Consider this example string:
mystr ="1. moody"
I want to capitalize the first letter that occurs in mystr. I am trying this regular expression in Ruby but still returns all the letters in mystr (moody) instead of the letter m only.
puts mystr.scan(/[a-zA-Z]{1}/)
Any help appreciated!
Do as below using String#sub
(arup~>~)$ pry --simple-prompt
>> s = "1. moody"
=> "1. moody"
>> s.sub(/[a-z]/i,&:upcase)
=> "1. Moody"
>>
If you want to modify the source string use s.sub!(/[a-z]/,&:upcase).
Just for completeness, although it doesn’t directly answer your question as posed but could be relevant, consider this variation:
mystr ="1. école"
The line mystr.sub(/[a-z]/i,&:upcase) (as in Arup Rakshit’s answer) will match the second letter of the word, producing
1. éCole
The line mystr.sub /\b\s?[a-zA-Z]{1}/, &:upcase (diego.greyrobot’s answer) won’t match at all and so the line will be unchanged.
There are two problems here. The first is that [a-zA-Z] doesn’t match accented characters, so é isn’t matched. The fix for this is to use the \p{Letter} character property:
mystr.sub /\p{Letter}/, &:upcase
This will match the character in question, but won’t change it. This is due to the second problem, which is that upcase (and downcase) only works on characters in the ASCII range. This is almost as easy to fix, but relies on using an external library such as unicode_utils:
require 'unicode_utils'
mystr.sub(/\p{Letter}/) { |c| UnicodeUtils.upcase(c)}
This results in:
1. École
which is probably what is wanted in this case.
This may not affect you if you are sure all your data is just ASCII, but is worth knowing for other situations.
The reason your attempt returns all the letters is because you are using the scan method which does just that, it returns all the characters which match the regex, in your case letters. For your use case you should use sub since you only want to substitute 1 letter.
I use http://rubular.com to practice my Ruby Regexes. Here's what I came up with http://rubular.com/r/fAQEDFVEVn
The regex is: /\b[a-z]/
It uses \b to find a word boundary, and finally we ask for one letter only with [a-zA-Z]
Finally we'll use sub to replace it with its upcased version:
"1. moody".sub /\b[a-z]/, &:upcase
=> "1. Moody"
Hope that helps.
Related
My code is about a robot who has 3 posible answers (it depends on what you put in the message)
So, inside this posible answers, one depends if the input it's a question, and to prove it, i think it has to identify the "?" symbol on the string.
May i have to use the "match" method or includes?
This code it's gonna be include in a loop, that may answer in 3 possible ways.
Example:
puts "whats your meal today?"
answer = gets.chomp
answer.includes? "?"
or
answer.match('?')
Take a look at String#end_with? I think that is what you should use.
Use String#match? Instead
String#chomp will only remove OS-specific newlines from a String, but neither String#chomp nor String#end_with? will handle certain edge cases like multi-line matches or strings where you have whitespace characters at the end. Instead, use a regular expression with String#match?. For example:
print "Enter a meal: "
answer = gets.chomp
answer.match? /\?\s*\z/m
The Regexp literal /\?\s*\z/m will return true value if the (possibly multi-line) String in your answer contains:
a literal question mark (which is why it's escaped)...
followed by zero or more whitespace characters...
anchored to the end-of-string with or without newline characters, e.g. \n or \r\n, although those will generally have been removed by #chomp already.
This will be more robust than your current solution, and will handle a wider variety of inputs while being more accurate at finding strings that end with a question mark without regard to trailing whitespace or line endings.
I'm trying to get my head around how to work with Classes in Ruby and would really appreciate some insight on this area. Currently, I've got a rather simple task to convert a string with the start of each word capitalized. For example:
Not Jaden-Cased: "How can mirrors be real if our eyes aren't real"
Jaden-Cased: "How Can Mirrors Be Real If Our Eyes Aren't Real"
This is my code currently:
class String
def toJadenCase
split
capitalize
end
end
#=> usual case: split.map(&:capitalize).join(' ')
Output:
Expected: "The Moment That Truth Is Organized It Becomes A Lie.",
instead got: "The moment that truth is organized it becomes a lie."
