I have a string of four blank lines which all up makes eight lines in total in the following:
str = "aaa\n\n\nbbb\n\nccc\ddd\n"
I want to return this all in one line. The output should be like this on a single line:
aaabbbcccddd
I used various trim functions to get the output but still I am failing.
What method do I have to use here?
The Ruby (and slightly less Perl-ish) way:
new_str = str.delete "\n"
...or if you want to do it in-place:
str.delete! "\n"
str.gsub(/\n/,'')
> str = "aaa\n\n\nbbb\n\nccc\ddd\n"
=> "aaa\n\n\nbbb\n\ncccddd\n"
> str.gsub("\n", "")
=> "aaabbbcccddd"
Related
I have some strings like this
../..//somestring
../somestring
../..somestring
./somestring
How to write a regular expression in ruby to extract "somestring" from above strings. Before some strings it can be any combination of . and /
Thanks for you help
Do this:
string.sub(/\A[.\/]+/, "")
"../../test/file/cases".sub(/\A[.\/]+/, "")
# => "test/file/cases"
Just match letters:
str = "../..//somestring" # or "../somestring", "../..somestring", "./somestring"
str[/[a-z]+/] # somestring
RE: your comment
If you just want to remove leading dots and slashes, use
str.gsub(/[.\/]/, '')
It looks like you're dealing with file paths. If so, there are more appropriate tools than regexps
File.basename('../..//somestring')
# "somestring"
regexp = /^[.\/]+(.*?)$/i
result = subject.scan(regexp)
http://rubular.com/r/tMYKPe78uc
Is it possible to display the characters around a regex match? I have the string below, and I want to substitute every occurrence of "change" while displaying the 3-5 characters before the match.
string = "val=change anotherval=change stringhere:change: foo=bar foofoo=barbar"
What I have so far
while line.match(/change/)
printf "\n\n Substitute the FIRST change below:\n"
printf "#{line}\n"
printf "\n\tSubstitute => "
substitution = gets.chomp
line = line.sub(/change/, "#{substitution}")
end
If you want to get down and dirty Perl style:
before_chars = $`[-3, 3]
This is the last three characters just before your pattern match.
You would likely use gsub! with block given in the following manner:
line = "val=change anotherval=change stringhere:change: foo=bar foofoo=barbar"
# line.gsub!(/(?<where>.{0,3})change/) {
line.gsub!(/(?<where>\S+)change/) {
printf "\n\n Substitute the change around #{Regexp.last_match[:where]} => \n"
substitution = gets.chomp
"#{Regexp.last_match[:where]}#{substitution}"
}
puts line
Yielding:
Substitute the change around val= =>
111
Substitute the change around anotherval= =>
222
Substitute the change around stringhere: =>
333
val=111 anotherval=222 stringhere:333: foo=bar foofoo=barbar
gsub! will substitute the matches in place, while more suitable pattern \S+ instead of commented out .{0,3} will give you an ability to print out the human-readable hint.
Alternative: Use $1 Match Variable
tadman's answer uses the special prematch variable ($`). Ruby will also store a capture group in a numbered variable, which is probably just as magical but possibly more intuitive. For example:
string = "val=change anotherval=change stringhere:change: foo=bar foofoo=barbar"
string.sub(/(.{3})?change/, "\\1#{substitution}")
$1
# => "al="
No matter what method you use, though, make sure you explicitly check your match variables for nils in the event that your last attempted match was unsuccessful.
Given the following indented text:
two spaces
four
six
non-leading spaces
I'd like to convert every 2 leading spaces to a tab, essentially converting from soft tabs to hard tabs. I'm looking for the following result (using an 'x' instead of "\t"):
xtwo spaces
xxfour
xxxsix
non-leading spaces
What is the most efficient or eloquent way to do this in ruby?
What I have so far seems to be working, but it doesn't feel right.
input.gsub!(/^ {2}/,"x")
res = []
input.split(/\n/).each do |line|
while line =~ /^x+ {2}/
line.gsub!(/^(x+) {2}/,"\\1x")
end
res << line
end
puts res.join("\n")
I noticed the answer using sed and \G:
perl -pe '1 while s/\G {2}/\t/gc' input.txt >output.txt
But I can't figure out how to mimic the pattern in Ruby. This is as far as I got:
rep = 1
while input =~ /^x* {2}/ && rep < 10
input.gsub!(/\G {2}/,"x")
rep += 1
end
puts input
Whats wrong with using (?:^ {2})|\G {2} in multi-line mode?
The first match will always be at the beginning of the line,
then \G will match succesively right next to that, or the match
will fail. The next match will always be the beginning of the line.. repeats.
In Perl its $str =~ s/(?:^ {2})|\G {2}/x/mg; or $str =~ s/(?:^ {2})|\G {2}/\t/mg;
Ruby http://ideone.com/oZ4Os
input.gsub!(/(?:^ {2})|\G {2}/m,"x")
Edit: Of course the anchors can be factored out and put into an alternation
http://ideone.com/1oDOJ
input.gsub!(/(?:^|\G) {2}/m,"x")
You can just use a single gsub for that:
str.gsub(/^( {2})+/) { |spaces| "\t" * (spaces.length / 2) }
How do i remove empty lines from a string?
I have tried
some_string = some_string.gsub(/^$/, "");
and much more, but nothing works.
Remove blank lines:
str.gsub /^$\n/, ''
Note: unlike some of the other solutions, this one actually removes blank lines and not line breaks :)
>> a = "a\n\nb\n"
=> "a\n\nb\n"
>> a.gsub /^$\n/, ''
=> "a\nb\n"
Explanation: matches the start ^ and end $ of a line with nothing in between, followed by a line break.
Alternative, more explicit (though less elegant) solution:
str.each_line.reject{|x| x.strip == ""}.join
squeeze (or squeeze!) does just that - without a regex.
str.squeeze("\n")
Replace multiple newlines with a single one:
fixedstr = str.gsub(/\n\n+/, "\n")
or
str.gsub!(/\n\n+/, "\n")
You could try to replace all occurrences of 2 or more line breaks with just one:
my_string.gsub(/\n{2,}/, '\n')
Originally
some_string = some_string.gsub(/\n/,'')
Updated
some_string = some_string.gsub(/^$\n/,'')
In Ruby, what regex will strip out all but a desired string if present in the containing string? I know about /[^abc]/ for characters, but what about strings?
Say I have the string "group=4&type_ids[]=2&type_ids[]=7&saved=1" and want to retain the pattern group=\d, if it is present in the string using only a regex?
Currently, I am splitting on & and then doing a select with matching condition =~ /group=\d/ on the resulting enumerable collection. It works fine, but I'd like to know the regex to do this more directly.
Simply:
part = str[/group=\d+/]
If you want only the numbers, then:
group_str = str[/group=(\d+)/,1]
If you want only the numbers as an integer, then:
group_num = str[/group=(\d+)/,1].to_i
Warning: String#[] will return nil if no match occurs, and blindly calling nil.to_i always returns 0.
You can try:
$str =~ s/.*(group=\d+).*/\1/;
Typically I wouldn't really worry too much about a complex regex. Simply break the string down into smaller parts and it becomes easier:
asdf = "group=4&type_ids[]=2&type_ids[]=7&saved=1"
asdf.split('&').select{ |q| q['group'] } # => ["group=4"]
Otherwise, you can use regex a bunch of different ways. Here's two ways I tend to use:
asdf.scan(/group=\d+/) # => ["group=4"]
asdf[/(group=\d+)/, 1] # => "group=4"
Try:
str.match(/group=\d+/)[0]