Extract number from string in Ruby

Question

I\'m using this code:

s = line.match( /ABCD(\\d{4})/ ).values_at( 1 )[0] 

To extract numbers from strings like:

ABCD1234
AB         


        

Answer 1

a.map {|x| x[/\d+/]}

Answer 2

There are many Ruby ways as per http://www.ruby-forum.com/topic/125709

1. line.scan(/\d/).join('')
2. line.gsub(/[^0-9]/, '')
3. line.gsub(/[^\d]/, '')
4. line.tr("^0-9", '')
5. line.delete("^0-9")
6. line.split(/[^\d]/).join
7. line.gsub(/\D/, '')

Try each on you console.

Also check the benchmark report in that post.

Answer 3

The simplest and fastest way is to just get all integers out of the string.

str = 'abc123def456'

str.delete("^0-9")
=> "123456"

Comparing benchmarks on a long string with some of the other solutions provided here, we can see this is orders of magnitude faster:

require 'benchmark'

@string = [*'a'..'z'].concat([*1..10_000].map(&:to_s)).shuffle.join

Benchmark.bm(10) do |x|
  x.report(:each_char) do
    @string.each_char{ |c| @string.delete!(c) if c.ord<48 or c.ord>57 }
  end
  x.report(:match) do |x|
    /\d+/.match(@string).to_s
  end
  x.report(:map) do |x|
    @string.split.map {|x| x[/\d+/]}
  end
  x.report(:gsub) do |x|
    @string.gsub(/\D/, '')
  end
  x.report(:delete) do
    @string.delete("^0-9")
  end
end

             user     system      total        real
each_char    0.020000   0.020000   0.040000 (  0.037325)
match        0.000000   0.000000   0.000000 (  0.001379)
map          0.000000   0.000000   0.000000 (  0.001414)
gsub         0.000000   0.000000   0.000000 (  0.000582)
delete       0.000000   0.000000   0.000000 (  0.000060)

Answer 4

Another solution may be to write:

myString = "sami103"
myString.each_char{ |c| myString.delete!(c) if c.ord<48 or c.ord>57 } #In this case, we are deleting all characters that do not represent numbers.

Now, if you type

myNumber = myString.to_i #or myString.to_f

This should return a

Answer 5

your_input = "abc1cd2"
your_input.split(//).map {|x| x[/\d+/]}.compact.join("").to_i

This should work.

Answer 6

there is even simpler solution

line.scan(/\d+/).first