I suggest you not pollute the core String class with the addition of an instance method. Instead, just add an argument to the method to hold the string. You can do that as follows, by downcasing the string then using gsub with a regular expression.
def to_jaden_case(str)
str.downcase.gsub(/(?<=\A| )[a-z]/) { |c| c.upcase }
end
to_jaden_case "The moMent That trUth is organized, it becomes a lie."
#=> "The Moment That Truth Is Organized, It Becomes A Lie."
Ruby's regex engine performs the following operations.
(?<=\A| ) : use a positive lookbehind to assert that the following match
is immediately preceded by the start of the string or a space
[a-z] : match a lowercase letter
(?<=\A| ) can be replaced with the negative lookbehind (?<![^ ]), which asserts that the match is not preceded by a character other than a space.
Notice that by using String#gsub with a regular expression (unlike the split-process-join dance), extra spaces are preserved.
When spaces are to be matched by a regular expression one often sees whitespaces (\s) matched instead. Here, for example, /(?<=\A|\s)[a-z]/ works fine, but sometimes matching whitespaces leads to problems, mainly because they also match newlines (\n) (as well as spaces, tabs and a few other characters). My advice is to match space characters if spaces are to be matched. If tabs are to be matched as well, use a character class ([ \t]).
Try:
def toJadenCase
self.split.map(&:capitalize).join(' ')
end
Trying to extract '4995' from the string '4.995,-' with regex in Ruby.
I tried with
/\d+/
Which seems to work from this Rubular screenshot: http://cl.ly/image/111c2x0N3s0C
but running it only outputs
4
You cannot match it in a single regex because it is not a single substring.
"4.995,-".gsub(/\D/, "") # => "4995"
I'm up-voting sawa's answer because it's a good answer.
But since you are new to regular expressions, you may want further explanation as to why his answer works for you.
When you are trying to match with the regexp /\d+/, what you are saying is "Match for me 1 or more consecutive digits." But your target string, 4.995,-, is not made up of only consecutive digits. It has a 4 and it has a 995. The first match of "1 or more consecutive digits" is 4. That's why what you're getting as a result is 4.
Try to look at your problem differently. Instead of saying, "Find me all the digits and extract those out," you could say, "Find me anything that's not a digit, and get rid of it." To do this, you can use ruby's search-and-replace function, gsub. gsub searches a target string for anything that matches a given regular expression, and then it replaces those matches with some replacement string that you also provide. Documentation on gsub can be found here
The regular expression for "non-digit" is /\D/. So, you can do a gsub that looks for any /\D/ and replaces it with a blank string.
'4.995,-'.gsub(/\D/,'')
Do as below using String#[] and String#tr:
"4.995,-"[/\d+.\d+/].tr('.','') # => "4995"
# more Rubyish way using #tr method only
"4.995,-".tr("^0-9",'') # => "4995"
p '4.995,-1'.delete('.')[/\d+/] #=> "4995"
Here's another way that, like #Arup's solution, works when a digit follows the first non-digit:
'4.995,-1'.sub('.','').to_i.to_s #=> "4995"
This works because
'4.995,-1'.sub('.','') #=> "4995,-1"
and to_i takes the first part part of a string that can be converted to a Fixnum.
Alternatively:
'4.995,-1'.to_f.to_s.sub('.','') #=> "4995"
I have a string such as this: "im# -33.870816,151.203654"
I want to extract the two numbers including the hyphen.
I tried this:
mystring = "im# -33.870816,151.203654"
/\D*(\-*\d+\.\d+),(\-*\d+\.\d+)/.match(mystring)
This gives me:
33.870816,151.203654
How do I get the hyphen?
I need to do this in ruby
Edit: I should clarify, the "im# " was just an example, there can be any set of characters before the numbers. the numbers are mostly well formed with the comma. I was having trouble with the hyphen (-)
Edit2: Note that the two nos are lattidue, longitude. That pattern is mostly fixed. However, in theory, the preceding string can be arbitrary. I don't expect it to have nos. or hyphen, but you never know.
