How to compute 'map' in Ruby without using blocks?

Question

I know I can do this in Ruby:

  ['a', 'b'].map do |s| s.to_sym end

and get this:

  [:a, :b]

I'm looking for a more concise way to do it, without using a block. Unfortunately, this doesn't work:

  ['a', 'b'].map #to_sym

Can I do better than with the initial code?

PreciousBodilyFluids · Accepted Answer · 2011-09-05 16:05:55Z

13

Read a bit about Symbol#to_proc:

['a', 'b'].map(&:to_sym)
# or
['a', 'b'].map &:to_sym
# Either will result in [:a, :b]

This works if you're using Ruby 1.8.7 or higher, or if you're using Rails - ActiveSupport will add this functionality for you.

answered Sep 5, 2011 at 16:05

PreciousBodilyFluids

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3 Comments

lucapette Over a year ago

It works on 1.8.7 too. 1.8.7 (main):0 > RUBY_DESCRIPTION => "ruby 1.8.7 (2011-06-30 patchlevel 352) [x86_64-linux]" 1.8.7 (main):0 > ['a', 'b'].map(&:to_sym) => [:a, :b]

PreciousBodilyFluids Over a year ago

lucapette: Yeah, I did a quick edit to add/fix some information.

lucapette Over a year ago

Yep I saw it. You know, with all this implementations out there I'm having issues with remembering stuff. When I read your reply I went to check I remembered it well ;)

lucapette · Accepted Answer · 2011-09-05 16:05:13Z

6

['a', 'b'].map(&:to_sym) is shorter

answered Sep 5, 2011 at 16:05

lucapette

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