987 (456)322-3456 (324)5688890 (340)HelloWorld 456748 |
ph_nbrs.each { |s| puts "#{s.ljust(15)} \#=> #{valid?(s)}" }
9347584987 #=> true
(456)322-3456 #=> true
(324)5688890 #=> true
(340)HelloWorld #=> false
456748 #=> false
String#match? made its debut in Ruby v2.4. There are many alternatives, including str.match(r) and str =~ r.
"9347584987" =~ /(?:\d.*){9}/ #=> 0
"(456)322-3456" =~ /(?:\d.*){9}/ #=> 1
"(324)5688890" =~ /(?:\d.*){9}/ #=> 1
"(340)HelloWorld" =~ /(?:\d.*){9}/ #=> nil
"456748" =~ /(?:\d.*){9}/ #=> nil
Pattern: (Rubular Demo)
^\(?\d{3}\)?\d{3}-?\d{4}$ # this makes the expected symbols optional
This pattern will ensure that an opening ( at the start of the string is followed by 3 numbers the a closing ).
^(\(\d{3}\)|\d{3})\d{3}-?\d{4}$
On principle, though, I agree with melpomene in advising that you remove all non-digital characters, test for 9 character length, then store/handle the phone numbers in a single/reliable/basic format.

How to use Regexp.union to match a character at the beginning of my string

I'm using Ruby 2.4. I want to match an optional "a" or "b" character, followed by an arbitrary amount of white space, and then one or more numbers, but my regex's are failing to match any of these:
2.4.0 :017 > MY_TOKENS = ["a", "b"]
=> ["a", "b"]
2.4.0 :018 > str = "40"
=> "40"
2.4.0 :019 > str =~ Regexp.new("^[#{Regexp.union(MY_TOKENS)}]?[[:space:]]*\d+[^a-z^0-9]*$")
=> nil
2.4.0 :020 > str =~ Regexp.new("^#{Regexp.union(MY_TOKENS)}?[[:space:]]*\d+[^a-z^0-9]*$")
=> nil
2.4.0 :021 > str =~ Regexp.new("^#{Regexp.union(MY_TOKENS)}?[[:space:]]*\d+$")
=> nil
I'm stumped as to what I'm doing wrong.
If they are single characters, just use MY_TOKENS.join inside the character class:
MY_TOKENS = ["a", "b"]
str = "40"
first_regex = /^[#{MY_TOKENS.join}]?[[:space:]]*\d+[^a-z0-9]*$/
# /^[ab]?[[:space:]]*\d+[^a-z0-9]*$/
puts str =~ first_regex
# 0
You can also integrate the Regexp.union, it might lead to some unexpected bugs though, because the flags of the outer regexp won't apply to the inner one :
second_regex = /^#{Regexp.union(MY_TOKENS)}?[[:space:]]*\d+[^a-z0-9]*$/
# /^(?-mix:a|b)?[[:space:]]*\d+[^a-z0-9]*$/
puts str =~ second_regex
# 0
The above regex looks a lot like what you did, but using // instead of Regexp.new prevents you from having to escape the backslashes.
You could use Regexp#source to avoid this behaviour :
third_regex = /^(?:#{Regexp.union(MY_TOKENS).source})?[[:space:]]*\d+[^a-z0-9]*$/
# /^(?:a|b)?[[:space:]]*\d+[^a-z0-9]*$/
puts str =~ third_regex
# 0
or simply build your regex union :
fourth_regex = /^(?:#{MY_TOKENS.join('|')})?[[:space:]]*\d+[^a-z0-9]*$/
# /^(?:a|b)?[[:space:]]*\d+[^a-z0-9]*$/
puts str =~ fourth_regex
# 0
The 3 last examples should work fine if MY_TOKENS has words instead of just characters.
first_regex, third_regex and fourth_regex should all work fine with /i flag.
As an example :
first_regex = /^[#{MY_TOKENS.join}]?[[:space:]]*\d+[^a-z0-9]*$/i
"A 40" =~ first_regex
# 0
I believe you want to match a string that may contain any of the alternatives you defined in the MY_TOKENS, then 0+ whitespaces and then 1 or more digits up to the end of the string.
Then you need to use
Regexp.new("\\A#{Regexp.union(MY_TOKENS)}?[[:space:]]*\\d+\\z").match?(s)
or
/\A#{Regexp.union(MY_TOKENS)}?[[:space:]]*\d+\z/.match?(s)
When you use a Regexp.new, you should rememeber to double escape backslashes to define a literal backslash (e.g. "\d" is a digit matching pattern). In a regex literal notation, you may use a single backslash (/\d/).
Do not forget to match the start of a string with \A and end of string with \z anchors.
Note that [...] creates a character class that matches any char that is defined inside it: [ab] matches an a or b, [program] will match one char, either p, r, o, g, r, a or m. If you have multicharacter sequences in the MY_TOKENS, you need to remove [...] from the pattern.
To make the regex case insensitive, pass a case insensitive modifier to the pattern and make sure you use .source property of the Regex.union created regex to remove flags (thanks, Eric):
Regexp.new("(?i)\\A#{Regexp.union(MY_TOKENS).source}?[[:space:]]*\\d+\\z")
or
/\A#{Regexp.union(MY_TOKENS).source}?[[:space:]]*\d+\z/i
The regex created is /(?i-mx:\Aa|b?[[:space:]]*\d+\z)/ where (?i-mx) means the case insensitive mode is on and multiline (dot matches line breaks and verbose modes are off).

