I have some parameters that I have to sort into different lists. The prefix determines which list should it belong to.
I use prefixes like: c, a, n, o and an additional hyphen (-) to determine whether to put it in include l it or exclude list.
I use the regex grouped as:
/^(-?)([o|a|c|n])(\w+)/
But here the third group (\w+) is not generic, and it should actually be dependent on the second group's result. I.e, if the prefix is:
'c' or 'a' -> /\w{3}/
'o' -> /\w{2}/
else -> /\w+/
Can I do this with a single regex? Currently I am using an if condition to do so.
Example input:
Valid:
"-cABS", "-aXYZ", "-oWE", "-oqr", "-ncanbeanyting", "nstillanything", "a123", "-conT" (will go to c_exclude_list)
Invalid:
"cmorethan3chars", "c1", "-a1234", "prefizisnotvalid", "somethingelse", "oABC"
Output: for each arg push to the correct list, ignore the invalid.
c_include_list, c_exclude_list, a_include_list, a_exclude_list etc.
You can use this pattern:
/(-?)\b([aocn])((?:(?<=[ac])\w{3}|(?<=o)\w{2}|(?<=n)\w+))\b/
The idea consists to use lookbehinds to check the previous character without including it in the capture group.
Since version 2.0, Ruby has switched from Oniguruma to Onigmo (a fork of Oniguruma), which adds support for conditional regex, among other features.
So you can use the following regex to customize the pattern based on the prefix:
^-(?:([ca])|(o)|(n))?(?(1)\w{3}|(?(2)\w{2}|(?(3)\w+)))$
Demo at rubular
Is a single, mind-bending regex the best way to deal with this problem?
Here's a simpler approach that does not employ a regex at all. I suspect that it would be at least as efficient as a single regex, considering that with the latter you must still assign matching strings to their respective arrays. I think it also reads better and would be easier to maintain. The code below should be easy to modify if I have misunderstood some fine points of the question.
Code
def devide_em_up(str)
h = { a_exclude: [], a_include: [], c_exclude: [], c_include: [],
o_exclude: [], o_include: [], other_exclude: [], other_include: [] }
str.split.each do |s|
exclude = (s[0] == ?-)
s = s[1..-1] if exclude
first = s[0]
s = s[1..-1] if 'cao'.include?(first)
len = s.size
case first
when 'a'
(exclude ? h[:a_exclude] : h[:a_include]) << s if len == 3
when 'c'
(exclude ? h[:c_exclude] : h[:c_include]) << s if len == 3
when 'o'
(exclude ? h[:o_exclude] : h[:o_include]) << s if len == 2
else
(exclude ? h[:other_exclude] : h[:other_include]) << s
end
end
h
end
Example
Let's try it:
str = "-cABS cABT -cDEF -aXYZ -oWE -oQR oQT -ncanbeany nstillany a123 " +
"-conT cmorethan3chars c1 -a1234 prefizisnotvalid somethingelse oABC"
devide_em_up(str)
#=> {:a_exclude=>["XYZ"], :a_include=>["123"],
# :c_exclude=>["ABS", "DEF"], :c_include=>["ABT"],
# :o_exclude=>["WE", "QR"], :o_include=>["QT"],
# :other_exclude=>["ncanbeany"], :other_include=>["nstillany"]}
Related
Where as str[] will replace a character, str.insert will insert a character at a position. But it requires two lines of code:
str = "COSO17123456"
str.insert 4, "-"
str.insert 7, "-"
=> "COSO-17-123456"
I was thinking how to do this in one line of code. I came up with the following solution:
str = "COSO17123456"
str.each_char.with_index.reduce("") { |acc,(c,i)| acc += c + ( (i == 3 || i == 5) ? "-" : "" ) }
=> "COSO-17-123456
Is there a built-in Ruby helper for this task? If not, should I stick with the insert option rather than combining several iterators?
Use each to iterate over an array of indices:
str = "COSO17123456"
[4, 7].each { |i| str.insert i, '-' }
str #=> "COSO-17-123456"
You can uses slices and .join:
> [str[0..3], str[4..5],str[6..-1]].join("-")
=> "COSO-17-123456"
Note that the index after the first one (between 3 and 4) will be different since you are not inserting earlier insertion first. ie, more natural (to me anyway...)
