I'm trying to take a string which is a simple math expression, remove all spaces, remove all duplicate operators, convert to single digit numbers, and then evaluate.
For example, an string like "2 7+*3*95" should be converted to "2+3*9" and then evaluated as 29.
Here's what I have so far:
expression.slice!(/ /) # Remove whitespace
expression.slice!(/\A([\+\-\*\/]+)/) # Remove operators from the beginning
expression.squeeze!("0123456789") # Single digit numbers (doesn't work)
expression.squeeze!("+-*/") # Removes duplicate operators (doesn't work)
expression.slice!(/([\+\-\*\/]+)\Z/) # Removes operators from the end
puts eval expression
Unfortunately this doesn't make single digit numbers nor remove duplicate operators quite like I expected. Any ideas?
"2 7+*3*95".gsub(/([0-9])[0-9 ]*/, '\1').gsub(/([\+\*\/\-])[ +\*\/\-]+/, '\1')
The first regex handles the single-digit thing and the second handles repeat operators. You could probably condense it into a single regex if you really wanted to.
This works for a quick-and-dirty solution, but you might be better served by a proper parser.
DATA.each { |expr|
expr.gsub!(%r'\s+', '')
expr.gsub!(%r'([*/+-])[*/+-]+', '\1')
expr.gsub!(%r'(\d)\d+', '\1')
expr.sub!(%r'\A[*/+-]+', '')
expr.sub!(%r'[*/+-]+\Z', '')
puts expr + ' = ' + eval(expr).to_s
}
__END__
2 7+*3*95
+-2 888+*3*95+8*-2/+435+-
Related
Based on "How to Delete Strings that Start with Certain Characters in Ruby", I know that the way to remove a string that starts with the character "#" is:
email = email.gsub( /(?:\s|^)#.*/ , "") #removes strings that start with "#"
I want to also remove strings that end in ".". Inspired by "Difference between \A \z and ^ $ in Ruby regular expressions" I came up with:
email = email.gsub( /(?:\s|$).*\./ , "")
Basically I used gsub to remove the dollar sign for the carrot and reversed the order of the part after the closing parentheses (making sure to escape the period). However, it is not doing the trick.
An example I'd like to match and remove is:
"a8&23q2aas."
You were so close.
email = email.gsub( /.*\.\s*$/ , "")
The difference lies in the fact that you didn't consider the relationship between string of reference and the regex tokens that describe the condition you wish to trigger. Here, you are trying to find a period (\.) which is followed only by whitespace (\s) or the end of the line ($). I would read the regex above as "Any characters of any length followed by a period, followed by any amount of whitespace, followed by the end of the line."
As commenters pointed out, though, there's a simpler way: String#end_with?.
I'd use:
words = %w[#a day in the life.]
# => ["#a", "day", "in", "the", "life."]
words.reject { |w| w.start_with?('#') || w.end_with?('.') }
# => ["day", "in", "the"]
Using a regex is overkill for this if you're only concerned with the starting or ending character, and, in fact, regular expressions will slow your code in comparison with using the built-in methods.
I would really like to stick to using gsub....
gsub is the wrong way to remove an element from an array. It could be used to turn the string into an empty string, but that won't remove that element from the array.
def replace_suffix(str,suffix)
str.end_with?(suffix)? str[0, str.length - suffix.length] : str
end
I'm receiving a string that contains two numbers in a handful of different formats:
"344, 345", "334,433", "345x532" and "432 345"
I need to split them into two separate numbers in an array using split, and then convert them using Integer(num).
What I've tried so far:
nums.split(/[\s+,x]/) # split on one or more spaces, a comma or x
However, it doesn't seem to match multiple spaces when testing. Also, it doesn't allow a space in the comma version shown above ("344, 345").
How can I match multiple delimiters?
You are using a character class in your pattern, and it matches only one character. [\s+,x] matches 1 whitespace, or a +, , or x. You meant to use (?:\s+|x).
However, perhaps, a mere \D+ (1 or more non-digit characters) should suffice:
"345, 456".split(/\D+/).map(&:to_i)
R1 = Regexp.union([", ", ",", "x", " "])
#=> /,\ |,|x|\ /
R2 = /\A\d+#{R1}\d+\z/
#=> /\A\d+(?-mix:,\ |,|x|\ )\d+\z/
def split_it(s)
return nil unless s =~ R2
s.split(R1).map(&:to_i)
end
split_it("344, 345") #=> [344, 345]
split_it("334,433") #=> [334, 433]
split_it("345x532") #=> [345, 532]
split_it("432 345") #=> [432, 345]
split_it("432&345") #=> nil
split_it("x32 345") #=> nil
Your original regex would work with a minor adjustment to move the '+' symbol outside the character class:
"344 ,x 345".split(/[\s,x]+/).map(&:to_i) #==> [344,345]
If the examples are actually the only formats that you'll encounter, this will work well. However, if you have to be more flexible and accommodate unknown separators between the numbers, you're better off with the answer given by Wiktor:
"344 ,x 345".split(/\D+/).map(&:to_i) #==> [344,345]
Both cases will return an array of Integers from the inputs given, however the second example is both more robust and easier to understand at a glance.
