CODEWITHSUNDEEP
ruby-on-railsruby
Mellon
Mellon

Reputation: 38902

Ruby array: change string element format

I have an array of string which contains the "firstname.lastname" strings:

customers = ["aaa.bbb", "ccc.ddd", "www.uuu", "iii.oooo", ...]

Now, I would like to transfer each element's string format in the array by removing the "." and use space instead. That's I want the array to be:

customers = ["aaa bbb", "ccc ddd", "www uuu", "iii oooo", ...]

What is the most efficient way to do it?

---------------- MORE -------------------

And how about my case here

Upvotes: 0

Views: 1644

Answers (3)

Paul Groves
Paul Groves

Reputation: 4051

customers.collect!{|name| name.gsub(/\./, " ")}

Update

@tadman has it, here's my benchmarking FWIW

require 'benchmark'

customers = []
500000.times { customers << "john.doe" }

Benchmark.bm do|b|
  b.report("+= ") do
    # customers.collect!{|name| name.gsub(/\./, " ")} # 2.414220
    # customers.each { |c| c.gsub!(/\./, ' ') } # 2.223308
    customers.each { |c| c.tr!('.', ' ') } # 0.379226
  end
end

Upvotes: 7

tadman
tadman

Reputation: 211740

You can always just modify each value in-place:

customers.each { |c| c.gsub!(/\./, ' ') }

The alternatives, like collect! are more appropriate when you're switching the kind of object, not just altering the content of an existing object.

Note that this will mangle the original input, so if the String values in customers are frozen or need to be preserved in their initial form this won't work. This is probably not the case, but it is important to keep this in mind.

Update: After some benchmarking I've come to the conclusion that the fastest way to perform this operation is:

customers.each { |c| c.tr!('.', ' ') }

For those that are curious, here's a benchmark scaffold to experiment with.

  #     user     system      total        real
  # 0.740000   0.020000   0.760000 (  0.991042)
  list.each { |c| c.gsub!(/\./, ' ') }

  # 0.680000   0.010000   0.690000 (  1.011136)
  list.collect! { |c| c.gsub(/\./, ' ') }

  # 0.490000   0.020000   0.510000 (  0.628354)
  list.collect!{ |c| c.split('.').join(' ') }

  # 0.090000   0.000000   0.090000 (  0.103968)
  list.collect!{ |c| c.tr!('.', ' ') }

Upvotes: 2

Thilo
Thilo

Reputation: 17735

Don't know if this is the most efficient, but it does the job:

customers.collect!{ |name| name.split('.').join(' ') }

Upvotes: 2

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