Question

我有一系列的体力,其中包括“firstname.lastname”号:

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", ...]

这样做的最有效方式是什么?

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

https/a>。

Answer 1

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

<>Update>

@tadman has, here sbaseing 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

Answer 2

Don t 知道这是否最有效<><> >,但它确实是:

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

Answer 3

你们总是能够改变各自的价值:

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.

请注意,这将吸引原始投入,这样,如果冻结<条码>底线的封顶价值,或需要以最初的形式保存这一作品。情况可能并非如此,但必须牢记这一点。

<>Update: 在经过一些基准之后,我得出以下结论:开展这一行动的最快方式是:

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

https://gist.github.com/1339615“rel=“nofollow”?

  #     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!( . ,    ) }

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