You will insert at the absolute index of the original string -- not the moving relative index as insertions are made.
If you want to insert at specific absolute index values, you can also use ..each_with_index and control the behavior character by character:
str2 = ""
tgts=[3,5]
str.split("").each_with_index { |c,idx| str2+=c; str2+='-' if tgts.include? idx }
Both of the above create a new string.
String#insert returns the string itself.
This means you can chain the method calls, which can be a prettier and more efficient if you only have to do it a couple of times like in your example:
str = "COSO17123456".insert(4, "-").insert(7, "-")
puts str
COSO-17-123456
Your reduce version can be therefore more concisely written as:
[4,7].reduce(str) { |str, idx| str.insert(idx, '-') }
I'll bring one more variation to the table, String#unpack:
new_str = str.unpack("A4A2A*").join('-')
# or with String#%
new_str = "%s-%s-%s" % str.unpack("A4A2A*")
Idea. Given the string, return all the matches (with overlaps) and the text before these matches.
Example. For the text atatgcgcatatat and the query atat there are three matches, and the desired output is atat, atatgcgcatat and atatgcgcatatat.
Problem. I use Ruby 2.2 and String#scan method to get multiple matches. I've tried to use lookahead, but the regex /(?=(.*?atat))/ returns every substring that ends with atat. There must be some regex magic to solve this problem, but I can't figure out the right spell.
I believe this is at least better than the OP's answer:
text = "atatgcgcatatat"
query = "atat"
res = []
text.scan(/(?=#{query})/){res.push($` + query)} #`
res # => ["atat", "atatgcgcatat", "atatgcgcatatat"]
Given the nature and purpose of regex, there is no way to do that. When a regex matches text, there is no way to include the same text in another match. Therefore, the best option that I can think of is to use a look-behind to find the ending position of each match:
(?<=atat)
With your example input of atatgcgcatatat, that would return the following three matches:
Position 4, Length 0
Position 12, Length 0
Position 14, Length 0
You could then loop through those results, get the position for each one, and then get the sub-string that starts at the beginning of the input string and ends at that position. If you don't know how to get the positions of each match, you may find the answers to this question helpful.
You could do this:
str = 'atatgcgcatatat'
target = 'atat'
[].tap do |a|
str.gsub(/(?=#{target})/) { a << str[0, $~.end(0)+target.size] }
end
#=> ["atat", "atatgcgcatat", "atatgcgcatatat"]
Notice that the string returned by gsub is discarded.
It seems, there's no way to solve the problem in just one go.
One possible solution is to use this knowledge to get indices of matches when using String#scan, and then return the array of sliced strings:
def find_by_end text, query
res = []
n = query.length
text.scan( /(?=(#{query}))/ ) do |m|
res << text.slice(0, $~.offset(0).first + n)
end
res
end
find_by_end "atatgcgcatatat", "atat" #=> ["atat", "atatgcgcatat", "atatgcgcatatat"]
A slightly different solution was proposed by #StevenDoggart. Here's a nice and short code which uses this hack to solve the problem:
"atatgcatatat".to_enum(:scan, /(?<=atat)/).map { $` } #`
#=> ["atat", "atatgcatat", "atatgcatatat"]
As #CasimiretHippolyte notes, reversing the string might help to solve the problem. It actually does, but it's hardly the prettiest solution:
"atatgcatatat".reverse.scan(/(?=(tata.*))/).flatten.map(&:reverse).reverse
#=> ["atat", "atatgcatat", "atatgcatatat"]
Is there a way / gem to create regular expressions with some basic search parameters.
e.g.
Search = ["\"German Shepherd\"","Collie","poodle", "Miniature Schnauzer"]
Such that the regexp will search (case insensitively) for:
"German Shepherd" - exactly
OR
"Collie"
OR
"poodle"
OR
"Miniature" AND "Schnauzer"
So in this case something like:
/German\ Shepherd|Collie|poodle|(?=.*Miniature)(?=.*Schnauzer).+/i
(Open to suggestions of better ways of doing the last bit...)
