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Remove first character in Excel To delete the first character from a string, you can use either the REPLACE function or a combination of RIGHT and LEN functions. Here, we simply take 1 character from the first position and replace it with an empty string ("").
The idea is to use the deleteCharAt() method of StringBuilder class to remove first and the last character of a string. The deleteCharAt() method accepts a parameter as an index of the character you want to remove.
To remove the first character of a string, we can use the built-in erase() function by passing the 0,1 as an arguments to it. Where 0 is the first character index, 1 is the number of characters we need to remove from that index. Note: The erase() function modifies the original string instead of creating a new string.
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 kind of favor using something like:
asdf = "[12,23,987,43" asdf[0] = '' p asdf # >> "12,23,987,43"
I'm always looking for the fastest and most readable way of doing things:
require 'benchmark'
N = 1_000_000
puts RUBY_VERSION
STR = "[12,23,987,43"
Benchmark.bm(7) do |b|
b.report('[0]') { N.times { "[12,23,987,43"[0] = '' } }
b.report('sub') { N.times { "[12,23,987,43".sub(/^\[+/, "") } }
b.report('gsub') { N.times { "[12,23,987,43".gsub(/^\[/, "") } }
b.report('[1..-1]') { N.times { "[12,23,987,43"[1..-1] } }
b.report('slice') { N.times { "[12,23,987,43".slice!(0) } }
b.report('length') { N.times { "[12,23,987,43"[1..STR.length] } }
end
Running on my Mac Pro:
1.9.3
user system total real
[0] 0.840000 0.000000 0.840000 ( 0.847496)
sub 1.960000 0.010000 1.970000 ( 1.962767)
gsub 4.350000 0.020000 4.370000 ( 4.372801)
[1..-1] 0.710000 0.000000 0.710000 ( 0.713366)
slice 1.020000 0.000000 1.020000 ( 1.020336)
length 1.160000 0.000000 1.160000 ( 1.157882)
Updating to incorporate one more suggested answer:
require 'benchmark'
N = 1_000_000
class String
def eat!(how_many = 1)
self.replace self[how_many..-1]
end
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
class Array
def eat!(how_many = 1)
self.replace self[how_many..-1]
end
end
puts RUBY_VERSION
STR = "[12,23,987,43"
Benchmark.bm(7) do |b|
b.report('[0]') { N.times { "[12,23,987,43"[0] = '' } }
b.report('sub') { N.times { "[12,23,987,43".sub(/^\[+/, "") } }
b.report('gsub') { N.times { "[12,23,987,43".gsub(/^\[/, "") } }
b.report('[1..-1]') { N.times { "[12,23,987,43"[1..-1] } }
b.report('slice') { N.times { "[12,23,987,43".slice!(0) } }
b.report('length') { N.times { "[12,23,987,43"[1..STR.length] } }
b.report('eat!') { N.times { "[12,23,987,43".eat! } }
b.report('reverse') { N.times { "[12,23,987,43".reverse.chop.reverse } }
end
Which results in:
2.1.2
user system total real
[0] 0.300000 0.000000 0.300000 ( 0.295054)
sub 0.630000 0.000000 0.630000 ( 0.631870)
gsub 2.090000 0.000000 2.090000 ( 2.094368)
[1..-1] 0.230000 0.010000 0.240000 ( 0.232846)
slice 0.320000 0.000000 0.320000 ( 0.320714)
length 0.340000 0.000000 0.340000 ( 0.341918)
eat! 0.460000 0.000000 0.460000 ( 0.452724)
reverse 0.400000 0.000000 0.400000 ( 0.399465)
And another using /^./ to find the first character:
require 'benchmark'
N = 1_000_000
class String
def eat!(how_many = 1)
self.replace self[how_many..-1]
end
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
class Array
def eat!(how_many = 1)
self.replace self[how_many..-1]
end
end
puts RUBY_VERSION
STR = "[12,23,987,43"
Benchmark.bm(7) do |b|
b.report('[0]') { N.times { "[12,23,987,43"[0] = '' } }
b.report('[/^./]') { N.times { "[12,23,987,43"[/^./] = '' } }
b.report('[/^\[/]') { N.times { "[12,23,987,43"[/^\[/] = '' } }
b.report('sub+') { N.times { "[12,23,987,43".sub(/^\[+/, "") } }
b.report('sub') { N.times { "[12,23,987,43".sub(/^\[/, "") } }
b.report('gsub') { N.times { "[12,23,987,43".gsub(/^\[/, "") } }
b.report('[1..-1]') { N.times { "[12,23,987,43"[1..-1] } }
b.report('slice') { N.times { "[12,23,987,43".slice!(0) } }
b.report('length') { N.times { "[12,23,987,43"[1..STR.length] } }
b.report('eat!') { N.times { "[12,23,987,43".eat! } }
b.report('reverse') { N.times { "[12,23,987,43".reverse.chop.reverse } }
end
Which results in:
# >> 2.1.5
# >> user system total real
# >> [0] 0.270000 0.000000 0.270000 ( 0.270165)
# >> [/^./] 0.430000 0.000000 0.430000 ( 0.432417)
# >> [/^\[/] 0.460000 0.000000 0.460000 ( 0.458221)
# >> sub+ 0.590000 0.000000 0.590000 ( 0.590284)
# >> sub 0.590000 0.000000 0.590000 ( 0.596366)
# >> gsub 1.880000 0.010000 1.890000 ( 1.885892)
# >> [1..-1] 0.230000 0.000000 0.230000 ( 0.223045)
# >> slice 0.300000 0.000000 0.300000 ( 0.299175)
# >> length 0.320000 0.000000 0.320000 ( 0.325841)
# >> eat! 0.410000 0.000000 0.410000 ( 0.409306)
# >> reverse 0.390000 0.000000 0.390000 ( 0.393044)
Here's another update on faster hardware and a newer version of Ruby:
2.3.1
user system total real
[0] 0.200000 0.000000 0.200000 ( 0.204307)
[/^./] 0.390000 0.000000 0.390000 ( 0.387527)
[/^\[/] 0.360000 0.000000 0.360000 ( 0.360400)
sub+ 0.490000 0.000000 0.490000 ( 0.492083)
sub 0.480000 0.000000 0.480000 ( 0.487862)
gsub 1.990000 0.000000 1.990000 ( 1.988716)
[1..-1] 0.180000 0.000000 0.180000 ( 0.181673)
slice 0.260000 0.000000 0.260000 ( 0.266371)
length 0.270000 0.000000 0.270000 ( 0.267651)
eat! 0.400000 0.010000 0.410000 ( 0.398093)
reverse 0.340000 0.000000 0.340000 ( 0.344077)
Why is gsub so slow?
After doing a search/replace, gsub has to check for possible additional matches before it can tell if it's finished. sub only does one and finishes. Consider gsub like it's a minimum of two sub calls.
Also, it's important to remember that gsub, and sub can also be handicapped by poorly written regex which match much more slowly than a sub-string search. If possible anchor the regex to get the most speed from it. There are answers here on Stack Overflow demonstrating that so search around if you want more information.
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.
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"
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)
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