String#delete interprets a-z as character range. However, I would like it to delete fa-zo.
"fojwfa-zowj".delete("fa-zo") #=> "-"
Desired result:
"fojwwj"
You could also use this little trick:
string = "fojwfa-zowj"
string[/fa-zo/] = ''
string
# => "fojwwj"
Notice however, that this modifies the string in place like #gsub!, which should be faster and should use less memory, but which could introduce side-effects if not considered well.
"fojwfa-zowj".gsub("fa-zo","") # => "fojwwj"
"fojwfa-zowj".tap{ |s| s.slice! "fa-zo" } # just for the Heaven of it
Related
I have a string as given below,
./component/unit
and need to split to get result as component/unit which I will use this as key for inserting hash.
I tried with .split(/.\//).last but its giving result as unit only not getting component/unit.
I think, this should help you:
string = './component/unit'
string.split('./')
#=> ["", "component/unit"]
string.split('./').last
#=> "component/unit"
Your regex was almost fine :
split(/\.\//)
You need to escape both . (any character) and / (regex delimiter).
As an alternative, you could just remove the first './' substring :
'./component/unit'.sub('./','')
#=> "component/unit"
All the other answers are fine, but I think you are not really dealing with a String here but with a URI or Pathname, so I would advise you to use these classes if you can. If so, please adjust the title, as it is not about do-it-yourself-regexes, but about proper use of the available libraries.
Link to the ruby doc:
https://docs.ruby-lang.org/en/2.1.0/URI.html
and
https://ruby-doc.org/stdlib-2.1.0/libdoc/pathname/rdoc/Pathname.html
An example with Pathname is:
require 'pathname'
pathname = Pathname.new('./component/unit')
puts pathname.cleanpath # => "component/unit"
# pathname.to_s # => "component/unit"
Whether this is a good idea (and/or using URI would be cool too) also depends on what your real problem is, i.e. what you want to do with the extracted String. As stated, I doubt a bit that you are really intested in Strings.
Using a positive lookbehind, you could do use regex:
reg = /(?<=\.\/)[\w+\/]+\w+\z/
Demo
str = './component'
str2 = './component/unit'
str3 = './component/unit/ruby'
str4 = './component/unit/ruby/regex'
[str, str2, str3, str4].each { |s| puts s[reg] }
#component
#component/unit
#component/unit/ruby
#component/unit/ruby/regex
How can I remove from a string all characters except white spaces, numbers, and some others?
Something like this:
oneLine.gsub(/[^ULDR0-9\<\>\s]/i,'')
I need only: 0-9 l d u r < > <space>
Also, is there a good document about the use of regex in Ruby, like a list of special characters with examples?
The regex you have is already working correctly. However, you do need to assign the result back to the string you're operating on. Otherwise, you're not changing the string (.gsub() does not modify the string in-place).
You can improve the regex a bit by adding a '+' quantifier (so consecutive characters can be replaced in one go). Also, you don't need to escape angle brackets:
oneLine = oneLine.gsub(/[^ULDR0-9<>\s]+/i, '')
A good resource with special consideration of Ruby regexes is the Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan. A good online tutorial by the same author is here.
Good old String#delete does this without a regular expression. The ^ means 'NOT'.
str = "12eldabc8urp pp"
p str.delete('^0-9ldur<> ') #=> "12ld8ur "
Just for completeness: you don't need a regular expression for this particular task, this can be done using simple string manipulation:
irb(main):005:0> "asdasd123".tr('^ULDRuldr0-9<>\t\r\n ', '')
=> "dd123"
There's also the tr! method if you want to replace the old value:
irb(main):009:0> oneLine = 'UasdL asd 123'
irb(main):010:0> oneLine.tr!('^ULDRuldr0-9<>\t\r\n ', '')
irb(main):011:0> oneLine
=> "UdL d 123"
This should be a bit faster as well (but performance shouldn't be a big concern in Ruby :)
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-"
I want to strip leading and trailing quotes, in Ruby, from a string. The quote character will occur 0 or 1 time. For example, all of the following should be converted to foo,bar:
"foo,bar"
"foo,bar
foo,bar"
foo,bar
You could also use the chomp function, but it unfortunately only works in the end of the string, assuming there was a reverse chomp, you could:
'"foo,bar"'.rchomp('"').chomp('"')
Implementing rchomp is straightforward:
class String
def rchomp(sep = $/)
self.start_with?(sep) ? self[sep.size..-1] : self
end
end
Note that you could also do it inline, with the slightly less efficient version:
'"foo,bar"'.chomp('"').reverse.chomp('"').reverse
EDIT: Since Ruby 2.5, rchomp(x) is available under the name delete_prefix, and chomp(x) is available as delete_suffix, meaning that you can use
'"foo,bar"'.delete_prefix('"').delete_suffix('"')
I can use gsub to search for the leading or trailing quote and replace it with an empty string:
s = "\"foo,bar\""
s.gsub!(/^\"|\"?$/, '')
As suggested by comments below, a better solution is:
s.gsub!(/\A"|"\Z/, '')
As usual everyone grabs regex from the toolbox first. :-)
As an alternate I'll recommend looking into .tr('"', '') (AKA "translate") which, in this use, is really stripping the quotes.
