Ruby/Rails 循环中的魔术第一个和最后一个指标？

Question

Ruby/Rails 在为基本事物添加糖时做了很多很酷的事情，我认为有一个非常常见的场景，我想知道是否有人做过帮助程序或类似的事情。

   a = Array.new(5, 1)

   a.each_with_index do |x, i|
     if i == 0
       print x+1
     elsif i == (a.length - 1)
       print x*10
     else
        print x
     end
   end

请原谅丑陋，但这得到了人们可能想要的......有没有一种红宝石方法可以对循环的第一个和最后一个做一些事情？

[EDIT] 我认为理想情况下这将是带有参数的 Array 的扩展（数组实例、所有元素函数、第一个元素函数、最后一个元素函数）......但我愿意其他想法。

Answer 1

我在这里看到很多非常接近的 hack，但都严重依赖于给定的具有固定大小的迭代器，而不是迭代器。我还想建议在迭代时保存前一个元素以了解迭代的第一个/最后一个元素。

previous = {}
elements.each do |element|
  unless previous.has_key?(:element)
    # will only execute the first time
  end

  # normal each block here

  previous[:element] = element
end

# the last element will be stored in previous[:element] 

Answer 2

正如许多人所指出的，each_with_index 似乎是解决这个问题的关键。我有这个我喜欢的代码块。

array.each_with_index do |item,index|
  if index == 0
    # first item
  elsif index == array.length-1
    # last item
  else
    # middle items
  end
  # all items
end

或者

array.each_with_index do |item,index|
  if index == 0
    # first item
  end
  # all items
  if index == array.length-1
    # last item
  end
end

或者通过数组扩展

class Array

  def each_with_position
    array.each_with_index do |item,index|
      if index == 0
        yield item, :first
      elsif index == array.length-1
        yield item, :last
      else
        yield item, :middle
      end
    end
  end

  def each_with_index_and_position
    array.each_with_index do |item,index|
      if index == 0
        yield item, index, :first
      elsif index == array.length-1
        yield item, index, :last
      else
        yield item, index, :middle
      end
    end
  end

  def each_with_position_and_index
    array.each_with_index do |item,index|
      if index == 0
        yield item, :first, index
      elsif index == array.length-1
        yield item, :last, index
      else
        yield item, :middle, index
      end
    end
  end

end

Answer 3

我无法抗拒 :) 虽然我想它不应该比这里的大多数其他答案慢很多，但它并没有针对性能进行调整。一切都与糖有关！

class Array
  class EachDSL
    attr_accessor :idx, :max

    def initialize arr
      self.max = arr.size
    end

    def pos
      idx + 1
    end

    def inside? range
      range.include? pos
    end

    def nth? i
      pos == i
    end

    def first?
      nth? 1
    end

    def middle?
      not first? and not last?
    end

    def last?
      nth? max
    end

    def inside range
      yield if inside? range
    end

    def nth i
      yield if nth? i
    end

    def first
      yield if first?
    end

    def middle
      yield if middle?
    end

    def last
      yield if last?
    end
  end

  def each2 &block
    dsl = EachDSL.new self
    each_with_index do |x,i|
      dsl.idx = i
      dsl.instance_exec x, &block
    end
  end
end

示例 1：

[1,2,3,4,5].each2 do |x|
  puts "#{x} is first"  if first?
  puts "#{x} is third"  if nth? 3
  puts "#{x} is middle" if middle?
  puts "#{x} is last"   if last?
  puts
end

# 1 is first
# 
# 2 is middle
# 
# 3 is third
# 3 is middle
# 
# 4 is middle
# 
# 5 is last

示例 2：

%w{some short simple words}.each2 do |x|
  first do
    puts "#{x} is first"
  end

  inside 2..3 do
    puts "#{x} is second or third"
  end

  middle do
    puts "#{x} is middle"
  end

  last do
    puts "#{x} is last"
  end
end

# some is first
# short is second or third
# short is middle
# simple is second or third
# simple is middle
# words is last

Answer 4

亲吻

arr.each.with_index do |obj, index|
  p 'first' if index == 0        
  p 'last' if index == arr.count-1                  
end

Answer 5

我不时需要这个功能，所以我为此设计了一个小类。

最新版本在：https://gist.github.com/3823837

示例：

("a".."m").to_a.each_pos do |e|
  puts "Char\tfirst?\tlast?\tprev\tnext\twrapped?\tindex\tposition" if e.first?
  print "#{e.item}\t"
  print "#{e.first?}\t"
  print "#{e.last?}\t"
  print "#{e.prev}\t"
  print "#{e.next}\t"
  print "#{e.wrapped?}\t\t"
  print "#{e.index}\t"
  puts  "#{e.position}\t"
end