If I understood the question properly, here you go:
regexps = ["\"German Shepherd\"","Collie","poodle", "Miniature Schnauzer"]
# those in quotes
greedy = regexps.select { |re| re =~ /\A['"].*['"]\z/ } # c'"mon, parser
# the rest unquoted
non_greedy = (regexps - greedy).map(&:split).flatten
# concatenating... ⇓⇓⇓ get rid of quotes
all = Regexp.union(non_greedy + greedy.map { |re| re[1...-1] })
#⇒ /Collie|poodle|Miniature|Schnauzer|German\ Shepherd/
UPD
I finally got what is to be done with Miniature Schnauzer (please see a comment below for further explanation.) That said, these words are to be permuted and joined with non-greedy .*?:
non_greedy = (regexps - greedy).map(&:split).map do |re|
# single word? YES : NO, permute and join
re.length < 2 ? re : re.permutation.map { |p| Regexp.new p.join('.*?') }
end.flatten
all = Regexp.union(non_greedy + greedy.map { |re| re[1...-1] })
#=> /Collie|poodle|(?-mix:Miniature.*?Schnauzer)|(?-mix:Schnauzer.*?Miniature)|German\ Shepherd/
Suppose I want to make sure a string x equals any combination of abcd (each character appearing one or zero times-->each character should not repeat, but the combination may appear in any order)
valid ex: bc .. abcd ... bcad ... b... d .. dc
invalid ex. abcdd, cc, bbbb, abcde (ofcourse)
my effort:
I tried various techniques:
the closest I came was
x =~ ^(((a)?(b)?(c)?(d)?))$
but this wont work if I do not type them in the same order as i have written:
works for: ab, acd, abcd, a, d, c
wont work for: bcda, cb, da (anything that is not in the above order)
you can test your solutions here : http://rubular.com/r/wCpD355bub
PS: the characters may not be in alphabetical order, it could be u c e t
If you can use things besides regexes, you can try:
str.chars.uniq.length == str.length && str.match(/^[a-d]+$/)
The general idea here is that you just strip any duplicated characters from the string, and if the length of the uniq array is not equal to the length of the source string, you have a duplicated character in the string. The regex then enforces the character set.
This can probably be improved, but it's pretty straightforward. It does create a couple of extra arrays, so you might want a different approach if this needs to be used in a performance-critical location.
If you want to stick to regexes, you could use:
str.match(/^[a-d]+$/) && !str.match(/([a-d]).*\1/)
That'll basically check that the string only contains the allowed characters, and that those characters are never repeated.
This is really not what regular expressions are meant to do, but if you really really want to.
Here is a regex that satisfies the conditions.
^([a-d])(?!(\1))([a-d])?(?!(\1|\3))([a-d])?(?!(\1|\3|\5))([a-d])?(?!(\1|\3|\5|\7))$
basically it goes through each character, making the group, then makes sure that that group isn't matched. Then checks the next character, and makes sure that group and the previous groups don't match.
You can reverse it (match the condition that would make it fail)
re = /^ # start of line
(?=.*([a-d]).*\1) # match if a letter appears more than once
| # or
(?=.*[^a-d]) # match if a non abcd char appears
/x
puts 'fail' if %w{bc abcd bcad b d dc}.any?{|s| s =~ re}
puts 'fail' unless %w{abcdd cc bbbb abcde}.all?{|s| s =~ re}
I don't think regexes are well suited to this problem, so here is another non-regex solution. It's recursive:
def match_chars_no_more_than_once(characters, string)
return true if string.empty?
if characters.index(string[0])
match_chars_no_more_than_once(characters.sub(string[0],''), string[1..-1])
else
false
end
end
%w{bc bdac hello acbbd cdda}.each do |string|
p [string, match_chars_no_more_than_once('abcd', string)]
end
Output:
["bc", true]
["bdac", true]
["hello", false]
["acbbd", false]
["cdda", false]
I am currently working on a Ruby Problem quiz but I'm not sure if my solution is right. After running the check, it shows that the compilation was successful but i'm just worried it is not the right answer.
The problem:
A string S consisting only of characters '(' and ')' is called properly nested if:
S is empty,
S has the form "(U)" where
U is a properly nested string,
S has
the form "VW" where V and W are
properly nested strings.
For example, "(()(())())" is properly nested and "())" isn't.
Write a function
def nesting(s)
that given a string S returns 1 if S
is properly nested and 0 otherwise.
Assume that the length of S does not
exceed 1,000,000. Assume that S
consists only of characters '(' and
')'.
For example, given S = "(()(())())"
the function should return 1 and given
S = "())" the function should return
0, as explained above.
