[ruby] What is the easiest way to remove the first character from a string?

Example:

[12,23,987,43

What is the fastest, most efficient way to remove the "[", using maybe a chop() but for the first character?

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"

I find a nice solution to be str.delete(str[0]) for its readability, though I cannot attest to it's performance.

Ruby 2.5+

As of Ruby 2.5 you can use delete_prefix or delete_prefix! to achieve this in a readable manner.

In this case "[12,23,987,43".delete_prefix("[").

More info here:

• Official docs

• 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"

• Docs
• https://bugs.ruby-lang.org/issues/13665

Edit:

Using the Tin Man's benchmark setup, it looks pretty quick too (under the last two entries delete_p and delete_p!). Doesn't quite pip the previous faves for speed, though is very readable.

2.5.0
              user     system      total        real
[0]       0.174766   0.000489   0.175255 (  0.180207)
[/^./]    0.318038   0.000510   0.318548 (  0.323679)
[/^\[/]   0.372645   0.001134   0.373779 (  0.379029)
sub+      0.460295   0.001510   0.461805 (  0.467279)
sub       0.498351   0.001534   0.499885 (  0.505729)
gsub      1.669837   0.005141   1.674978 (  1.682853)
[1..-1]   0.199840   0.000976   0.200816 (  0.205889)
slice     0.279661   0.000859   0.280520 (  0.285661)
length    0.268362   0.000310   0.268672 (  0.273829)
eat!      0.341715   0.000524   0.342239 (  0.347097)
reverse   0.335301   0.000588   0.335889 (  0.340965)
delete_p  0.222297   0.000832   0.223129 (  0.228455)
delete_p!  0.225798   0.000747   0.226545 (  0.231745)

Inefficient alternative:

str.reverse.chop.reverse

For example : a = "One Two Three"

1.9.2-p290 > a = "One Two Three"
 => "One Two Three" 

1.9.2-p290 > a = a[1..-1]
 => "ne Two Three" 

1.9.2-p290 > a = a[1..-1]
 => "e Two Three" 

1.9.2-p290 > a = a[1..-1]
 => " Two Three" 

1.9.2-p290 > a = a[1..-1]
 => "Two Three" 

1.9.2-p290 > a = a[1..-1]
 => "wo Three" 

In this way you can remove one by one first character of the string.

Thanks to @the-tin-man for putting together the benchmarks!

Alas, I don't really like any of those solutions. Either they require an extra step to get the result ([0] = '', .strip!) or they aren't very semantic/clear about what's happening ([1..-1]: "Um, a range from 1 to negative 1? Yearg?"), or they are slow or lengthy to write out (.gsub, .length).

What we are attempting is a 'shift' (in Array parlance), but returning the remaining characters, rather than what was shifted off. Let's use our Ruby to make this possible with strings! We can use the speedy bracket operation, but give it a good name, and take an arg to specify how much we want to chomp off the front:

class String
  def eat!(how_many = 1)
    self.replace self[how_many..-1]
  end
end

But there is more we can do with that speedy-but-unwieldy bracket operation. While we are at it, for completeness, let's write a #shift and #first for String (why should Array have all the fun??), taking an arg to specify how many characters we want to remove from the beginning:

class String
  def first(how_many = 1)
    self[0...how_many]
  end

  def shift(how_many = 1)
    shifted = first(how_many)
    self.replace self[how_many..-1]
    shifted
  end
  alias_method :shift!, :shift
end

Ok, now we have a good clear way of pulling characters off the front of a string, with a method that is consistent with Array#first and Array#shift (which really should be a bang method??). And we can easily get the modified string as well with #eat!. Hm, should we share our new eat!ing power with Array? Why not!

class Array
  def eat!(how_many = 1)
    self.replace self[how_many..-1]
  end
end

Now we can:

> str = "[12,23,987,43" #=> "[12,23,987,43"
> str.eat!              #=> "12,23,987,43"
> str                   #=> "12,23,987,43"

> str.eat!(3)           #=> "23,987,43"
> str                   #=> "23,987,43"

> str.first(2)          #=> "23"
> str                   #=> "23,987,43"

> str.shift!(3)         #=> "23,"
> str                   #=> "987,43"

> arr = [1,2,3,4,5]     #=> [1, 2, 3, 4, 5] 
> arr.eat!              #=> [2, 3, 4, 5] 
> arr                   #=> [2, 3, 4, 5] 

That's better!

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)

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 :) )

We can use slice to do this:

val = "abc"
 => "abc" 
val.slice!(0)
 => "a" 
val
 => "bc" 

Using slice! we can delete any character by specifying its index.

Using regex:

str = 'string'
n = 1  #to remove first n characters

str[/.{#{str.size-n}}\z/] #=> "tring"

list = [1,2,3,4] list.drop(1)

# => [2,3,4]

List drops one or more elements from the start of the array, does not mutate the array, and returns the array itself instead of the dropped element.

str = "[12,23,987,43"

str[0] = ""

I prefer this:

str = "[12,23,987,43"
puts str[1..-1]
>> 12,23,987,43

class String
  def bye_felicia()
    felicia = self.strip[0] #first char, not first space.
    self.sub(felicia, '')
  end
end