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[1, 2, 3]から[[1, 2], [2, 3]]を得る

Ruby

Last updated at 2012-12-26Posted at 2012-12-05

a = [1, 2, 3]

# v1
(a.size-1).times.map {|i|
  [a[i], a[i+1]]
}

# v2
a[0..-2].zip(a[1..-1]) # a.size > 0 でないと駄目

# v3 by akicho8さん
a.each_cons(2).to_a # akicho8

もっと読みやすい方法を知りたい．→each_consがよさそう

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