# Char  first?  last?  prev  next  wrapped?  index  position
# a     true    false        b     false     0      1
# b     false   false  a     c     true      1      2
# c     false   false  b     d     true      2      3
# d     false   false  c     e     true      3      4
# e     false   false  d     f     true      4      5
# f     false   false  e     g     true      5      6
# g     false   false  f     h     true      6      7
# h     false   false  g     i     true      7      8
# i     false   false  h     j     true      8      9
# j     false   false  i     k     true      9      10
# k     false   false  j     l     true      10     11
# l     false   false  k     m     true      11     12
# m     false   true   l           false     12     13



{
  a: "0",
  b: "1",
  c: "2",
  d: "3",
  e: "4",
  f: "5",
  g: "6",
  h: "7",
  i: "8",
  j: "9",
  k: "10",
  l: "11",
  m: "12",
}.each_pos do |(k, v), e|
  puts "KV\tChar\t\tfirst?\tlast?\tprev\t\tnext\t\twrapped?\tindex\tposition" if e.first?
  print "#{k} => #{v}\t"
  print "#{e.item}\t"
  print "#{e.first?}\t"
  print "#{e.last?}\t"
  print "#{e.prev || "\t"}\t"
  print "#{e.next || "\t"}\t"
  print "#{e.wrapped?}\t\t"
  print "#{e.index}\t"
  puts  "#{e.position}\t"
end

# KV      Char        first?  last?   prev        next        wrapped?  index position
# a => 0  [:a, "0"]   true    false               [:b, "1"]   false     0     1
# b => 1  [:b, "1"]   false   false   [:a, "0"]   [:c, "2"]   true      1     2
# c => 2  [:c, "2"]   false   false   [:b, "1"]   [:d, "3"]   true      2     3
# d => 3  [:d, "3"]   false   false   [:c, "2"]   [:e, "4"]   true      3     4
# e => 4  [:e, "4"]   false   false   [:d, "3"]   [:f, "5"]   true      4     5
# f => 5  [:f, "5"]   false   false   [:e, "4"]   [:g, "6"]   true      5     6
# g => 6  [:g, "6"]   false   false   [:f, "5"]   [:h, "7"]   true      6     7
# h => 7  [:h, "7"]   false   false   [:g, "6"]   [:i, "8"]   true      7     8
# i => 8  [:i, "8"]   false   false   [:h, "7"]   [:j, "9"]   true      8     9
# j => 9  [:j, "9"]   false   false   [:i, "8"]   [:k, "10"]  true      9     10
# k => 10 [:k, "10"]  false   false   [:j, "9"]   [:l, "11"]  true      10    11
# l => 11 [:l, "11"]  false   false   [:k, "10"]  [:m, "12"]  true      11    12
# m => 12 [:m, "12"]  false   true    [:l, "11"]              false     12    13

实际类：

module Enumerable
  # your each_with_position method
  def each_pos &block
    EachWithPosition.each(self, &block)
  end
end

class EachWithPosition
  attr_reader :index

  class << self
    def each *a, &b
      handler = self.new(*a, :each, &b)
    end
  end

  def initialize collection, method, &block
    @index = 0
    @item, @prev, @next = nil
    @collection = collection
    @callback = block
    self.send(method)
  end

  def count
    @collection.count
  end
  alias_method :length, :count
  alias_method :size, :count

  def rest
    count - position
  end

  def first?
    @index == 0
  end

  def last?
    @index == (count - 1)
  end

  def wrapped?
    !first? && !last?
  end
  alias_method :inner?, :wrapped?

  def position
    @index + 1
  end

  def prev
    @prev
  end

  def next
    @next
  end

  def current
    @item
  end
  alias_method :item, :current
  alias_method :value, :current

  def call
    if @callback.arity == 1
      @callback.call(self)
    else
      @callback.call(@item, self)
    end
  end

  def each
    @collection.each_cons(2) do |e, n|
      @prev = @item
      @item = e
      @next = n

      self.call
      @index += 1

      # fix cons slice behaviour
      if last?
        @prev, @item, @next = @item, @next, nil
        self.call
        @index += 1
      end
    end
  end
end