How to split a string without getting an empty string inserted in the array

I'm having trouble splitting a character from a string using a regular expression, assuming there is a match.
I want to split off either an "m" or an "f" character from the first part of a string assuming the next character is one or more numbers followed by optional space characters, followed by a string from an array I have.
I tried:
2.4.0 :006 > MY_SEPARATOR_TOKENS = ["-", " to "]
=> ["-", " to "]
2.4.0 :008 > str = "M14-19"
=> "M14-19"
2.4.0 :011 > str.split(/^(m|f)\d+[[:space:]]*#{Regexp.union(MY_SEPARATOR_TOKENS)}/i)
=> ["", "M", "19"]
Notice the extraneous "" element at the beginning of my array and also notice that the last expression is just "19" whereas I would want everything else in the string ("14-19").
How do I adjust my regular expression so that only the parts of the expression that get split end up in the array?
I find match to be a bit more elegant when extracting characters from regular expressions in Ruby:
string = "M14-19"
string.match(/\A(?<m>[M|F])(?<digits>\d{2}(-| to )\d{2})/)[1, 2]
=> ["M", "14-19"]
# also can extract the symbols from match
extract_string = string.match(/\A(?<m>[M|F])(?<digits>\d{2}(-| to )\d{2})/)
[[extract_string[:m], extract_string[:digits]]
=> ["M", "14-19"]
string = 'M14 to 14'
extract_string = string.match(/\A(?<m>[M|F])(?<digits>\d{2}(-| to )\d{2})/)[1, 2]
=> ["M", "14 to 14"]
TOKENS = ["-", " to "]
r = /
(?<=\A[mMfF]) # match the beginning of the string and then one
# of the 4 characters in a positive lookbehind
(?= # begin positive lookahead
\d+ # match one or more digits
[[:space:]]* # match zero or more spaces
(?:#{TOKENS.join('|')}) # match one of the tokens
) # close the positive lookahead
/x # free-spacing regex definition mode
(?:#{TOKENS.join('|')}) is replaced by (?:-| to ).
This can of course be written in the usual way.
r = /(?<=\A[mMfF])(?=\d+[[:space:]]*(?:#{TOKENS.join('|')}))/
When splitting on r you are splitting between two characters (between a positive lookbehind and a positive lookahead) so no characters are consumed.
"M14-19".split r
#=> ["M", "14-19"]
"M14 to 19".split r
#=> ["M", "14 to 19"]
"M14 To 19".split r
#=> ["M14 To 19"]
If it is desired that ["M", "14 To 19"] be returned in the last example, change [mMfF] to [mf] and /x to /xi.
You have a bug brewing in your code. Don't get in the habit of doing this:
#{Regexp.union(MY_SEPARATOR_TOKENS)}
You're setting yourself up with a very hard to debug problem.
Here's what's happening:
regex = Regexp.union(%w(a b)) # => /a|b/
/#{regex}/ # => /(?-mix:a|b)/
/#{regex.source}/ # => /a|b/
/(?-mix:a|b)/ is an embedded sub-pattern with its set of the regex flags m, i and x which are independent of the surrounding pattern's settings.
Consider this situation:
'CAT'[/#{regex}/i] # => nil
We'd expect that the regular expression i flag would match because it's ignoring case, but the sub-expression still only allows only lowercase, causing the match to fail.
Using the bare (a|b) or adding source succeeds because the inner expression gets the main expression's i:
'CAT'[/(a|b)/i] # => "A"
'CAT'[/#{regex.source}/i] # => "A"
See "How to embed regular expressions in other regular expressions in Ruby" for additional discussion of this.
The empty element will always be there if you get a match, because the captured part appears at the beginning of the string and the string between the start of the string and the match is added to the resulting array, be it an empty or non-empty string. Either shift/drop it once you get a match, or just remove all empty array elements with .reject { |c| c.empty? } (see How do I remove blank elements from an array?).
Then, 14- is eaten up (consumed) by the \d+[[:space:]]... pattern part - put it into a (?=...) lookahead that will just check for the pattern match, but won't consume the characters.
Use something like
MY_SEPARATOR_TOKENS = ["-", " to "]
s = "M14-19"
puts s.split(/^(m|f)(?=\d+[[:space:]]*#{Regexp.union(MY_SEPARATOR_TOKENS)})/i).drop(1)
#=> ["M", "14-19"]
See Ruby demo

Why does /[<>]/ not return both angle brackets with String#match?

