arrays - Ruby:排序数组，跳过第一个元素

Question

我想通过跳过第一个数组的第一个字符串对字符串数组的数组进行排序，但我不知道如何使用内置的 sort 方法进行排序。我可以复制没有第一个元素的整个数组，然后对结果数组进行排序，但是没有更优雅的方法来做到这一点吗？

ar = [["zzzz", "skip", "this"], ["EFP3","eins","eins"], ["EFP10","zwei","zwei"], ["EFP1","drei","drei"]]
ar.sort!{ |a,b|
  if a == ar.first   # why doesn't 
    next             # this
  end                # work ?

  # compare length, otherwise it would be e.g. 10 < 3
  if a[0].length == b[0].length
    a[0] <=> b[0]
  else
    a[0].length <=> b[0].length
  end
}

我想要这样的结果:

["zzzz", "skip", "this"], ["EFP1","drei","drei"], ["EFP3","eins","eins"], ["EFP10","zwei","zwei"]

按“EFP#”排序

编辑:如果重要的话，我使用的是 Ruby 1.8。

Answer 1

ar[1..-1].sort { whatever you want }