Answer 6

有时 for 循环只是您的最佳选择

if(array.count > 0)
   first= array[0]
   #... do something with the first

   cx = array.count -2 #so we skip the last record on a 0 based array
   for x in 1..cx
     middle = array[x]
     #... do something to the middle
   end

   last = array[array.count-1]
   #... do something with the last item. 
end

我知道这个问题已经回答了，但是这个方法没有副作用，并且不检查第 13、14、15 .. 10,001、10,001... 记录是第一条记录，还是最后一条记录。

以前的答案会导致任何数据结构类中的分配失败。

Answer 7

有趣的问题，也是我思考过的一个问题。

我认为您必须创建三个不同的块/过程/无论它们被称为什么，然后创建一个方法来调用正确的块/过程/什么。 （抱歉含糊不清 - 我还不是黑带元程序员）[编辑：不过，我是从最底层的人那里抄来的）

class FancyArray
  def initialize(array)
    @boring_array = array
    @first_code = nil
    @main_code = nil
    @last_code = nil
  end

  def set_first_code(&code)
    @first_code = code
  end

  def set_main_code(&code)
    @main_code = code
  end

  def set_last_code(&code)
    @last_code = code
  end

  def run_fancy_loop
    @boring_array.each_with_index do |item, i|
      case i
      when 0 then @first_code.call(item)
      when @boring_array.size - 1 then @last_code.call(item)
      else @main_code.call(item)
      end
    end
  end
end

fancy_array = FancyArray.new(["Matti Nykanen", "Erik Johnsen", "Michael Edwards"])
fancy_array.set_first_code {|item| puts "#{item} came first in ski jumping at the 1988 Winter Olympics"}
fancy_array.set_main_code {|item| puts "#{item} did not come first or last in ski jumping at the 1988 Winter Olympics"}
fancy_array.set_last_code {|item| puts "#{item} came last in ski jumping at the 1988 Winter Olympics"}
fancy_array.run_fancy_loop

生产

Matti Nykanen came first in ski jumping at the 1988 Winter Olympics
Erik Johnsen did not come first or last in ski jumping at the 1988 Winter Olympics
Michael Edwards came last in ski jumping at the 1988 Winter Olympics

编辑：Svante 的answer（附有 molf 的建议）对相关问题的说明显示了如何将多个代码块传递给单个方法：

class FancierArray < Array
  def each_with_first_last(first_code, main_code, last_code)
    each_with_index do |item, i|
      case i
        when 0 then first_code.call(item)
        when size - 1 then last_code.call(item)
        else main_code.call(item)
      end
    end
  end
end

fancier_array = FancierArray.new(["Matti Nykanen", "Erik Johnsen", "Michael Edwards"])
fancier_array.each_with_first_last(
  lambda {|person| puts "#{person} came first in ski jumping at the 1988 Winter Olympics"},
  lambda {|person| puts "#{person} did not come first or last in ski jumping at the 1988 Winter Olympics"},
  lambda {|person| puts "#{person} came last in ski jumping at the 1988 Winter Olympics"})

Answer 8

或者一个很小的领域特定语言：

a = [1, 2, 3, 4]

FirstMiddleLast.iterate(a) do
  first do |e|
    p [e, 'first']
  end
  middle do |e|
    p [e, 'middle']
  end
  last do |e|
    p [e, 'last']
  end
end

# => [1, "first"]
# => [2, "middle"]
# => [3, "middle"]
# => [4, "last"]

以及实现它的代码：

class FirstMiddleLast

  def self.iterate(array, &block)
    fml = FirstMiddleLast.new(array)
    fml.instance_eval(&block)
    fml.iterate
  end

  attr_reader :first, :middle, :last

  def initialize(array)
    @array = array
  end

  def first(&block)
    @first = block
  end

  def middle(&block)
    @middle = block
  end

  def last(&block)
    @last = block
  end

  def iterate
    @first.call(@array.first) unless @array.empty?
    if @array.size > 1
      @array[1..-2].each do |e|
        @middle.call(e)
      end
      @last.call(@array.last)
    end
  end

end

我开始想，“如果你能将多个块传递给一个 Ruby 函数，那么你就可以对这个问题有一个巧妙而简单的解决方案。”然后我意识到 DSL 玩的小技巧几乎就像传递多个块一样。

Answer 9

如果你能做到这一点呢？

%w(a b c d).each.with_position do |e, position|
  p [e, position]    # => ["a", :first]
                     # => ["b", :middle]
                     # => ["c", :middle]
                     # => ["d", :last]
end

还是这个？

%w(a, b, c, d).each_with_index.with_position do |(e, index), position|
  p [e, index, position]    # => ["a,", 0, :first]
                            # => ["b,", 1, :middle]
                            # => ["c,", 2, :middle]
                            # => ["d", 3, :last]
end