Another approach would be
remove_quotations('"foo,bar"')
def remove_quotations(str)
if str.start_with?('"')
str = str.slice(1..-1)
end
if str.end_with?('"')
str = str.slice(0..-2)
end
end
It is without RegExps and start_with?/end_with? are nicely readable.
It frustrates me that strip only works on whitespace. I need to strip all kinds of characters! Here's a String extension that will fix that:
class String
def trim sep=/\s/
sep_source = sep.is_a?(Regexp) ? sep.source : Regexp.escape(sep)
pattern = Regexp.new("\\A(#{sep_source})*(.*?)(#{sep_source})*\\z")
self[pattern, 2]
end
end
Output
'"foo,bar"'.trim '"' # => "foo,bar"
'"foo,bar'.trim '"' # => "foo,bar"
'foo,bar"'.trim '"' # => "foo,bar"
'foo,bar'.trim '"' # => "foo,bar"
' foo,bar'.trim # => "foo,bar"
'afoo,bare'.trim /[aeiou]/ # => "foo,bar"
Assuming that quotes can only appear at the beginning or end, you could just remove all quotes, without any custom method:
'"foo,bar"'.delete('"')
I wanted the same but for slashes in url path, which can be /test/test/test/ (so that it has the stripping characters in the middle) and eventually came up with something like this to avoid regexps:
'/test/test/test/'.split('/').reject(|i| i.empty?).join('/')
Which in this case translates obviously to:
'"foo,bar"'.split('"').select{|i| i != ""}.join('"')
or
'"foo,bar"'.split('"').reject{|i| i.empty?}.join('"')
Regexs can be pretty heavy and lead to some funky errors. If you are not dealing with massive strings and the data is pretty uniform you can use a simpler approach.
If you know the strings have starting and leading quotes you can splice the entire string:
string = "'This has quotes!'"
trimmed = string[1..-2]
puts trimmed # "This has quotes!"
This can also be turned into a simple function:
# In this case, 34 is \" and 39 is ', you can add other codes etc.
def trim_chars(string, char_codes=[34, 39])
if char_codes.include?(string[0]) && char_codes.include?(string[-1])
string[1..-2]
else
string
end
end
You can strip non-optional quotes with scan:
'"foo"bar"'.scan(/"(.*)"/)[0][0]
# => "foo\"bar"
How can I remove the very first "1" from any string if that string starts with a "1"?
"1hello world" => "hello world"
"112345" => "12345"
I'm thinking of doing
string.sub!('1', '') if string =~ /^1/
but I' wondering there's a better way. Thanks!
Why not just include the regex in the sub! method?
string.sub!(/^1/, '')
As of Ruby 2.5 you can use delete_prefix or delete_prefix! to achieve this in a readable manner.
In this case "1hello world".delete_prefix("1").
More info here:
https://blog.jetbrains.com/ruby/2017/10/10-new-features-in-ruby-2-5/
https://bugs.ruby-lang.org/issues/12694
'invisible'.delete_prefix('in') #=> "visible"
'pink'.delete_prefix('in') #=> "pink"
N.B. you can also use this to remove items from the end of a string with delete_suffix and delete_suffix!
'worked'.delete_suffix('ed') #=> "work"
'medical'.delete_suffix('ed') #=> "medical"
https://bugs.ruby-lang.org/issues/13665
I've answered in a little more detail (with benchmarks) here: What is the easiest way to remove the first character from a string?
if you're going to use regex for the match, you may as well use it for the replacement
string.sub!(%r{^1},"")
BTW, the %r{} is just an alternate syntax for regular expressions. You can use %r followed by any character e.g. %r!^1!.
Careful using sub!(/^1/,'') ! In case the string doesn't match /^1/ it will return nil. You should probably use sub (without the bang).
This answer might be more optimised: What is the easiest way to remove the first character from a string?
string[0] = '' if string[0] == '1'
I'd like to post a tiny improvement to the otherwise excellent answer by Zach. The ^ matches the beginning of every line in Ruby regex. This means there can be multiple matches per string. Kenji asked about the beginning of the string which means they have to use this regex instead:
string.sub!(/\A1/, '')
Compare this - multiple matches with this - one match.