I have this string:
string = "SEGUNDA A SEXTA\n05:24 \n05:48\n06:12\n06:36\n07:00\n07:24\n07:48\n\n08:12 \n08:36\n09:00\n09:24\n09:48\n10:12\n10:36\n11:00 \n11:24\n11:48\n12:12\n12:36\n13:00\n13:24\n13:48 \n14:12\n14:36\n15:00\n15:24\n15:48\n16:12\n16:36 \n17:00\n17:24\n17:48\n18:12\n18:36\n19:00\n19:48 \n20:36\n21:24\n22:26\n23:15\n00:00\n"
And I'd like to replace all \n\n occurrences to only one \n and if it's possible I'd like to remove also all " " (spaces) between the numbers and the newline character \n
I'm trying to do:
string.gsub(/\n\n/, '\n')
but it is replacing \n\n by \\n
Can anyone help me?
The real reason is because single quoted sting doesn't escape special characters (like \n).
string.gsub(/\n/, '\n')
It replaces one single character \n with two characters '\' and 'n'
You can see the difference by printing the string:
[302] pry(main)> puts '\n'
\n
=> nil
[303] pry(main)> puts "\n"
=> nil
[304] pry(main)> string = '\n'
=> "\\n"
[305] pry(main)> string = "\n"
=> "\n"
I think you're looking for:
string.gsub( / *\n+/, "\n" )
This searches for zero or more spaces followed by one or more newlines, and replaces the match with a single newline.
Related
I have some text that has this particular character:
When I call the string split() method (with just ' ' as the input) , the gets removed. What should I do to keep the ?
That's the expected behavior when passing ' '. According to the docs:
If pattern is a single space, str is split on whitespace, with leading and trailing whitespace and runs of contiguous whitespace characters ignored.
With "whitespace" being space (" ") and \t, \n, \v, \f, \r.
"foo bar\nbaz \f qux".split(' ')
#=> ["foo", "bar", "baz", "qux"]
To split on space (U+0020) only, you have to use a regexp:
"foo bar\nbaz \f qux".split(/ /)
#=> ["foo", "bar\nbaz", "\f", "qux"]
I'm trying to do a regex with lookbehind that changes \n to but not if it's a \\n.
My closest attempt has no effect:
text.gsub /(?<!\\)\n/, ''
Unfortunately, no number of backslashes in the lookbehind seem to fix the problem. How can I address this?
You need to double the backslash before the n in the regex, otherwise it's looking for a newline instead of a literal backslash followed by n:
irb(main):001:0> puts "hello\\nthere\\\\n".gsub(/(?<!\\)\\n/, ' ')
hello there\\n
You don't need anything special. "\n" is a single character. It does not include a "\" or "n" character.
text.gsub(/\n/, "")
But instead of that, you should do:
text.gsub("\n", "")
or
text.tr("\n", "")
But I would do:
text.tr($/, "")
I have a string that looks like this.
mystring="The Body of a\r\n\t\t\t\tSpider"
I want to replace all the \r, \n, \t etc with a whitespace.
The code I wrote for this is :
mystring.gsub(/\\./, " ")
But this isn't doing anything to the string.
Help.
\r, \n and \t are escape sequences representing carriage return, line feed and tab. Although they are written as two characters, they are interpreted as a single character:
"\r\n\t".codepoints #=> [13, 10, 9]
Because it is such a common requirement, there's a shortcut \s to match all whitespace characters:
mystring.gsub(/\s/, ' ')
#=> "The Body of a Spider"
Or \s+ to match multiple whitespace characters:
mystring.gsub(/\s+/, ' ')
#=> "The Body of a Spider"
/\s/ is equivalent to /[ \t\r\n\f]/
String#tr is designed for stream symbol substitution. It appears to be a bit quickier, than String#gsub:
mystring.tr "\r", ' '
It hasan insplace version also (this will replace all carriage returns, line feed and spaces with space):
mystring.tr! "\s\r\n\t\f", ' '
Stefen's Answer is really very Cool as always comeup with very short and clean solutions. But here what I tried to remove all special characters. [Posted as just optional solution] ;)
> a = "The Body of a\r\n\t\t\t\tSpider"
=> "The Body of a\r\n\t\t\t\tSpider"
> a.gsub(/[^0-9A-Za-z]/, ' ')
=> "The Body of a Spider"
you can use strip , then add a space to your string
mystring.strip . " "
If you literally has \r\n\t in your string:
mystring="The Body of a\r\n\t\t\t\tSpider"
mystring.split(/[\r\t\n]/)
Why doesn't this code work:
"hello \nworld".each_line(separator = '\n') {|s| p s}
while this works?
"hello \nworld".each_line(separator = $/) {|s| p s}
A 10 second google yielded this:
$/ is the input record separator, newline by default.
The first one doesn't work because you used single quotes. Backslash escape sequences are ignored in single quoted strings. Use double quotes instead:
"hello \nworld".each_line(separator = "\n") {|s| p s}
First, newline is the default. All you need is
"hello \nworld".each_line {|s| p s}
Secondly, single quotes behave differently than double quotes. '\n' means a literal backslash followed by the letter n, whereas "\n" means the newline character.
Last, the special variable $/ is the record separator which is "\n" by default, which is why you don't need to specify the separator in the above example.
Simple gsub! your string with valid "\n" new line character:
text = 'hello \nworld'
text.gsub!('\n', "\n")
After that \n character will act like newline character.
Here is a string str:
str = "line1
line2
line3"
We would like to add string "\n" to the end of each line:
str = "line1 \n
line2 \n
line3 \n"
A method is defined:
def mod_line(str)
s = ""
str.each_line do |l|
s += l + '\\n'
end
end
The problem is that '\n' is a line feed and was not added to the end of the str even with escape \. What's the right way to add '\n' literally to each line?
String#gsub/String#gsub! plus a very simple regular expression can be used to achieve that:
str = "line1
line2
line3"
str.gsub!(/$/, ' \n')
puts str
Output:
line1 \n
line2 \n
line3 \n
The platform-independent solution:
str.gsub(/\R/) { " \\n#{$~}" }
It will search for line-feeds/carriage-returns and replace them with themselves, prepended by \n.
\n needs to be interpreted as a special character. You need to put it in double quotes.
"\n"
Your attempt:
'\\n'
only escapes the backslash, which is actually redundant. With or without escaping on the backslash, it gives you a backslash followed by the letter n.
Also, your method mod_line returns the result of str.each_line, which is the original string str. You need to return the modified string s:
def mod_line(str)
...
s
end
And by the way, be aware that each line of the original string already has "\n" at the end of each line, so you are adding the second "\n" to each line (making it two lines).
This is the closest I got to it.
def mod_line(str)
s = ""
str.each_line do |l|
s += l
end
p s
end
Using p instead of puts leaves the \n on the end of each line.