How do I a string against a regex such that it will return true if the whole string matches (not a substring)?
eg:
test( \ee\ , "street" ) #=> returns false
test( \ee\ , "ee" ) #=> returns true!
Thank you.
You can match the beginning of the string with \A and the end with \Z. In ruby ^ and $ match also the beginning and end of the line, respectively:
>> "a\na" =~ /^a$/
=> 0
>> "a\na" =~ /\Aa\Z/
=> nil
>> "a\na" =~ /\Aa\na\Z/
=> 0
This seems to work for me, although it does look ugly (probably a more attractive way it can be done):
!(string =~ /^ee$/).nil?
Of course everything inside // above can be any regex you want.
Example:
>> string = "street"
=> "street"
>> !(string =~ /^ee$/).nil?
=> false
>> string = "ee"
=> "ee"
>> !(string =~ /^ee$/).nil?
=> true
Note: Tested in Rails console with ruby (1.8.7) and rails (3.1.1)
So, what you are asking is how to test whether the two strings are equal, right? Just use string equality! This passes every single one of the examples that both you and Tomas cited:
'ee' == 'street' # => false
'ee' == 'ee' # => true
"a\na" == 'a' # => false
"a\na" == "a\na" # => true
Related
How do I a string against a regex such that it will return true if the whole string matches (not a substring)?
eg:
test( \ee\ , "street" ) #=> returns false
test( \ee\ , "ee" ) #=> returns true!
Thank you.
You can match the beginning of the string with \A and the end with \Z. In ruby ^ and $ match also the beginning and end of the line, respectively:
>> "a\na" =~ /^a$/
=> 0
>> "a\na" =~ /\Aa\Z/
=> nil
>> "a\na" =~ /\Aa\na\Z/
=> 0
This seems to work for me, although it does look ugly (probably a more attractive way it can be done):
!(string =~ /^ee$/).nil?
Of course everything inside // above can be any regex you want.
Example:
>> string = "street"
=> "street"
>> !(string =~ /^ee$/).nil?
=> false
>> string = "ee"
=> "ee"
>> !(string =~ /^ee$/).nil?
=> true
Note: Tested in Rails console with ruby (1.8.7) and rails (3.1.1)
So, what you are asking is how to test whether the two strings are equal, right? Just use string equality! This passes every single one of the examples that both you and Tomas cited:
'ee' == 'street' # => false
'ee' == 'ee' # => true
"a\na" == 'a' # => false
"a\na" == "a\na" # => true
I need to create regular expression for 2 and only 2 letters. I understood it has to be the following /[a-z]{2}/i, but it matches any string with 2 or more letters. Here is what I get:
my_reg_exp = /[a-z]{2}/i
my_reg_exp.match('aa') # => #<MatchData "aa">
my_reg_exp.match('AA') # => #<MatchData "AA">
my_reg_exp.match('a') # => nil
my_reg_exp.match('aaa') # => #<MatchData "aa">
Any suggestion?
You can add the anchors like this:
my_reg_exp = /^[a-z]{2}$/i
Test:
my_reg_exp.match('aaa')
#=> nil
my_reg_exp.match('aa')
#=> #<MatchData "aa">
Hao's solution matches isn't locale sensitive. If this is important for your use case:
/\a[[:alpha:]]{2}\z/
2.0.0-p451 :005 > 'aba' =~ /\A[[:alpha:]]{2}\Z/
=> nil
2.0.0-p451 :006 > 'ab' =~ /\A[[:alpha:]]{2}\Z/
=> 0
2.0.0-p451 :007 > 'xy' =~ /\A[[:alpha:]]{2}\Z/
=> 0
2.0.0-p451 :008 > 'zxy' =~ /\A[[:alpha:]]{2}\Z/
=> nil
Per usual, if you need further assistance, leave a comment.
You can use /\b[a-z]{2}\b/i to match a two-letter string. /b Matches a word-break.
This means you can scan a string to find all occurrences:
'Foo is a bar'.scan(/\b[a-z]{2}\b/i) #=> ["is"]
Or find the first match in a string using:
'a bc def'[/\b[a-z]{2}\b/i] # => "bc"
I have an Excel column containing part numbers. Here is a sample
As you can see, it can be many different datatypes: Float, Int, and String. I am using roo gem to read the file. The problem is that roo interprets integer cells as Float, adding a trailing zero to them (16431 => 16431.0). I want to trim this trailing zero. I cannot use to_i because it will trim all the trailing numbers of the cells that require a decimal in them (the first row in the above example) and will cut everything after a string char in the String rows (the last row in the above example).