How about this?
arr = "im# -33.2222,151.200".split(/[, ]/)[1..-1]
and arr is ["-33.2222", "151.200"], (using the split method).
now
arr[0].to_f is -33.2222 and arr[1].to_f is 151.2
EDIT: stripped "im#" part with [1..-1] as suggested in comments.
EDIT2: also, this work regardless of what the first characters are.
If you want to capture the two numbers with the hyphen you can use this regex:
> str = "im# -33.870816,151.203654"
> str.match(/([\d.,-]+)/).captures
=> ["33.870816,151.203654"]
Edit: now it captures hyphen.
This one captures each number separetely: http://rubular.com/r/NNP2OTEdiL
Note: Using String#scan will match all ocurrences of given pattern, in this case
> str.scan /\b\s?([-\d.]+)/
=> [["-33.870816"], ["151.203654"]] # Good, but flattened version is better
> str.scan(/\b\s?([-\d.]+)/).flatten
=> ["-33.870816", "151.203654"]
I recommend you playing around a little with Rubular. There's also some docs about regegular expressions with Ruby:
http://www.ruby-doc.org/docs/ProgrammingRuby/html/language.html#UJ
http://www.regular-expressions.info/ruby.html
http://www.ruby-doc.org/core-1.9.3/Regexp.html
Your regex doesn't work because the hyphen is caught by \D, so you have to modify it to catch only the right set of characters.
[^0-9-]* would be a good option.
I'm not sure how to use regular expressions in a function so that I could grab all the words in a sentence starting with a particular letter. I know that I can do:
word =~ /^#{letter}/
to check if the word starts with the letter, but how do I go from word to word. Do I need to convert the string to an array and then iterate through each word or is there a faster way using regex? I'm using ruby so that would look like:
matching_words = Array.new
sentance.split(" ").each do |word|
matching_words.push(word) if word =~ /^#{letter}/
end
Scan may be a good tool for this:
#!/usr/bin/ruby1.8
s = "I think Paris in the spring is a beautiful place"
p s.scan(/\b[it][[:alpha:]]*/i)
# => ["I", "think", "in", "the", "is"]
\b means 'word boundary."
[:alpha:] means upper or lowercase alpha (a-z).
You can use \b. It matches word boundaries--the invisible spot just before and after a word. (You can't see them, but oh they're there!) Here's the regex:
/\b(a\w*)\b/
The \w matches a word character, like letters and digits and stuff like that.
You can see me testing it here: http://rubular.com/regexes/13347
Similar to Anon.'s answer:
/\b(a\w*)/g
and then see all the results with (usually) $n, where n is the n-th hit. Many libraries will return /g results as arrays on the $n-th set of parenthesis, so in this case $1 would return an array of all the matching words. You'll want to double-check with whatever library you're using to figure out how it returns matches like this, there's a lot of variation on global search returns, sadly.
As to the \w vs [a-zA-Z], you can sometimes get faster execution by using the built-in definitions of things like that, as it can easily have an optimized path for the preset character classes.
The /g at the end makes it a "global" search, so it'll find more than one. It's still restricted by line in some languages / libraries, though, so if you wish to check an entire file you'll sometimes need /gm, to make it multi-line
If you want to remove results, like your title (but not question) suggests, try:
/\ba\w*//g
which does a search-and-replace in most languages (/<search>/<replacement>/). Sometimes you need a "s" at the front. Depends on the language / library. In Ruby's case, use:
string.gsub(/(\b)a\w*(\b)/, "\\1\\2")
to retain the non-word characters, and optionally put any replacement text between \1 and \2. gsub for global, sub for the first result.
/\ba[a-z]*\b/i
will match any word starting with 'a'.
The \b indicates a word boundary - we want to only match starting from the beginning of a word, after all.
Then there's the character we want our word to start with.
Then we have as many as possible letter characters, followed by another word boundary.
To match all words starting with t, use:
\bt\w+
That will match test but not footest; \b means "word boundary".
Personally i think that regex is overkill for this application, simply running a select is more than capable of solving this particular problem.
"this is a test".split(' ').select{ |word| word[0,1] == 't' }
result => ["this", "test"]
or if you are determined to use regex then go with grep
"this is a test".split(' ').grep(/^t/)
result => ["this", "test"]
Hope this helps.