it doesn't seem to match multiple spaces when testing
Yeah, character class (square brackets) doesn't work like this. You apply quantifiers on the class itself, not on its characters. You could use | operator instead. Something like this:
.split(%r[\s+|,\s*|x])
This is my expected result.
Input a string and get three returned string.
I have no idea how to finish it with Regex in Ruby.
this is my roughly idea.
match(/(.*?)(_)(.*?)(\d+)/)
Input and expected output
# "R224_OO2003" => R224, OO, 2003
# "R2241_OOP2003" => R2244, OOP, 2003
If the example description I gave in my comment on the question is correct, you need a very straightforward regex:
r = /(.+)_(.+)(\d{4})/
Then:
"R224_OO2003".scan(r).flatten #=> ["R224", "OO", "2003"]
"R2241_OOP2003".scan(r).flatten #=> ["R2241", "OOP", "2003"]
Assuming that your three parts consist of (R and one or more digits), then an underbar, then (one or more non-whitespace characters), before finally (a 4-digit numeric date), then your regex could be something like this:
^(R\d+)_(\S+)(\d{4})$
The ^ indicates start of string, and the $ indicates end of string. \d+ indicates one or more digits, while \S+ says one or more non-whitespace characters. The \d{4} says exactly four digits.
To recover data from the matches, you could either use the pre-defined globals that line up with your groups, or you could could use named captures.
To use the match globals just use $1, $2, and $3. In general, you can figure out the number to use by counting the left parentheses of the specific group.
To use the named captures, include ? right after the left paren of a particular group. For example:
x = "R2241_OOP2003"
match_data = /^(?<first>R\d+)_(?<second>\S+)(?<third>\d{4})$/.match(x)
puts match_data['first'], match_data['second'], match_data['third']
yields
R2241
OOP
2003
as expected.
As long as your pattern covers all possibilities, then you just need to use the match object to return the 3 strings:
my_match = "R224_OO2003".match(/(.*?)(_)(.*?)(\d+)/)
#=> #<MatchData "R224_OO2003" 1:"R224" 2:"_" 3:"OO" 4:"2003">
puts my_match[0] #=> "R224_OO2003"
puts my_match[1] #=> "R224"
puts my_match[2] #=> "_"
puts my_match[3] #=> "00"
puts my_match[4] #=> "2003"
A MatchData object contains an array of each match group starting at index [1]. As you can see, index [0] returns the entire string. If you don't want the capture the "_" you can leave it's parentheses out.
Also, I'm not sure you are getting what you want with the part:
(.*?)
this basically says one or more of any single character followed by zero or one of any single character.
I'm trying to write a regular expressions that will match a set of characters without regard to order. For example:
str = "act"
str.scan(/Insert expression here/)
would match:
cat
act
tca
atc
tac
cta
but would not match ca, ac or cata.
I read through a lot of similar questions and answers here on StackOverflow, but have not found one that matches my objectives exactly.
To clarify a bit, I'm using ruby and do not want to allow repeat characters.
Here is your solution
^(?:([act])(?!.*\1)){3}$
See it here on Regexr
^ # matches the start of the string
(?: # open a non capturing group
([act]) # The characters that are allowed and a capturing group
(?!.*\1) # That character is matched only if it does not occur once more, Lookahead assertion
){3} # Defines the amount of characters
$
The only special think is the lookahead assertion, to ensure the character is not repeated.
^ and $ are anchors to match the start and the end of the string.
[act]{3} or ^[act]{3}$ will do it in most regular expression dialects. If you can narrow down the system you're using, that will help you get a more specific answer.
Edit: as mentioned by #georgydyer in the comments below, it's unclear from your question whether or not repeated characters are allowed. If not, you can adapt the answer from this question and get:
^(?=[act]{3}$)(?!.*(.).*\1).*$
That is, a positive lookahead to check a match, and then a negative lookahead with a backreference to exclude repeated characters.
