Example: "[12,23,987,43"
What is the fastest, most efficient way to remove the "[" ?
Like, maybe a chop() but for the first char.
+1
A:
Easy way:
str = "[12,23,987,43"
removed = str[1..str.length]
Awesome way:
class String
def reverse_chop()
self[1..self.length]
end
end
"[12,23,987,43".reverse_chop()
(Note: prefer the easy way :) )
Pablo Fernandez
2010-09-01 01:17:52
If you want to preserve the "chop" semantics you could just `"[12,23,987,43".reverse.chop.reverse`
Chris Heald
2010-09-01 01:24:22
that is a pretty big performance overhead just to strip one char
Pablo Fernandez
2010-09-01 02:16:16
why not use [1..-1] rather than [1..self.length] ?
banister
2010-09-01 03:59:31
+2
A:
If you always want to strip leading brackets:
"[12,23,987,43".gsub(/^\[/, "")
If you just want to remove the first character, and you know it won't be in a multibyte character set:
"[12,23,987,43"[1..-1]
or
"[12,23,987,43".slice(1..-1)
Chris Heald
2010-09-01 01:20:12
+2
A:
I kind of favor using something like:
asdf = "[12,23,987,43" asdf[0] = '' p asdf # >> "12,23,987,43"
Greg
2010-09-01 02:13:23
It's important to note that this will only work in Ruby 1.9. In Ruby 1.8, this will remove the first *byte* from the string, *not* the first character, which is not what the OP wants.
Jörg W Mittag
2010-09-01 08:02:36
It's important to note that this will only work in Ruby 1.9. In Ruby 1.8, this will remove the first *byte* from the string, *not* the first character, which is not what the OP wants.
Jörg W Mittag
2010-09-01 08:03:49
+1
A:
Similar to Pablo's answer above, but a shade cleaner :
str[1..-1]
Will return the array from 1 to the last character.
'Hello World'[1..-1]
=> "ello World"
Jason Stirk
2010-09-01 02:25:35