Currently, I have a a method that checks the last two characters of the cell and trims them if they are ".0"
def trim(row)
if row[0].to_s[-2..-1] == ".0"
row[0] = row[0].to_s[0..-3]
end
end
This works, but it feels terrible and hacky. What is the proper way of getting my Excel file contents into a Ruby data structure?
def trim num
i, f = num.to_i, num.to_f
i == f ? i : f
end
trim(2.5) # => 2.5
trim(23) # => 23
or, from string:
def convert x
Float(x)
i, f = x.to_i, x.to_f
i == f ? i : f
rescue ArgumentError
x
end
convert("fjf") # => "fjf"
convert("2.5") # => 2.5
convert("23") # => 23
convert("2.0") # => 2
convert("1.00") # => 1
convert("1.10") # => 1.1
For those using Rails, ActionView has the number_with_precision method that takes a strip_insignificant_zeros: true argument to handle this.
number_with_precision(13.00, precision: 2, strip_insignificant_zeros: true)
# => 13
number_with_precision(13.25, precision: 2, strip_insignificant_zeros: true)
# => 13.25
See the number_with_precision documentation for more information.
This should cover your needs in most cases: some_value.gsub(/(\.)0+$/, '').
It trims all trailing zeroes and a decimal point followed only by zeroes. Otherwise, it leaves the string alone.
It's also very performant, as it is entirely string-based, requiring no floating point or integer conversions, assuming your input value is already a string:
Loading development environment (Rails 3.2.19)
irb(main):001:0> '123.0'.gsub(/(\.)0+$/, '')
=> "123"
irb(main):002:0> '123.000'.gsub(/(\.)0+$/, '')
=> "123"
irb(main):003:0> '123.560'.gsub(/(\.)0+$/, '')
=> "123.560"
irb(main):004:0> '123.'.gsub(/(\.)0+$/, '')
=> "123."
irb(main):005:0> '123'.gsub(/(\.)0+$/, '')
=> "123"
irb(main):006:0> '100'.gsub(/(\.)0+$/, '')
=> "100"
irb(main):007:0> '127.0.0.1'.gsub(/(\.)0+$/, '')
=> "127.0.0.1"
irb(main):008:0> '123xzy45'.gsub(/(\.)0+$/, '')
=> "123xzy45"
irb(main):009:0> '123xzy45.0'.gsub(/(\.)0+$/, '')
=> "123xzy45"
irb(main):010:0> 'Bobby McGee'.gsub(/(\.)0+$/, '')
=> "Bobby McGee"
irb(main):011:0>
Numeric values are returned as type :float
def convert_cell(cell)
if cell.is_a?(Float)
i = cell.to_i
cell == i.to_f ? i : cell
else
cell
end
end
convert_cell("foobar") # => "foobar"
convert_cell(123) # => 123
convert_cell(123.4) # => 123.4
I don't remember where I learned the !~ method of the String class. However I know it compares a string to a regex and check whether the string not match the regex. See my below example.
C:\>irb
irb(main):001:0> "abba" =~ /(\w)(\w)\2\1/i
=> 0
irb(main):002:0> "xxxx" =~ /(\w)(\w)\2\1/i
=> 0
irb(main):003:0> "asdf" =~ /(\w)(\w)\2\1/i
=> nil
irb(main):004:0> "asdf" !~ /(\w)(\w)\2\1/i
=> true
irb(main):005:0> "asdf" !~ /asdf/i
=> false
irb(main):006:0>
I want to find more information of the method but I can't find it in the rdoc of both String and Regexp. Anyone can give some help?
Thanks.
Since this is the method you can find it here in the Methods filter.
I've found this description.
obj !~ other → true or false
Returns true if two objects do not match (using the =~ method), otherwise false.
(Update: this question's main focus is to test the "nothing else" part)
Given a string s, which can contain anything, what is the most correct regexp in Ruby to check whether it is a single digit and nothing else? (a single digit and only a single digit).
Use /\A\d\z/
irb(main):001:0> "asd\n7\n" =~ /\A\d\Z/
=> nil # works as false
irb(main):002:0> "asd\n7\n" =~ /\A\d\z/
=> nil # works as false
irb(main):083:0> "7\n"=~/\A\d\Z/
=> 0 # \Z fails, need \z
irb(main):084:0> "7\n"=~/\A\d\z/
=> nil # works as false
irb(main):005:0> "7" =~ /\A\d\Z/
=> 0 # works as true
irb(main):006:0> "7" =~ /\A\d\z/
=> 0 # works as true
http://www.zenspider.com/Languages/Ruby/QuickRef.html :
\z end of a string
\Z end of a string, or before newline at the end
Try /\A\d(?![\S\W])/?
irb(main):016:0> "7" =~ /\A\d(?![\S\W])/
=> 0
irb(main):017:0> "7\n" =~ /\A\d(?![\S\W])/
=> nil
irb(main):018:0> "aljda\n7\n" =~ /\A\d(?![\S\W])/
=> nil
irb(main):022:0> "85" =~ /\A\d(?![\S\W])/
=> nil
irb(main):023:0> "b5" =~ /\A\d(?![\S\W])/
=> nil
s.scan(/\b\d\b/)
irb(main):001:0> "7\n" =~ /\b\d\z/
=> nil
irb(main):002:0> "7" =~ /\b\d\z/
=> 0
^\d$
^ is start of the string, \d is a digit and $ is the end of the string
NOTE: as stated in comments ^ and $ work for both lines and strings so if you plan to have a multi-line input you should use \A and \Z