Here's how I'd go about it:
regex = /\b(?:#{ Regexp.union(str.split('').permutation.map{ |a| a.join }).source })\b/
# => /(?:act|atc|cat|cta|tac|tca)/
%w[
cat act tca atc tac cta
ca ac cata
].each do |w|
puts '"%s" %s' % [w, w[regex] ? 'matches' : "doesn't match"]
end
That outputs:
"cat" matches
"act" matches
"tca" matches
"atc" matches
"tac" matches
"cta" matches
"ca" doesn't match
"ac" doesn't match
"cata" doesn't match
I use the technique of passing an array into Regexp.union for a lot of things; I works especially well with the keys of a hash, and passing the hash into gsub for rapid search/replace on text templates. This is the example from the gsub documentation:
'hello'.gsub(/[eo]/, 'e' => 3, 'o' => '*') #=> "h3ll*"
Regexp.union creates a regex, and it's important to use source instead of to_s when extracting the actual pattern being generated:
puts regex.to_s
=> (?-mix:\b(?:act|atc|cat|cta|tac|tca)\b)
puts regex.source
=> \b(?:act|atc|cat|cta|tac|tca)\b
Notice how to_s embeds the pattern's flags inside the string. If you don't expect them you can accidentally embed that pattern into another, which won't behave as you expect. Been there, done that and have the dented helmet as proof.
If you really want to have fun, look into the Perl Regexp::Assemble module available on CPAN. Using that, plus List::Permutor, lets us generate more complex patterns. On a simple string like this it won't save much space, but on long strings or large arrays of desired hits it can make a huge difference. Unfortunately, Ruby has nothing like this, but it is possible to write a simple Perl script with the word or array of words, and have it generate the regex and pass it back:
use List::Permutor;
use Regexp::Assemble;
my $regex_assembler = Regexp::Assemble->new;
my $perm = new List::Permutor split('', 'act');
while (my #set = $perm->next) {
$regex_assembler->add(join('', #set));
}
print $regex_assembler->re, "\n";
(?-xism:(?:a(?:ct|tc)|c(?:at|ta)|t(?:ac|ca)))
See "Is there an efficient way to perform hundreds of text substitutions in Ruby?" for more information about using Regexp::Assemble with Ruby.
I will assume several things here:
- You are looking for permutations of given characters
- You are using ruby
str = "act"
permutations = str.split(//).permutation.map{|p| p.join("")}
# and for the actual test
permutations.include?("cat")
It is no regex though.
No doubt - the regex that uses positive/negative lookaheads and backreferences is slick, but if you're only dealing with three characters, I'd err on the side of verbosity by explicitly enumerating the character permutations like #scones suggested.
"act".split('').permutation.map(&:join)
=> ["act", "atc", "cat", "cta", "tac", "tca"]
And if you really need a regex out of it for scanning a larger string, you can always:
Regexp.union "act".split('').permutation.map(&:join)
=> /\b(act|atc|cat|cta|tac|tca)\b/
Obviously, this strategy doesn't scale if your search string grows, but it's much easier to observe the intent of code like this in my opinion.
EDIT: Added word boundaries for false positive on cata based on #theTinMan's feedback.
I don't know whether it's really easy and I'm out of my mind....
In Ruby's regular expressions, how to match strings which do not contain two consecutive underscores, i.e., "__".
Ex:
Matches: "abcd", "ab_cd", "a_b_cd", "%*##_#+"
Does not match: "ab__cd", "a_b__cd"
-thanks
EDIT: I can't use reverse logic, i.e., checking for "__" strings and excluding them, since need to use with Ruby on Rails "validates_format_of()" which expects a regular expression with which it will match.
You could use negative lookahead:
^((?!__).)*$
The beginning-of-string ^ and end of string $ are important, they force a check of "not followed by double underscore" on every position.
/^([^_]*(_[^_])?)*_?$/
Tests:
regex=/^([^_]*(_[^_])?)*_?$/
# Matches
puts "abcd" =~ regex
puts "ab_cd" =~ regex
puts "a_b_cd" =~ regex
puts "%*##_#+" =~ regex
puts "_" =~ regex
puts "_a_" =~ regex
# Non-matches
puts "__" =~ regex
puts "ab__cd" =~ regex
puts "a_b__cd" =~ regex
But regex is overkill for this task. A simple string test is much easier:
puts ('a_b'['__'])
Would altering your logic still be valid?
You could check if the string contains two underscores with the regular expression [_]{2} and then just ignore it?
Negative lookahead
\b(?!\w*__\w*)\w+\b
Search for two consecutive underscores in the next word from the beginning of the word, and match that word if it is not found.
Edit: To accommodate anything other than whitespaces in the match:
(?!\S*__\S*)\S+
If you wish to accommodate a subset of symbols, you can write something like the following, but then it will match _cd from a_b__cd among other things.
(?![a-zA-Z0-9_%*##+]*__[a-zA-Z0-9_%*##+]*)[a-zA-Z0-9_%*##+]+