ruby - Iterate over n to last item in a Ruby enumerable

Question

I am trying to find the cleanest way to do something special for the first item in an array, then do something different for all the rest. So far I have something like this:

puts @model.children.first.full_name
@model.children[1..@model.children.count].each do |child|
  puts child.short_name
end

I don't like how I have to specify the count in the range. I would like to find a shorter way to specify "from the second item to the last". How can this be done?

Answer 1

Ruby 使用 splat 运算符有一个很酷的方法*。这是一个例子：

a = [1,2,3,4]
first, *rest = *a
puts first # 1
puts rest # [2,3,4]
puts a # [1,2,3,4]

这是您重写的代码：

first, *rest = @model.children
puts first.full_name
rest.each do |child|
  puts child.short_name
end

我希望这有帮助！

Answer 2

您可以采用这种方法：

@model.children[1..-1].each do |child|
  puts child.short_name
end

Answer 3

您可能会使用drop：

puts @model.children.first.full_name
@model.children.drop(1).each do |child|
  puts child.short_name
end

Answer 4

像这样的东西？：

puts @model.children.first.full_name
@model.children[1..-1].each do |child|
  puts child.short_name
end

Answer 5

你的标题说你想省略n元素，但 OP 文本只要求特别对待第一个元素。如果您对更通用的 nanswer 感兴趣，这里有一个简洁的选项：

n = 1 # how many elements to omit
@model.children.drop( n ).map( &:short_name ).each &method( :puts )