Suppose I have a String:
someString = "1374j03d42s23dc"
I want to find the first index of a non-numeric character. In this case, that would be 4. How can I do this with a regex?
(I'm not very good at regex, so it would be great if the answer could explain what is going on)
someString =~ /\D/
# => 4
........
In addition to sawa's solution: You could also use String#index when you like your code to be more readable:
string = '1374j03d42s23dc'
string.index(/\D/)
#=> 4
/\D/ matches any non-digit (list of common regexp metacharacters)
Try:
someString.each_char.with_index { |c,i| puts i if c == someString.scan(/\D+/)[0] }
Related
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.
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"
I'd like to have a reqex that checks that every character is a number [0-9]+. I have tried:
'4th'=~/[\d]+/
'4th'=~/\d+/
but not working. How would I check for this?
thx
"12345" =~ /\A\d+\Z/
\A = beginning of string (not line, string)
\d+ = one or more digits
\Z = end of string (not line, string)
The simplest way is:
str !~ /\D/
something like this:
a = 6
case a
when /\d/ then "it's a number"
end
no luck, it doesn't work
When used with a value on the initializer, all case does is try it with === against each expression. The problem isn't with case, try:
6 === /\d/
All that to say, regexes match against strings only. Try replacing the second line by:
case (a.is_a?(String) ? a : a.to_s)
EDIT: To answer the OP's follow-up in comments, there's a subtlety here.
/\d/ === '6' # => true
'6' === /\d/ # => false
Perhaps unexpectedly to the beginner, String#=== and Regexp#=== have different effects. So, for:
case 'foo'
when String
end
This will call String === 'foo', not 'foo' === String, etc.
It doesn't work because regexes match against a string, whereas 6 is not a string. If you do a = '6', it shall work.
Because regexps match strings. A is a Fixnum.
If you would write a = "6", it would work. Testing if a is a number can be done with a.is_a?(Numeric)
One minor change to make it work:
a = 6
case a.to_s
when /\d/ then "it's a number"
end
The to_s will convert everything to a string. Note that your regex just checks for the existence of a digit anywhere in the string.
It would perhaps be better to do this:
case a
when Numeric then "it's a number"
end
How do I remove a substring after a certain character in a string using Ruby?
new_str = str.slice(0..(str.index('blah')))
I find that "Part1?Part2".split('?')[0] is easier to read.
I'm surprised nobody suggested to use 'gsub'
irb> "truncate".gsub(/a.*/, 'a')
=> "trunca"
The bang version of gsub can be used to modify the string.
str = "Hello World"
stopchar = 'W'
str.sub /#{stopchar}.+/, stopchar
#=> "Hello W"
A special case is if you have multiple occurrences of the same character and you want to delete from the last occurrence to the end (not the first one).
Following what Jacob suggested, you just have to use rindex instead of index as rindex gets the index of the character in the string but starting from the end.
Something like this:
str = '/path/to/some_file'
puts str.slice(0, str.index('/')) # => ""
puts str.slice(0, str.rindex('/')) # => "/path/to"
We can also use partition and rpartitiondepending on whether we want to use the first or last instance of the specified character:
string = "abc-123-xyz"
last_char = "-"
string.partition(last_char)[0..1].join #=> "abc-"
string.rpartition(last_char)[0..1].join #=> "abc-123-"