Solution:
def nesting ( s )
# write your code here
if s == '(()(())())' && s.length <= 1000000
return 1
elsif s == ' ' && s.length <= 1000000
return 1
elsif
s == '())'
return 0
end
end
Here are descriptions of two algorithms that should accomplish the goal. I'll leave it as an exercise to the reader to turn them into code (unless you explicitly ask for a code solution):
Start with a variable set to 0 and loop through each character in the string: when you see a '(', add one to the variable; when you see a ')', subtract one from the variable. If the variable ever goes negative, you have seen too many ')' and can return 0 immediately. If you finish looping through the characters and the variable is not exactly 0, then you had too many '(' and should return 0.
Remove every occurrence of '()' in the string (replace with ''). Keep doing this until you find that nothing has been replaced (check the return value of gsub!). If the string is empty, the parentheses were matched. If the string is not empty, it was mismatched.
You're not supposed to just enumerate the given examples. You're supposed to solve the problem generally. You're also not supposed to check that the length is below 1000000, you're allowed to assume that.
The most straight forward solution to this problem is to iterate through the string and keep track of how many parentheses are open right now. If you ever see a closing parenthesis when no parentheses are currently open, the string is not well-balanced. If any parentheses are still open when you reach the end, the string is not well-balanced. Otherwise it is.
Alternatively you could also turn the specification directly into a regex pattern using the recursive regex feature of ruby 1.9 if you were so inclined.
My algorithm would use stacks for this purpose. Stacks are meant for solving such problems
Algorithm
Define a hash which holds the list of balanced brackets for
instance {"(" => ")", "{" => "}", and so on...}
Declare a stack (in our case, array) i.e. brackets = []
Loop through the string using each_char and compare each character with keys of the hash and push it to the brackets
Within the same loop compare it with the values of the hash and pop the character from brackets
In the end, if the brackets stack is empty, the brackets are balanced.
def brackets_balanced?(string)
return false if string.length < 2
brackets_hash = {"(" => ")", "{" => "}", "[" => "]"}
brackets = []
string.each_char do |x|
brackets.push(x) if brackets_hash.keys.include?(x)
brackets.pop if brackets_hash.values.include?(x)
end
return brackets.empty?
end
You can solve this problem theoretically. By using a grammar like this:
S ← LSR | LR
L ← (
R ← )
The grammar should be easily solvable by recursive algorithm.
That would be the most elegant solution. Otherwise as already mentioned here count the open parentheses.
Here's a neat way to do it using inject:
class String
def valid_parentheses?
valid = true
self.gsub(/[^\(\)]/, '').split('').inject(0) do |counter, parenthesis|
counter += (parenthesis == '(' ? 1 : -1)
valid = false if counter < 0
counter
end.zero? && valid
end
end
> "(a+b)".valid_parentheses? # => true
> "(a+b)(".valid_parentheses? # => false
> "(a+b))".valid_parentheses? # => false
> "(a+b))(".valid_parentheses? # => false
You're right to be worried; I think you've got the very wrong end of the stick, and you're solving the problem too literally (the info that the string doesn't exceed 1,000,000 characters is just to stop people worrying about how slow their code would run if the length was 100times that, and the examples are just that - examples - not the definitive list of strings you can expect to receive)
I'm not going to do your homework for you (by writing the code), but will give you a pointer to a solution that occurs to me:
The string is correctly nested if every left bracket has a right-bracket to the right of it, or a correctly nested set of brackets between them. So how about a recursive function, or a loop, that removes the string matches "()". When you run out of matches, what are you left with? Nothing? That was a properly nested string then. Something else (like ')' or ')(', etc) would mean it was not correctly nested in the first place.
Define method:
def check_nesting str
pattern = /\(\)/
while str =~ pattern do
str = str.gsub pattern, ''
end
str.length == 0
end
And test it:
>ruby nest.rb (()(())())
true
>ruby nest.rb (()
false
>ruby nest.rb ((((()))))
true
>ruby nest.rb (()
false
>ruby nest.rb (()(((())))())
true
>ruby nest.rb (()(((())))()
false
Your solution only returns the correct answer for the strings "(()(())())" and "())". You surely need a solution that works for any string!
As a start, how about counting the number of occurrences of ( and ), and seeing if they are equal?