在 MRI >= 1.8.7 中，只需要这个猴子补丁：

class Enumerable::Enumerator

  def with_position(&block)
    state = :init
    e = nil
    begin
      e_last = e
      e = self.next
      case state
      when :init
        state = :first
      when :first
        block.call(e_last, :first)
        state = :middle
      when :middle
        block.call(e_last, :middle)
      end
    rescue StopIteration
      case state
      when :first
        block.call(e_last, :first)
      when :middle
        block.call(e_last, :last)
      end
      return
    end while true
  end

end

它有一个小的状态引擎，因为它必须向前看一次迭代。

诀窍是 each, each_with_index, &c.如果没有给定块，则返回一个枚举器。 Enumerators 做了 Enumerable 所做的一切，甚至更多。但对我们来说，重要的是我们可以对 Enumerator 进行猴子修补，以添加另一种迭代方式，“包装”现有的迭代，无论它是什么。

Answer 10

如果你不介意“最后”动作发生在中间的东西之前，那么这个猴子补丁：

class Array

  def for_first
    return self if empty?
    yield(first)
    self[1..-1]
  end

  def for_last
    return self if empty?
    yield(last)
    self[0...-1]
  end

end

允许这样做：

%w(a b c d).for_first do |e|
  p ['first', e]
end.for_last do |e|
  p ['last', e]
end.each do |e|
  p ['middle', e]
end

# => ["first", "a"]
# => ["last", "d"]
# => ["middle", "b"]
# => ["middle", "c"]

Answer 11

将数组划分为多个范围，每个范围内的元素应该表现不同。将由此创建的每个范围映射到一个块。

class PartitionEnumerator
    include RangeMaker

    def initialize(array)
        @array = array
        @handlers = {}
    end

    def add(range, handler)
        @handlers[range] = handler
    end

    def iterate
        @handlers.each_pair do |range, handler|
          @array[range].each { |value| puts handler.call(value) }
        end
    end
end

可以手动创建范围，但下面的这些助手更容易：

module RangeMaker
  def create_range(s)
    last_index = @array.size - 1
    indexes = (0..last_index)
    return (indexes.first..indexes.first) if s == :first
    return (indexes.second..indexes.second_last) if s == :middle
    return (indexes.last..indexes.last) if s == :last
  end  
end

class Range
  def second
    self.first + 1
  end

  def second_last
    self.last - 1
  end
end

用法：

a = [1, 2, 3, 4, 5, 6]

e = PartitionEnumerator.new(a)
e.add(e.create_range(:first), Proc.new { |x| x + 1 } )
e.add(e.create_range(:middle), Proc.new { |x| x * 10 } )
e.add(e.create_range(:last), Proc.new { |x| x } )

e.iterate

Answer 12

如果你愿意添加一些样板，你可以在数组类中添加这样的东西：

class Array
  def each_fl
    each_with_index do |x,i|
      yield [i==0 ? :first : (i==length-1 ? :last : :inner), x]
    end
  end
end

然后在任何你需要的地方，你都会得到以下语法：

[1,2,3,4].each_fl do |t,x|
  case t
    when :first
      puts "first: #{x}"
    when :last
      puts "last: #{x}"
    else
      puts "otherwise: #{x}"
  end
end

对于以下输出：

first: 1
otherwise: 2
otherwise: 3
last: 4

Answer 13

如果您知道数组中的项目是唯一的（与本例不同），您可以这样做：

a = [1,2,3,4,5]

a.each_with_index do |x, i|
  if x == a.first
    print x+1
  elsif x == a.last
    print x*10
  else
    print x
  end
end

Answer 14

如果您愿意，您可以抓取第一个和最后一个元素并以不同方式处理它们。

first = array.shift
last = array.pop
process_first_one
array.each { |x| process_middle_bits }
process_last_one

Answer 15

Ruby 中没有“在（第一次|最后一次）时间执行此操作”语法。但如果你想要简洁，你可以这样做：

a.each_with_index do |x, i|
  print (i > 0 ? (i == a.length - 1 ? x*10 : x) : x+1)
end

结果如你所愿：

irb(main):001:0> a = Array.new(5,1)
=> [1, 1, 1, 1, 1]
irb(main):002:0> a.each_with_index do |x,i|
irb(main):003:1*   puts (i > 0 ? (i == a.length - 1 ? x*10 : x) : x+1)
irb(main):004:1> end
2
1
1
1
10

Answer 16

如果第一次和最后一次迭代的代码与其他迭代的代码没有共同之处，您也可以这样做：

do_something( a.first )
a[1..-2].each do |x|
  do_something_else( x )
end
do_something_else_else( a.last )

如果不同的情况有一些共同的代码，你的方法很好。