I expect this example to match the two characters <and >:
a = "<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>"
a.match /[<>]/
# => #<MatchData "<">
It matches only the first character. Why?
#match only returns the first match as you have seen as MatchData, #scan will return all matches.
>> a="<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>"
=> "<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>"
>> a.scan /[<>]/
=> ["<", ">"]
Problem
You are misunderstanding your expression. /[<>]/ means:
Match a single character from the character class, which may be either < or >.
Ruby is correctly giving you exactly what you've asked for in your pattern.
Solution
If you're expecting the entire string between the two characters, you need a different pattern. For example:
"<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>".match /<.*?>/
#=> #<MatchData "<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>">
Alternatively, if you just want to match all the instances of < or > in your string, then you should use String#scan with a character class or alternation. In this particular case, the results will be identical either way. For example:
"<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>".scan /<|>/
#=> ["<", ">"]
"<1acf457f477b41d4a363e89a1d0a6e57#Open-Xchange>".scan /[<>]/
#=> ["<", ">"]

Finding the first duplicate character in the string Ruby

I am trying to call the first duplicate character in my string in Ruby.
I have defined an input string using gets.
How do I call the first duplicate character in the string?
This is my code so far.
string = "#{gets}"
print string
How do I call a character from this string?
Edit 1:
This is the code I have now where my output is coming out to me No duplicates 26 times. I think my if statement is wrongly written.
string "abcade"
puts string
for i in ('a'..'z')
if string =~ /(.)\1/
puts string.chars.group_by{|c| c}.find{|el| el[1].size >1}[0]
else
puts "no duplicates"
end
end
My second puts statement works but with the for and if loops, it returns no duplicates 26 times whatever the string is.
The following returns the index of the first duplicate character:
the_string =~ /(.)\1/
Example:
'1234556' =~ /(.)\1/
=> 4
To get the duplicate character itself, use $1:
$1
=> "5"
Example usage in an if statement:
if my_string =~ /(.)\1/
# found duplicate; potentially do something with $1
else
# there is no match
end
s.chars.map { |c| [c, s.count(c)] }.drop_while{|i| i[1] <= 1}.first[0]
With the refined form from Cary Swoveland :
s.each_char.find { |c| s.count(c) > 1 }
Below method might be useful to find the first word in a string
def firstRepeatedWord(string)
h_data = Hash.new(0)
string.split(" ").each{|x| h_data[x] +=1}
h_data.key(h_data.values.max)
end
I believe the question can be interpreted in either of two ways (neither involving the first pair of adjacent characters that are the same) and offer solutions to each.
Find the first character in the string that is preceded by the same character
I don't believe we can use a regex for this (but would love to be proved wrong). I would use the method suggested in a comment by #DaveNewton:
require 'set'
def first_repeat_char(str)
str.each_char.with_object(Set.new) { |c,s| return c unless s.add?(c) }
nil
end
first_repeat_char("abcdebf") #=> b
first_repeat_char("abcdcbe") #=> c
first_repeat_char("abcdefg") #=> nil
Find the first character in the string that appears more than once
r = /
(.) # match any character in capture group #1
.* # match any character zero of more times
? # do the preceding lazily
\K # forget everything matched so far
\1 # match the contents of capture group 1
/x
"abcdebf"[r] #=> b
"abccdeb"[r] #=> b
"abcdefg"[r] #=> nil
This regex is fine, but produces the warning, "regular expression has redundant nested repeat operator '*'". You can disregard the warning or suppress it by doing something clunky, like:
r = /([^#{0.chr}]).*?\K\1/
where ([^#{0.chr}]) means "match any character other than 0.chr in capture group 1".
Note that a positive lookbehind cannot be used here, as they cannot contain variable-length matches (i.e., .*).
You could probably make your string an array and use detect. This should return the first char where the count is > 1.
string.split("").detect {|x| string.count(x) > 1}
I'll use positive lookahead with String#[] method :
"abcccddde"[/(.)(?=\1)/] #=> c
As a variant:
str = "abcdeff"
p str.chars.group_by{|c| c}.find{|el| el[1].size > 1}[0]
prints "f"

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